id stringlengths 11 15 | title stringlengths 7 171 | problem stringlengths 9 4.33k | solution stringlengths 6 19k | problem_wikitext stringlengths 9 4.42k | solution_wikitext stringlengths 7 19.1k | proof_title stringlengths 9 171 | theorem_url stringlengths 34 198 | proof_url stringlengths 36 198 | categories listlengths 0 9 | theorem_references listlengths 0 36 | proof_references listlengths 0 253 |
|---|---|---|---|---|---|---|---|---|---|---|---|
proofwiki-6100 | Cardinal Equal to Collection of All Dominated Ordinals | Let $S$ be a set.
Let $\preccurlyeq$ denote the dominance relation.
Let $\On$ denote the class of all ordinals.
Let $x = \set {y \in \On: y \preccurlyeq S}$
Then:
{{begin-itemize}}
{{item|(1):|$x$ is an element of the class of all cardinals}}
{{item|(2):|There is no injection $f$ such that $f : x \to S$}}
{{end-itemize... | === $x$ is an ordinal ===
$x$ is clearly a subset of the class of all ordinals.
Moreover, suppose $y \in x$ and $z \in y$.
Then $z \subseteq y$ by the fact that $y$ is an ordinal.
$y \in x$ means that $f : y \to S$ for some injective mapping $f$ by the definition of dominance.
But then, $f \restriction_z : z \to S$ is ... | Let $S$ be a [[Definition:Set|set]].
Let $\preccurlyeq$ denote the [[Definition:Dominate (Set Theory)|dominance relation]].
Let $\On$ denote the [[Definition:Class of All Ordinals|class of all ordinals]].
Let $x = \set {y \in \On: y \preccurlyeq S}$
Then:
{{begin-itemize}}
{{item|(1):|$x$ is an element of the [[De... | === $x$ is an ordinal ===
$x$ is clearly a subset of the [[Definition:Class of All Ordinals|class of all ordinals]].
Moreover, suppose $y \in x$ and $z \in y$.
Then $z \subseteq y$ by the fact that $y$ is an [[Definition:Ordinal|ordinal]].
$y \in x$ means that $f : y \to S$ for some [[Definition:Injection|injectiv... | Cardinal Equal to Collection of All Dominated Ordinals | https://proofwiki.org/wiki/Cardinal_Equal_to_Collection_of_All_Dominated_Ordinals | https://proofwiki.org/wiki/Cardinal_Equal_to_Collection_of_All_Dominated_Ordinals | [
"Cardinals"
] | [
"Definition:Set",
"Definition:Dominate (Set Theory)",
"Definition:Class of All Ordinals",
"Definition:Class of All Cardinals",
"Definition:Injection"
] | [
"Definition:Class of All Ordinals",
"Definition:Ordinal",
"Definition:Injection",
"Definition:Mapping",
"Definition:Dominate (Set Theory)",
"Definition:Injection",
"Restriction of Injection is Injection",
"Definition:Transitive Class",
"Definition:Ordinal",
"Definition:Transitive Class",
"Defini... |
proofwiki-6101 | Class of All Cardinals is Proper Class | The class $\NN$ of all cardinal numbers is a proper class. | {{AimForCont}} $\NN$ is a small class.
By Class of All Cardinals is Subclass of Class of All Ordinals:
:$\NN \subseteq \On$
Therefore, $\bigcup \NN$ is an ordinal by Union of Set of Ordinals is Ordinal.
{{mistake|The above theorem has been handled only for a set of ordinals, not a class}}
Take $x = \set {y \in \On: y \... | The [[Definition:Class (Class Theory)|class]] $\NN$ of all [[Definition:Cardinal Number|cardinal numbers]] is a [[Definition:Proper Class|proper class]]. | {{AimForCont}} $\NN$ is a [[Definition:Small Class|small class]].
By [[Class of All Cardinals is Subclass of Class of All Ordinals]]:
:$\NN \subseteq \On$
Therefore, $\bigcup \NN$ is an [[Definition:Ordinal|ordinal]] by [[Union of Set of Ordinals is Ordinal]].
{{mistake|The above theorem has been handled only for a ... | Class of All Cardinals is Proper Class | https://proofwiki.org/wiki/Class_of_All_Cardinals_is_Proper_Class | https://proofwiki.org/wiki/Class_of_All_Cardinals_is_Proper_Class | [
"Class of All Cardinals",
"Class Theory"
] | [
"Definition:Class (Class Theory)",
"Definition:Cardinal Number",
"Definition:Class (Class Theory)/Proper Class"
] | [
"Definition:Small Class",
"Class of All Cardinals is Subclass of Class of All Ordinals",
"Definition:Ordinal",
"Union of Set of Ordinals is Ordinal",
"Cardinal Equal to Collection of All Dominated Ordinals",
"Definition:Cardinal Number",
"Definition:Injection",
"Definition:Cardinal Number",
"Set is ... |
proofwiki-6102 | Class of Infinite Cardinals is Proper Class | The class of infinite cardinals $\NN’$ is a proper class. | {{AimForCont}} $\NN'$ is a small class.
By Union of Small Classes is Small, $\NN’ \cup \omega$ is a small class.
By definition of the class of infinite cardinals, $\NN \subseteq \NN' \cup \omega$.
But by Axiom of Subsets Equivalents, this means that $\NN$ is a small class.
This contradicts Class of All Cardinals is Pro... | The [[Definition:Class of Infinite Cardinal Class|class of infinite cardinals]] $\NN’$ is a [[Definition:Proper Class|proper class]]. | {{AimForCont}} $\NN'$ is a [[Definition:Small Class|small class]].
By [[Union of Small Classes is Small]], $\NN’ \cup \omega$ is a [[Definition:Small Class|small class]].
By definition of the [[Definition:Class of Infinite Cardinals|class of infinite cardinals]], $\NN \subseteq \NN' \cup \omega$.
But by [[Axiom of ... | Class of Infinite Cardinals is Proper Class | https://proofwiki.org/wiki/Class_of_Infinite_Cardinals_is_Proper_Class | https://proofwiki.org/wiki/Class_of_Infinite_Cardinals_is_Proper_Class | [
"Class of All Cardinals"
] | [
"Definition:Class of Infinite Cardinal Class",
"Definition:Class (Class Theory)/Proper Class"
] | [
"Definition:Small Class",
"Union of Small Classes is Small",
"Definition:Small Class",
"Definition:Class of Infinite Cardinals",
"Axiom of Subsets Equivalents",
"Definition:Small Class",
"Class of All Cardinals is Proper Class",
"Definition:Small Class"
] |
proofwiki-6103 | Ordinal in Aleph iff Cardinal in Aleph | Let $x$ and $y$ be ordinals.
Then:
:$x \in \aleph_y \iff \card x \in \aleph_y$
where $\aleph$ denotes the aleph mapping. | By the definition of the aleph mapping, $\aleph_y$ is an element of the class of infinite cardinals.
By Cardinal Inequality implies Ordinal Inequality, it follows that:
:$x \in \aleph_y \iff \card x \in \aleph_y$
{{qed}} | Let $x$ and $y$ be [[Definition:Ordinal|ordinals]].
Then:
:$x \in \aleph_y \iff \card x \in \aleph_y$
where $\aleph$ denotes the [[Definition:Aleph Mapping|aleph mapping]]. | By the definition of the [[Definition:Aleph Mapping|aleph mapping]], $\aleph_y$ is an [[Definition:Element|element]] of the [[Definition:Class of Infinite Cardinals|class of infinite cardinals]].
By [[Cardinal Inequality implies Ordinal Inequality]], it follows that:
:$x \in \aleph_y \iff \card x \in \aleph_y$
{{qed}... | Ordinal in Aleph iff Cardinal in Aleph | https://proofwiki.org/wiki/Ordinal_in_Aleph_iff_Cardinal_in_Aleph | https://proofwiki.org/wiki/Ordinal_in_Aleph_iff_Cardinal_in_Aleph | [
"Aleph Mapping"
] | [
"Definition:Ordinal",
"Definition:Aleph Mapping"
] | [
"Definition:Aleph Mapping",
"Definition:Element",
"Definition:Class of Infinite Cardinals",
"Cardinal Inequality implies Ordinal Inequality"
] |
proofwiki-6104 | Aleph Product is Aleph | Let $x$ be an ordinal.
Then:
:$\left|{\aleph_x \times \aleph_x}\right| = \aleph_x$
where $\aleph$ denotes the aleph mapping. | {{begin-eqn}}
{{eqn | l = \left\vert{\aleph_x \times \aleph_x }\right\vert
| r = \left\vert{\aleph_x}\right\vert
| c = Non-Finite Cardinal is equal to Cardinal Product
}}
{{eqn | r = \aleph_x
| c = Aleph is Infinite Cardinal
}}
{{end-eqn}}
{{qed}} | Let $x$ be an [[Definition:Ordinal|ordinal]].
Then:
:$\left|{\aleph_x \times \aleph_x}\right| = \aleph_x$
where $\aleph$ denotes the [[Definition:Aleph Mapping|aleph mapping]]. | {{begin-eqn}}
{{eqn | l = \left\vert{\aleph_x \times \aleph_x }\right\vert
| r = \left\vert{\aleph_x}\right\vert
| c = [[Non-Finite Cardinal is equal to Cardinal Product]]
}}
{{eqn | r = \aleph_x
| c = [[Aleph is Infinite Cardinal]]
}}
{{end-eqn}}
{{qed}} | Aleph Product is Aleph | https://proofwiki.org/wiki/Aleph_Product_is_Aleph | https://proofwiki.org/wiki/Aleph_Product_is_Aleph | [
"Aleph Mapping"
] | [
"Definition:Ordinal",
"Definition:Aleph Mapping"
] | [
"Non-Finite Cardinal is equal to Cardinal Product",
"Aleph is Infinite Cardinal"
] |
proofwiki-6105 | Aleph is Infinite Cardinal | Let $x$ be an ordinal.
Then $\aleph_x$ is an infinite cardinal where $\aleph$ denotes the aleph mapping. | Let $\On$ denote the class of all ordinals.
By definition of the aleph mapping:
:$\aleph: \On \to \NN'$
where $\NN'$ denotes the class of infinite cardinals.
The theorem statement is an immediate consequence of this fact.
{{qed}}
Category:Aleph Mapping
fw7gnjchle5si1169w9p1j8nfp5fjpi | Let $x$ be an [[Definition:Ordinal|ordinal]].
Then $\aleph_x$ is an [[Definition:Infinite Cardinal|infinite cardinal]] where $\aleph$ denotes the [[Definition:Aleph Mapping|aleph mapping]]. | Let $\On$ denote the [[Definition:Class of All Ordinals|class of all ordinals]].
By definition of the [[Definition:Aleph Mapping|aleph mapping]]:
:$\aleph: \On \to \NN'$
where $\NN'$ denotes the [[Definition:Class of Infinite Cardinals|class of infinite cardinals]].
The theorem statement is an immediate consequence o... | Aleph is Infinite Cardinal | https://proofwiki.org/wiki/Aleph_is_Infinite_Cardinal | https://proofwiki.org/wiki/Aleph_is_Infinite_Cardinal | [
"Aleph Mapping"
] | [
"Definition:Ordinal",
"Definition:Infinite Cardinal",
"Definition:Aleph Mapping"
] | [
"Definition:Class of All Ordinals",
"Definition:Aleph Mapping",
"Definition:Class of Infinite Cardinals",
"Category:Aleph Mapping"
] |
proofwiki-6106 | Surjection from Aleph to Ordinal | Let $x$ and $y$ be ordinals.
Suppose that:
:$0 < y < \aleph_{x+1}$
Then there is a surjection:
:$f : \aleph_x \to y$ | :$y < \aleph_{x+1}$, then $y < \aleph_x \lor y \sim \aleph_x$ by Ordinal Less than Successor Aleph.
In either case, $\left|{ y }\right| \le \aleph_x$ by Ordinal in Aleph iff Cardinal in Aleph and Equivalent Sets have Equal Cardinal Numbers.
The existence of the surjection follows from Surjection iff Cardinal Inequality... | Let $x$ and $y$ be [[Definition:Ordinal|ordinals]].
Suppose that:
:$0 < y < \aleph_{x+1}$
Then there is a [[Definition:Surjection|surjection]]:
:$f : \aleph_x \to y$ | :$y < \aleph_{x+1}$, then $y < \aleph_x \lor y \sim \aleph_x$ by [[Ordinal Less than Successor Aleph]].
In either case, $\left|{ y }\right| \le \aleph_x$ by [[Ordinal in Aleph iff Cardinal in Aleph]] and [[Equivalent Sets have Equal Cardinal Numbers]].
The existence of the [[Definition:Surjection|surjection]] follow... | Surjection from Aleph to Ordinal | https://proofwiki.org/wiki/Surjection_from_Aleph_to_Ordinal | https://proofwiki.org/wiki/Surjection_from_Aleph_to_Ordinal | [
"Aleph Mapping"
] | [
"Definition:Ordinal",
"Definition:Surjection"
] | [
"Ordinal Less than Successor Aleph",
"Ordinal in Aleph iff Cardinal in Aleph",
"Equivalent Sets have Equal Cardinal Numbers",
"Definition:Surjection",
"Surjection iff Cardinal Inequality"
] |
proofwiki-6107 | Ordinal Less than Successor Aleph | Let $x$ and $y$ be ordinals.
Then:
:$y < \aleph_{x + 1} \iff y < \aleph_x \lor y \sim \aleph_x$ | === Sufficient Condition ===
{{begin-eqn}}
{{eqn|l = y
|o = <
|r = \aleph_{x + 1}
|c =
}}
{{eqn|ll= \leadsto
|l = \card y
|o = <
|r = \aleph_{x + 1}
|c = Ordinal in Aleph iff Cardinal in Aleph
}}
{{end-eqn}}
But $\card y$ is a cardinal number, so it is either finite or an element of ... | Let $x$ and $y$ be [[Definition:Ordinal|ordinals]].
Then:
:$y < \aleph_{x + 1} \iff y < \aleph_x \lor y \sim \aleph_x$ | === Sufficient Condition ===
{{begin-eqn}}
{{eqn|l = y
|o = <
|r = \aleph_{x + 1}
|c =
}}
{{eqn|ll= \leadsto
|l = \card y
|o = <
|r = \aleph_{x + 1}
|c = [[Ordinal in Aleph iff Cardinal in Aleph]]
}}
{{end-eqn}}
But $\card y$ is a [[Definition:Cardinal Number|cardinal number]], so ... | Ordinal Less than Successor Aleph | https://proofwiki.org/wiki/Ordinal_Less_than_Successor_Aleph | https://proofwiki.org/wiki/Ordinal_Less_than_Successor_Aleph | [
"Aleph Mapping"
] | [
"Definition:Ordinal"
] | [
"Ordinal in Aleph iff Cardinal in Aleph",
"Definition:Cardinal Number",
"Definition:Finite Set",
"Definition:Element",
"Definition:Class of Infinite Cardinals",
"Ordinal is Finite iff Natural Number",
"Definition:Infinite Set",
"Aleph is Infinite",
"Definition:Ordinal",
"Definition:Aleph Mapping",... |
proofwiki-6108 | Aleph is Infinite | Let $x$ be an ordinal.
:$\aleph_x \ge \omega$
where:
:$\aleph$ denotes the aleph mapping
:$\omega$ denotes the minimally inductive set. | Since $0 \le x$, it follows that $\aleph_0 \le \aleph_x$ by definition of the aleph mapping.
But $\aleph_0 = \omega$ by Aleph-Null.
Therefore, $\omega \le \aleph_x$.
{{qed}}
Category:Aleph Mapping
p6e8lhletljz11ngl6ronn8dye4jd29 | Let $x$ be an [[Definition:Ordinal|ordinal]].
:$\aleph_x \ge \omega$
where:
:$\aleph$ denotes the [[Definition:Aleph Mapping|aleph mapping]]
:$\omega$ denotes the [[Definition:Minimally Inductive Set|minimally inductive set]]. | Since $0 \le x$, it follows that $\aleph_0 \le \aleph_x$ by definition of the [[Definition:Aleph Mapping|aleph mapping]].
But $\aleph_0 = \omega$ by [[Aleph-Null]].
Therefore, $\omega \le \aleph_x$.
{{qed}}
[[Category:Aleph Mapping]]
p6e8lhletljz11ngl6ronn8dye4jd29 | Aleph is Infinite | https://proofwiki.org/wiki/Aleph_is_Infinite | https://proofwiki.org/wiki/Aleph_is_Infinite | [
"Aleph Mapping"
] | [
"Definition:Ordinal",
"Definition:Aleph Mapping",
"Definition:Minimally Inductive Set"
] | [
"Definition:Aleph Mapping",
"Aleph-Null",
"Category:Aleph Mapping"
] |
proofwiki-6109 | Aleph-Null | Let $\omega$ denote the minimally inductive set.
:$\omega = \aleph_0$
where $\aleph$ denotes the aleph mapping. | For all $n \in \omega$, $n \notin \NN'$ by the definition of the class of infinite cardinals.
Therefore, $\omega \le \aleph_0$.
{{qed|lemma}}
Moreover, $\omega \in \NN'$ by Minimally Inductive Set is Infinite Cardinal.
Therefore, $\aleph_x = \omega$ for some ordinal $x$.
It follows that $\aleph_0 \le \aleph_x$ since $0... | Let $\omega$ denote the [[Definition:Minimally Inductive Set|minimally inductive set]].
:$\omega = \aleph_0$
where $\aleph$ denotes the [[Definition:Aleph Mapping|aleph mapping]]. | For all $n \in \omega$, $n \notin \NN'$ by the definition of the [[Definition:Class of Infinite Cardinals|class of infinite cardinals]].
Therefore, $\omega \le \aleph_0$.
{{qed|lemma}}
Moreover, $\omega \in \NN'$ by [[Minimally Inductive Set is Infinite Cardinal]].
Therefore, $\aleph_x = \omega$ for some [[Definit... | Aleph-Null | https://proofwiki.org/wiki/Aleph-Null | https://proofwiki.org/wiki/Aleph-Null | [
"Aleph Mapping"
] | [
"Definition:Minimally Inductive Set",
"Definition:Aleph Mapping"
] | [
"Definition:Class of Infinite Cardinals",
"Minimally Inductive Set is Infinite Cardinal",
"Definition:Ordinal",
"Definition:Aleph Mapping",
"Category:Aleph Mapping"
] |
proofwiki-6110 | Minimally Inductive Set is Infinite Cardinal | $\omega$, the minimally inductive set, is an element of the class of infinite cardinals $\NN'$. | By {{Corollary|Cardinal Number Less than Ordinal}}:
:$\card \omega \le \omega$
Moreover, for any $n \in \omega$, by Cardinal of Finite Ordinal:
:$\card n < \card {n + 1} \le \card \omega$
Thus by Cardinal of Finite Ordinal:
:$n \in \card \omega$
Therefore:
:$\omega = \card \omega$
By {{Corollary|Cardinal of Cardinal Eq... | $\omega$, the [[Definition:Minimally Inductive Set|minimally inductive set]], is an [[Definition:Element|element]] of the [[Definition:Class of Infinite Cardinals|class of infinite cardinals]] $\NN'$. | By {{Corollary|Cardinal Number Less than Ordinal}}:
:$\card \omega \le \omega$
Moreover, for any $n \in \omega$, by [[Cardinal of Finite Ordinal]]:
:$\card n < \card {n + 1} \le \card \omega$
Thus by [[Cardinal of Finite Ordinal]]:
:$n \in \card \omega$
Therefore:
:$\omega = \card \omega$
By {{Corollary|Cardinal o... | Minimally Inductive Set is Infinite Cardinal | https://proofwiki.org/wiki/Minimally_Inductive_Set_is_Infinite_Cardinal | https://proofwiki.org/wiki/Minimally_Inductive_Set_is_Infinite_Cardinal | [
"Cardinals",
"Minimally Inductive Set"
] | [
"Definition:Minimally Inductive Set",
"Definition:Element",
"Definition:Class of Infinite Cardinals"
] | [
"Cardinal of Finite Ordinal",
"Cardinal of Finite Ordinal",
"Category:Cardinals",
"Category:Minimally Inductive Set"
] |
proofwiki-6111 | Set of All Mappings is Small Class | Let $S$ and $T$ be small classes.
It follows that the set of all mappings $S^T$ is a small class. | The set of all mappings $S^T$ is equal to the collection of all mappings $f : S \to T$.
Each of these mappings $f$ is a subset of $S \times T$.
Thus, $S^T \subseteq \powerset {S \times T}$.
Therefore, by Cartesian Product is Small and the axiom of powers, $S^T$ is a small class.
{{qed}} | Let $S$ and $T$ be [[Definition:Small Class|small classes]].
It follows that the [[Definition:Set of All Mappings|set of all mappings]] $S^T$ is a [[Definition:Small Class|small class]]. | The [[Definition:Set of All Mappings|set of all mappings]] $S^T$ is equal to the collection of all [[Definition:Mapping|mappings]] $f : S \to T$.
Each of these [[Definition:Mapping|mappings]] $f$ is a [[Definition:Subset|subset]] of $S \times T$.
Thus, $S^T \subseteq \powerset {S \times T}$.
Therefore, by [[Cartesi... | Set of All Mappings is Small Class | https://proofwiki.org/wiki/Set_of_All_Mappings_is_Small_Class | https://proofwiki.org/wiki/Set_of_All_Mappings_is_Small_Class | [
"Cardinals",
"Zermelo-Fraenkel Class Theory"
] | [
"Definition:Small Class",
"Definition:Set of All Mappings",
"Definition:Small Class"
] | [
"Definition:Set of All Mappings",
"Definition:Mapping",
"Definition:Mapping",
"Definition:Subset",
"Cartesian Product is Small",
"Axiom:Axiom of Powers/Set Theory",
"Definition:Small Class"
] |
proofwiki-6112 | Set of All Mappings of Cartesian Product | Let $R$, $S$, and $T$ be sets.
Then:
:$R^{S \times T} \sim \paren {R^S}^T$
where $R^{S \times T}$ denotes the set of all mappings from $S \times T$ to $R$. | Define the mapping $F: \paren {R^S}^T \to R^{S \times T}$ as follows:
:$\map {\map F f} {x, y} = \map {\paren {\map f x} } y$ for all $x \in S , y \in T$.
Suppose $\map F {f_1} = \map F {f_2}$.
Then $\map {\paren {\map {f_1} x} } y = \map {\paren {\map {f_2} x} } y$ for all $x \in S , y \in T$ by the definition of $F$.... | Let $R$, $S$, and $T$ be [[Definition:Set|sets]].
Then:
:$R^{S \times T} \sim \paren {R^S}^T$
where $R^{S \times T}$ denotes the [[Definition:Set of All Mappings|set of all mappings]] from $S \times T$ to $R$. | Define the [[Definition:Mapping|mapping]] $F: \paren {R^S}^T \to R^{S \times T}$ as follows:
:$\map {\map F f} {x, y} = \map {\paren {\map f x} } y$ for all $x \in S , y \in T$.
Suppose $\map F {f_1} = \map F {f_2}$.
Then $\map {\paren {\map {f_1} x} } y = \map {\paren {\map {f_2} x} } y$ for all $x \in S , y \in T... | Set of All Mappings of Cartesian Product | https://proofwiki.org/wiki/Set_of_All_Mappings_of_Cartesian_Product | https://proofwiki.org/wiki/Set_of_All_Mappings_of_Cartesian_Product | [
"Cardinals",
"Cardinality"
] | [
"Definition:Set",
"Definition:Set of All Mappings"
] | [
"Definition:Mapping",
"Equality of Mappings",
"Definition:Injection",
"Definition:Surjection",
"Definition:Bijection",
"Definition:Set Equivalence"
] |
proofwiki-6113 | Covariant Hom Functor is Functor | Let $\mathbf{Set}$ be the category of sets.
Let $\mathbf C$ be a locally small category.
Let $C \in \mathbf C_0$ be an object of $\mathbf C$.
Let $\operatorname{Hom}_\mathbf C \paren {C, \cdot}: \mathbf C \to \mathbf{Set}$ be the covariant hom functor based at $C$.
Then $\operatorname{Hom}_\mathbf C \paren {C, \cdot}$ ... | For brevity and readability, let us write $\operatorname{Hom}$ for $\operatorname{Hom}_\mathbf C$.
For any object $D$ of $\mathbf C$ and morphism $f: C \to D \in \operatorname{Hom} \paren {C, D}$, we have:
:$\operatorname{Hom} \paren {C, \operatorname{id}_D} \paren f = \operatorname{id}_D \circ f = f$
and so:
:$\operat... | Let $\mathbf{Set}$ be the [[Definition:Category of Sets|category of sets]].
Let $\mathbf C$ be a [[Definition:Locally Small Category|locally small category]].
Let $C \in \mathbf C_0$ be an [[Definition:Object|object]] of $\mathbf C$.
Let $\operatorname{Hom}_\mathbf C \paren {C, \cdot}: \mathbf C \to \mathbf{Set}$ be... | For brevity and readability, let us write $\operatorname{Hom}$ for $\operatorname{Hom}_\mathbf C$.
For any [[Definition:Object|object]] $D$ of $\mathbf C$ and [[Definition:Morphism (Category Theory)|morphism]] $f: C \to D \in \operatorname{Hom} \paren {C, D}$, we have:
:$\operatorname{Hom} \paren {C, \operatorname{id}... | Covariant Hom Functor is Functor | https://proofwiki.org/wiki/Covariant_Hom_Functor_is_Functor | https://proofwiki.org/wiki/Covariant_Hom_Functor_is_Functor | [
"Category of Sets",
"Functors"
] | [
"Definition:Category of Sets",
"Definition:Locally Small Category",
"Definition:Object",
"Definition:Covariant Hom Functor",
"Definition:Functor/Covariant"
] | [
"Definition:Object",
"Definition:Morphism",
"Definition:Identity Morphism",
"Definition:Composable Morphisms",
"Definition:Composable Morphisms",
"Equality of Mappings"
] |
proofwiki-6114 | Inverse Morphism is Unique | Let $\mathbf C$ be a metacategory.
Let $f: C \to D$ be an isomorphism of $\mathbf C$.
Then $f$ admits a unique inverse morphism $g: D \to C$. | Since $f$ is an isomorphism, it admits at least one inverse morphism.
Now let $g, g': D \to C$ be two inverse morphisms for $f$.
Then:
{{begin-eqn}}
{{eqn|l = g
|r = g \circ \operatorname{id}_D
|c = Axiom $(C2)$ for metacategories
}}
{{eqn|r = g \circ \left({f \circ g'}\right)
|c = $g'$ is an inverse mor... | Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
Let $f: C \to D$ be an [[Definition:Isomorphism (Category Theory)|isomorphism]] of $\mathbf C$.
Then $f$ admits a [[Definition:Unique|unique]] [[Definition:Inverse Morphism|inverse morphism]] $g: D \to C$. | Since $f$ is an [[Definition:Isomorphism (Category Theory)|isomorphism]], it admits at least one [[Definition:Inverse Morphism|inverse morphism]].
Now let $g, g': D \to C$ be two [[Definition:Inverse Morphism|inverse morphisms]] for $f$.
Then:
{{begin-eqn}}
{{eqn|l = g
|r = g \circ \operatorname{id}_D
|c ... | Inverse Morphism is Unique | https://proofwiki.org/wiki/Inverse_Morphism_is_Unique | https://proofwiki.org/wiki/Inverse_Morphism_is_Unique | [
"Morphisms"
] | [
"Definition:Metacategory",
"Definition:Isomorphism (Category Theory)",
"Definition:Unique",
"Definition:Inverse Morphism"
] | [
"Definition:Isomorphism (Category Theory)",
"Definition:Inverse Morphism",
"Definition:Inverse Morphism",
"Definition:Metacategory",
"Definition:Inverse Morphism",
"Definition:Metacategory",
"Definition:Inverse Morphism",
"Definition:Metacategory"
] |
proofwiki-6115 | Cofinal Ordinal Relation is Reflexive | Let $x$ be an ordinal.
Then $x$ is cofinal to itself.
That is:
:$\operatorname{cof} \left({x, x}\right)$ | {{MissingLinks|to $\le$, mostly. Some of the results have their own pages already}}
Each of the conditions for cofinal ordinals shall be verified:
:$x \le x$ follows by Set is Subset of Itself.
The mapping $f: x \to x$ can simply be the identity mapping $I_x$:
:$I_x: x \to x$
Moreover, $a < b \implies I_x \left({a}\rig... | Let $x$ be an [[Definition:Ordinal|ordinal]].
Then $x$ is [[Definition:Cofinal Relation on Ordinals|cofinal]] to itself.
That is:
:$\operatorname{cof} \left({x, x}\right)$ | {{MissingLinks|to $\le$, mostly. Some of the results have their own pages already}}
Each of the conditions for [[Definition:Cofinal Relation on Ordinals|cofinal]] ordinals shall be verified:
:$x \le x$ follows by [[Set is Subset of Itself]].
The [[Definition:Mapping|mapping]] $f: x \to x$ can simply be the [[Definit... | Cofinal Ordinal Relation is Reflexive | https://proofwiki.org/wiki/Cofinal_Ordinal_Relation_is_Reflexive | https://proofwiki.org/wiki/Cofinal_Ordinal_Relation_is_Reflexive | [
"Ordinals"
] | [
"Definition:Ordinal",
"Definition:Cofinal Relation on Ordinals"
] | [
"Definition:Cofinal Relation on Ordinals",
"Set is Subset of Itself",
"Definition:Mapping",
"Definition:Identity Mapping",
"Definition:Identity Mapping",
"Definition:Strictly Increasing",
"Definition:Identity Mapping",
"Existential Generalisation"
] |
proofwiki-6116 | Cofinal Ordinal Relation is Transitive | Let $x$, $y$, and $z$ be ordinals.
Let $\operatorname {cof}$ denote the cofinal relation.
Then:
:$\map {\operatorname {cof} } {x, y} \land \map {\operatorname {cof} } {y, z} \implies \map {\operatorname {cof} } {x, z}$ | {{tidy|Clarify and structure}}
The conditions for $z$ being cofinal with $x$ shall be verified individually.
Let $\le$ denote the subset relation.
Since $x \le y$ and $y \le z$, it follows that $x \le z$ by Subset Relation is Transitive
Let $f : y \to x$ be the strictly increasing mapping that satisfies:
:$\forall a \i... | Let $x$, $y$, and $z$ be [[Definition:Ordinal|ordinals]].
Let $\operatorname {cof}$ denote the [[Definition:Cofinal Relation on Ordinals|cofinal relation]].
Then:
:$\map {\operatorname {cof} } {x, y} \land \map {\operatorname {cof} } {y, z} \implies \map {\operatorname {cof} } {x, z}$ | {{tidy|Clarify and structure}}
The conditions for $z$ being [[Definition:Cofinal Relation on Ordinals|cofinal]] with $x$ shall be verified individually.
Let $\le$ denote the [[Ordering on Ordinal is Subset Relation|subset relation]].
Since $x \le y$ and $y \le z$, it follows that $x \le z$ by [[Subset Relation is Tr... | Cofinal Ordinal Relation is Transitive | https://proofwiki.org/wiki/Cofinal_Ordinal_Relation_is_Transitive | https://proofwiki.org/wiki/Cofinal_Ordinal_Relation_is_Transitive | [
"Ordinals"
] | [
"Definition:Ordinal",
"Definition:Cofinal Relation on Ordinals"
] | [
"Definition:Cofinal Relation on Ordinals",
"Ordering on Ordinal is Subset Relation",
"Subset Relation is Transitive",
"Definition:Strictly Increasing/Mapping",
"Definition:Strictly Increasing/Mapping",
"Composite of Strictly Increasing Mappings is Strictly Increasing",
"Definition:Strictly Increasing/Ma... |
proofwiki-6117 | Cofinal to Zero iff Ordinal is Zero | Let $x$ be an ordinal.
Let $\operatorname{cof}$ denote the cofinal relation.
Let $0$ denote the zero ordinal.
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$\map {\operatorname{cof} } {x, 0}$}}
{{item|(2):|$\map {\operatorname{cof} } {0, x}$}}
{{item|(3):|$x {{=}} 0$}}
{{end-itemize}} | === $(1) \implies (3)$ ===
If $\map {\operatorname{cof} } {x, 0}$, then there is a function $f : x \to 0$.
If $x \ne 0$, then $x$ has an element $a$.
But then, $\map f a \in 0$, which contradicts the definition of the empty set.
Therefore, $x = 0$.
{{qed|lemma}} | Let $x$ be an [[Definition:Ordinal|ordinal]].
Let $\operatorname{cof}$ denote the [[Definition:Cofinal Relation on Ordinals|cofinal relation]].
Let $0$ denote the [[Definition:Zero (Ordinal)|zero ordinal]].
{{TFAE}}
{{begin-itemize}}
{{item|(1):|$\map {\operatorname{cof} } {x, 0}$}}
{{item|(2):|$\map {\operatorname... | === $(1) \implies (3)$ ===
If $\map {\operatorname{cof} } {x, 0}$, then there is a function $f : x \to 0$.
If $x \ne 0$, then $x$ has an element $a$.
But then, $\map f a \in 0$, which contradicts the definition of the [[Definition:Empty Set|empty set]].
Therefore, $x = 0$.
{{qed|lemma}} | Cofinal to Zero iff Ordinal is Zero | https://proofwiki.org/wiki/Cofinal_to_Zero_iff_Ordinal_is_Zero | https://proofwiki.org/wiki/Cofinal_to_Zero_iff_Ordinal_is_Zero | [
"Ordinals"
] | [
"Definition:Ordinal",
"Definition:Cofinal Relation on Ordinals",
"Definition:Zero (Ordinal)"
] | [
"Definition:Empty Set"
] |
proofwiki-6118 | Condition for Cofinal Nonlimit Ordinals | Let $x$ and $y$ be nonlimit ordinals.
Let $\operatorname{cof}$ denote the cofinal relation.
Let $\le$ denote the subset relation.
{{explain|This statement could be worded a little more carefully. It is, from examining the link, clear that the subset relation and the ordering relation are the same thing, but it grates t... | Both $x$ and $y$ are non-empty, so by the definition of a limit ordinal:
:$x = z^+$ for some $z$.
:$y = w^+$ for some $w$.
:$\bigcup z^+ \le \bigcup w^+$ follows by Set Union Preserves Subsets/General Result.
:$z \le w$ follows by Union of Successor Ordinal.
Define the function $f : x \to y$ as follows:
:$\map f a = \b... | Let $x$ and $y$ be [[Definition:Limit Ordinal|nonlimit]] [[Definition:Ordinal|ordinals]].
Let $\operatorname{cof}$ denote the [[Definition:Cofinal Relation on Ordinals|cofinal relation]].
Let $\le$ denote the [[Ordering on Ordinal is Subset Relation|subset relation]].
{{explain|This statement could be worded a little... | Both $x$ and $y$ are [[Definition:Non-Empty Set|non-empty]], so by the definition of a [[Definition:Limit Ordinal|limit ordinal]]:
:$x = z^+$ for some $z$.
:$y = w^+$ for some $w$.
:$\bigcup z^+ \le \bigcup w^+$ follows by [[Set Union Preserves Subsets/General Result]].
:$z \le w$ follows by [[Union of Successor O... | Condition for Cofinal Nonlimit Ordinals | https://proofwiki.org/wiki/Condition_for_Cofinal_Nonlimit_Ordinals | https://proofwiki.org/wiki/Condition_for_Cofinal_Nonlimit_Ordinals | [
"Ordinals"
] | [
"Definition:Limit Ordinal",
"Definition:Ordinal",
"Definition:Cofinal Relation on Ordinals",
"Ordering on Ordinal is Subset Relation"
] | [
"Definition:Non-Empty Set",
"Definition:Limit Ordinal",
"Set Union Preserves Subsets/General Result",
"Union of Successor Ordinal",
"Proof by Cases"
] |
proofwiki-6119 | Nonlimit Ordinal Cofinal to One | Let $x$ be a nonlimit non-empty ordinal.
Let $\operatorname{cof}$ denote the cofinal relation.
Let $1$ denote the ordinal one.
Then:
:$\operatorname{cof} \left({x, 1}\right)$ | Since $1 = 0^+$, $1$ is not a limit ordinal.
Let $\le$ denote the subset relation.
It follows that $0 < 1$ by Ordinal is Less than Successor.
Moreover, $1 \le x$ follows by the fact that $0 < x$ and Successor of Element of Ordinal is Subset.
Thus we have $0 < 1 \le x$ and so by Condition for Cofinal Nonlimit Ordinals:
... | Let $x$ be a [[Definition:Limit Ordinal|nonlimit]] [[Definition:Non-Empty Set|non-empty]] [[Definition:Ordinal|ordinal]].
Let $\operatorname{cof}$ denote the [[Definition:Cofinal Relation on Ordinals|cofinal relation]].
Let $1$ denote the [[Definition:One (Ordinal)|ordinal one]].
Then:
:$\operatorname{cof} \left... | Since $1 = 0^+$, $1$ is not a [[Definition:Limit Ordinal|limit ordinal]].
Let $\le$ denote the [[Ordering on Ordinal is Subset Relation|subset relation]].
It follows that $0 < 1$ by [[Ordinal is Less than Successor]].
Moreover, $1 \le x$ follows by the fact that $0 < x$ and [[Successor of Element of Ordinal is Subs... | Nonlimit Ordinal Cofinal to One | https://proofwiki.org/wiki/Nonlimit_Ordinal_Cofinal_to_One | https://proofwiki.org/wiki/Nonlimit_Ordinal_Cofinal_to_One | [
"Ordinals"
] | [
"Definition:Limit Ordinal",
"Definition:Non-Empty Set",
"Definition:Ordinal",
"Definition:Cofinal Relation on Ordinals",
"Definition:One (Ordinal)"
] | [
"Definition:Limit Ordinal",
"Ordering on Ordinal is Subset Relation",
"Ordinal is Less than Successor",
"Successor of Element of Ordinal is Subset",
"Condition for Cofinal Nonlimit Ordinals"
] |
proofwiki-6120 | Cofinal Limit Ordinals | Let $x$ and $y$ be ordinals.
Let $\mathrm {cof}$ denote the cofinal relation.
Let $K_{II}$ denote the class of all limit ordinals.
Then:
:$\map {\mathrm {cof} } {x, y} \implies \paren {x \in K_{II} \iff y \in K_{II} }$ | === Necessary Condition ===
Suppose $y \in K_{II}$.
$x \ne 0$ by Cofinal to Zero iff Ordinal is Zero.
If $x = z^+$ for some $z$, then $z = \bigcup x$ by Union of Successor Ordinal.
Therefore, $z$ would be the least upper bound of $x$.
Since $\map {\mathrm {cof} } {x, y}$, it follows by the definition of cofinal that:
:... | Let $x$ and $y$ be [[Definition:Ordinal|ordinals]].
Let $\mathrm {cof}$ denote the [[Definition:Cofinal Relation on Ordinals|cofinal relation]].
Let $K_{II}$ denote the class of all [[Definition:Limit Ordinal|limit ordinals]].
Then:
:$\map {\mathrm {cof} } {x, y} \implies \paren {x \in K_{II} \iff y \in K_{II} }$ | === Necessary Condition ===
Suppose $y \in K_{II}$.
$x \ne 0$ by [[Cofinal to Zero iff Ordinal is Zero]].
If $x = z^+$ for some $z$, then $z = \bigcup x$ by [[Union of Successor Ordinal]].
Therefore, $z$ would be the [[Union of Ordinals is Least Upper Bound|least upper bound]] of $x$.
Since $\map {\mathrm {cof} ... | Cofinal Limit Ordinals | https://proofwiki.org/wiki/Cofinal_Limit_Ordinals | https://proofwiki.org/wiki/Cofinal_Limit_Ordinals | [
"Ordinals"
] | [
"Definition:Ordinal",
"Definition:Cofinal Relation on Ordinals",
"Definition:Limit Ordinal"
] | [
"Cofinal to Zero iff Ordinal is Zero",
"Union of Successor Ordinal",
"Union of Ordinals is Least Upper Bound",
"Definition:Cofinal Relation on Ordinals",
"Definition:Limit Ordinal",
"Successor of Ordinal Smaller than Limit Ordinal is also Smaller",
"Union of Ordinals is Least Upper Bound",
"Cofinal to... |
proofwiki-6121 | Smallest Positive Integer Combination is Greatest Common Divisor | Let $a, b \in \Z_{>0}$ be (strictly) positive integers.
Let $d \in \Z_{>0}$ be the smallest positive integer such that:
: $d = a s + b t$
where $s, t \in \Z$.
Then:
:$(1): \quad d \divides a \land d \divides b$
:$(2): \quad c \divides a \land c \divides b \implies c \divides d$
where $\divides$ denotes divisibility.
Th... | Let $D$ be the subset of $\Z_{>0}$ defined as:
:$D = \set {a s + b t: s, t \in \Z, a s + b t > 0}$
Setting $s = 1$ and $t = 0$ it is clear that $a = \paren {a \times 1 + b \times 0} \in D$.
So $D \ne \O$.
By Set of Integers Bounded Below by Integer has Smallest Element, $D$ has a smallest element $d$, say.
Thus $d = a ... | Let $a, b \in \Z_{>0}$ be [[Definition:Strictly Positive Integer|(strictly) positive integers]].
Let $d \in \Z_{>0}$ be the [[Definition:Smallest Element|smallest]] [[Definition:Strictly Positive Integer|positive integer]] such that:
: $d = a s + b t$
where $s, t \in \Z$.
Then:
:$(1): \quad d \divides a \land d \div... | Let $D$ be the [[Definition:Subset|subset]] of $\Z_{>0}$ defined as:
:$D = \set {a s + b t: s, t \in \Z, a s + b t > 0}$
Setting $s = 1$ and $t = 0$ it is clear that $a = \paren {a \times 1 + b \times 0} \in D$.
So $D \ne \O$.
By [[Set of Integers Bounded Below by Integer has Smallest Element]], $D$ has a [[Definiti... | Smallest Positive Integer Combination is Greatest Common Divisor/Proof 1 | https://proofwiki.org/wiki/Smallest_Positive_Integer_Combination_is_Greatest_Common_Divisor | https://proofwiki.org/wiki/Smallest_Positive_Integer_Combination_is_Greatest_Common_Divisor/Proof_1 | [
"Greatest Common Divisor",
"Integer Combinations",
"Smallest Positive Integer Combination is Greatest Common Divisor"
] | [
"Definition:Strictly Positive/Integer",
"Definition:Smallest Element",
"Definition:Strictly Positive/Integer",
"Definition:Divisor (Algebra)/Integer",
"Definition:Greatest Common Divisor/Integers"
] | [
"Definition:Subset",
"Set of Integers Bounded Below by Integer has Smallest Element",
"Definition:Smallest Element",
"Division Theorem",
"Definition:Smallest Element"
] |
proofwiki-6122 | Smallest Positive Integer Combination is Greatest Common Divisor | Let $a, b \in \Z_{>0}$ be (strictly) positive integers.
Let $d \in \Z_{>0}$ be the smallest positive integer such that:
: $d = a s + b t$
where $s, t \in \Z$.
Then:
:$(1): \quad d \divides a \land d \divides b$
:$(2): \quad c \divides a \land c \divides b \implies c \divides d$
where $\divides$ denotes divisibility.
Th... | From Bézout's Identity we have:
Let $a, b \in \Z$ such that $a$ and $b$ are not both zero.
Let $\gcd \set {a, b}$ be the greatest common divisor of $a$ and $b$.
Then:
:$\exists x, y \in \Z: a x + b y = \gcd \set {a, b}$
Furthermore, $\gcd \set {a, b}$ is the smallest positive integer combination of $a$ and $b$.
In this... | Let $a, b \in \Z_{>0}$ be [[Definition:Strictly Positive Integer|(strictly) positive integers]].
Let $d \in \Z_{>0}$ be the [[Definition:Smallest Element|smallest]] [[Definition:Strictly Positive Integer|positive integer]] such that:
: $d = a s + b t$
where $s, t \in \Z$.
Then:
:$(1): \quad d \divides a \land d \div... | From [[Bézout's Identity]] we have:
Let $a, b \in \Z$ such that $a$ and $b$ are not both [[Definition:Zero (Number)|zero]].
Let $\gcd \set {a, b}$ be the [[Definition:Greatest Common Divisor of Integers|greatest common divisor]] of $a$ and $b$.
Then:
:$\exists x, y \in \Z: a x + b y = \gcd \set {a, b}$
Furthermore,... | Smallest Positive Integer Combination is Greatest Common Divisor/Proof 2 | https://proofwiki.org/wiki/Smallest_Positive_Integer_Combination_is_Greatest_Common_Divisor | https://proofwiki.org/wiki/Smallest_Positive_Integer_Combination_is_Greatest_Common_Divisor/Proof_2 | [
"Greatest Common Divisor",
"Integer Combinations",
"Smallest Positive Integer Combination is Greatest Common Divisor"
] | [
"Definition:Strictly Positive/Integer",
"Definition:Smallest Element",
"Definition:Strictly Positive/Integer",
"Definition:Divisor (Algebra)/Integer",
"Definition:Greatest Common Divisor/Integers"
] | [
"Bézout's Identity",
"Definition:Zero (Number)",
"Definition:Greatest Common Divisor/Integers",
"Definition:Smallest Element",
"Definition:Positive/Integer",
"Definition:Integer Combination",
"Definition:Zero (Number)",
"Definition:Greatest Common Divisor/Integers"
] |
proofwiki-6123 | Matrix is Nonsingular iff Determinant has Multiplicative Inverse | Let $R$ be a commutative ring with unity.
Let $\mathbf A \in R^{n \times n}$ be a square matrix of order $n$.
Then $\mathbf A$ is nonsingular {{iff}} its determinant is a invertible in $R$.
If $R$ is one of the standard number fields $\Q$, $\R$ or $\C$, this translates into:
:$\mathbf A$ is nonsingular {{iff}} its dete... | === Necessary Condition ===
{{:Matrix is Nonsingular iff Determinant has Multiplicative Inverse/Necessary Condition}} | Let $R$ be a [[Definition:Commutative and Unitary Ring|commutative ring with unity]].
Let $\mathbf A \in R^{n \times n}$ be a [[Definition:Square Matrix|square matrix]] of [[Definition:Order of Square Matrix|order]] $n$.
Then $\mathbf A$ is [[Definition:Nonsingular Matrix|nonsingular]] {{iff}} its [[Definition:Deter... | === [[Matrix is Nonsingular iff Determinant has Multiplicative Inverse/Necessary Condition|Necessary Condition]] ===
{{:Matrix is Nonsingular iff Determinant has Multiplicative Inverse/Necessary Condition}} | Matrix is Nonsingular iff Determinant has Multiplicative Inverse | https://proofwiki.org/wiki/Matrix_is_Nonsingular_iff_Determinant_has_Multiplicative_Inverse | https://proofwiki.org/wiki/Matrix_is_Nonsingular_iff_Determinant_has_Multiplicative_Inverse | [
"Matrix is Nonsingular iff Determinant has Multiplicative Inverse",
"Inverse Matrices",
"Determinants"
] | [
"Definition:Commutative and Unitary Ring",
"Definition:Matrix/Square Matrix",
"Definition:Matrix/Square Matrix/Order",
"Definition:Nonsingular Matrix",
"Definition:Determinant/Matrix",
"Definition:Unit of Ring",
"Definition:Standard Number Field",
"Definition:Nonsingular Matrix",
"Definition:Determi... | [
"Matrix is Nonsingular iff Determinant has Multiplicative Inverse/Necessary Condition"
] |
proofwiki-6124 | Definition:Smith-Volterra-Cantor Set | Let $G$ be a Cantor collection.
Let $g_0 \in G$ such that $\map \mu {g_0} = b$.
Let $p$ be a natural number.
Then there are two nonempty disjoint sets $N_p$ and $P$ such that:
:$g_0 = N_p \bigcup P$
where $N_p$ is nowhere dense in the relative topology on $g_0$ and $\map \mu P \le b / p$. | Let $n = p + 1$ for $G, g_0, p, b$ as given in the hypothesis.
We construct our Cantor set by removing a diminishing proportion of sets available.
The construction will be accomplished using two techniques.
First, a ``conversational" approach, then one with more mathematical symbolism, or ``rigor". | Let $G$ be a Cantor collection.
Let $g_0 \in G$ such that $\map \mu {g_0} = b$.
Let $p$ be a [[Definition:Natural Number|natural number]].
Then there are two nonempty disjoint sets $N_p$ and $P$ such that:
:$g_0 = N_p \bigcup P$
where $N_p$ is nowhere dense in the relative topology on $g_0$ and $\map \mu P \le b ... | Let $n = p + 1$ for $G, g_0, p, b$ as given in the hypothesis.
We construct our Cantor set by removing a diminishing proportion of sets available.
The construction will be accomplished using two techniques.
First, a ``conversational" approach, then one with more mathematical symbolism, or ``rigor". | Definition:Smith-Volterra-Cantor Set | https://proofwiki.org/wiki/Definition:Smith-Volterra-Cantor_Set | https://proofwiki.org/wiki/Definition:Smith-Volterra-Cantor_Set | [
"Definitions/Cantor Set"
] | [
"Definition:Natural Numbers"
] | [] |
proofwiki-6125 | Prime Groups of Same Order are Isomorphic | Two prime groups of the same order are isomorphic to each other. | Let $G_1$ and $G_2$ be prime groups, both of finite order $p$.
From Prime Group is Cyclic, both $G_1$ and $G_2$ are cyclic.
The result follows directly from Cyclic Groups of Same Order are Isomorphic.
{{Qed}} | Two [[Definition:Prime Group|prime groups]] of the same [[Definition:Order of Group|order]] are [[Definition:Group Isomorphism|isomorphic]] to each other. | Let $G_1$ and $G_2$ be [[Definition:Prime Group|prime groups]], both of [[Definition:Finite Group|finite order]] $p$.
From [[Prime Group is Cyclic]], both $G_1$ and $G_2$ are [[Definition:Cyclic Group|cyclic]].
The result follows directly from [[Cyclic Groups of Same Order are Isomorphic]].
{{Qed}} | Prime Groups of Same Order are Isomorphic | https://proofwiki.org/wiki/Prime_Groups_of_Same_Order_are_Isomorphic | https://proofwiki.org/wiki/Prime_Groups_of_Same_Order_are_Isomorphic | [
"Prime Groups",
"Group Isomorphisms"
] | [
"Definition:Prime Group",
"Definition:Order of Structure",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism"
] | [
"Definition:Prime Group",
"Definition:Finite Group",
"Prime Group is Cyclic",
"Definition:Cyclic Group",
"Cyclic Groups of Same Order are Isomorphic"
] |
proofwiki-6126 | Infinite Cyclic Group is Unique up to Isomorphism | All infinite cyclic groups are isomorphic.
That is, up to isomorphism, there is only one infinite cyclic group. | Let $G_1$ and $G_2$ be infinite cyclic groups.
From Infinite Cyclic Group is Isomorphic to Integers we have:
:$G_1 \cong \struct {\Z, +} \cong G_2$
where $\struct {\Z, +}$ is the additive group of integers.
From Isomorphism is Equivalence Relation it follows that:
:$G_1 \cong G_2$
{{Qed}} | All [[Definition:Infinite Cyclic Group|infinite cyclic groups]] are [[Definition:Group Isomorphism|isomorphic]].
That is, up to [[Definition:Group Isomorphism|isomorphism]], there is only one [[Definition:Infinite Cyclic Group|infinite cyclic group]]. | Let $G_1$ and $G_2$ be [[Definition:Infinite Cyclic Group|infinite cyclic groups]].
From [[Infinite Cyclic Group is Isomorphic to Integers]] we have:
:$G_1 \cong \struct {\Z, +} \cong G_2$
where $\struct {\Z, +}$ is the [[Definition:Additive Group of Integers|additive group of integers]].
From [[Isomorphism is Equiva... | Infinite Cyclic Group is Unique up to Isomorphism | https://proofwiki.org/wiki/Infinite_Cyclic_Group_is_Unique_up_to_Isomorphism | https://proofwiki.org/wiki/Infinite_Cyclic_Group_is_Unique_up_to_Isomorphism | [
"Infinite Cyclic Group",
"Group Isomorphisms"
] | [
"Definition:Infinite Cyclic Group",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Infinite Cyclic Group"
] | [
"Definition:Infinite Cyclic Group",
"Infinite Cyclic Group is Isomorphic to Integers",
"Definition:Additive Group of Integers",
"Isomorphism is Equivalence Relation"
] |
proofwiki-6127 | Intersection of Subgroups is Subgroup/General Result | Let $\mathbb S$ be a set of subgroups of $\struct {G, \circ}$, where $\mathbb S \ne \O$.
Then the intersection $\ds \bigcap \mathbb S$ of the elements of $\mathbb S$ is itself a subgroup of $G$.
Also, $\ds \bigcap \mathbb S$ is the largest subgroup of $\struct {G, \circ}$ contained in each element of $\mathbb S$. | Let $\ds H = \bigcap \mathbb S$.
Let $H_k$ be an arbitrary element of $\mathbb S$.
Then:
{{begin-eqn}}
{{eqn | l = a, b
| o = \in
| r = H
| c =
}}
{{eqn | ll= \leadsto
| q = \forall k
| l = a, b
| o = \in
| r = H_k
| c = {{Defof|Intersection of Set of Sets}}
}}
{{eqn | l... | Let $\mathbb S$ be a [[Definition:Set|set]] of [[Definition:Subgroup|subgroups]] of $\struct {G, \circ}$, where $\mathbb S \ne \O$.
Then the [[Definition:Intersection of Set of Sets|intersection]] $\ds \bigcap \mathbb S$ of the [[Definition:Element|elements]] of $\mathbb S$ is itself a [[Definition:Subgroup|subgroup]... | Let $\ds H = \bigcap \mathbb S$.
Let $H_k$ be an arbitrary [[Definition:Element|element]] of $\mathbb S$.
Then:
{{begin-eqn}}
{{eqn | l = a, b
| o = \in
| r = H
| c =
}}
{{eqn | ll= \leadsto
| q = \forall k
| l = a, b
| o = \in
| r = H_k
| c = {{Defof|Intersection of ... | Intersection of Subgroups is Subgroup/General Result | https://proofwiki.org/wiki/Intersection_of_Subgroups_is_Subgroup/General_Result | https://proofwiki.org/wiki/Intersection_of_Subgroups_is_Subgroup/General_Result | [
"Subgroups",
"Set Intersection"
] | [
"Definition:Set",
"Definition:Subgroup",
"Definition:Set Intersection/Set of Sets",
"Definition:Element",
"Definition:Subgroup",
"Definition:Subgroup",
"Definition:Element"
] | [
"Definition:Element",
"Definition:Group",
"One-Step Subgroup Test",
"Definition:Subgroup",
"Definition:Subgroup",
"Definition:Subgroup",
"Definition:Element",
"Definition:Subgroup",
"Definition:Subgroup",
"Definition:Element"
] |
proofwiki-6128 | Union of Subgroups/Corollary 2 | Let $H \vee K$ be the join of $H$ and $K$.
Then $H \vee K = H \cup K$ {{iff}} $H \subseteq K$ or $K \subseteq H$. | From the definition of join, $H \vee K$ is the smallest subgroup of $G$ containing $H \cup K$.
The result follows from Union of Subgroups.
{{qed}} | Let $H \vee K$ be the [[Definition:Join of Subgroups|join]] of $H$ and $K$.
Then $H \vee K = H \cup K$ {{iff}} $H \subseteq K$ or $K \subseteq H$. | From the definition of [[Definition:Join of Subgroups|join]], $H \vee K$ is the smallest [[Definition:Subgroup|subgroup]] of $G$ containing $H \cup K$.
The result follows from [[Union of Subgroups]].
{{qed}} | Union of Subgroups/Corollary 2 | https://proofwiki.org/wiki/Union_of_Subgroups/Corollary_2 | https://proofwiki.org/wiki/Union_of_Subgroups/Corollary_2 | [
"Union of Subgroups"
] | [
"Definition:Join of Subgroups"
] | [
"Definition:Join of Subgroups",
"Definition:Subgroup",
"Union of Subgroups"
] |
proofwiki-6129 | Existence of Unique Subgroup Generated by Subset/Singleton Generator | Let $a \in G$.
Then $H = \gen a = \set {a^n: n \in \Z}$ is the unique smallest subgroup of $G$ such that $a \in H$.
That is:
: $K \le G: a \in K \implies H \subseteq K$ | From Powers of Element form Subgroup, $H = \set {a^n: n \in \Z}$ is a subgroup of $G$.
Let $K \le G: a \in K$.
Then $\forall n \in \Z: a^n \in K$.
Thus, $H \subseteq K$.
{{qed}} | Let $a \in G$.
Then $H = \gen a = \set {a^n: n \in \Z}$ is the [[Definition:Unique|unique]] smallest [[Definition:Subgroup|subgroup]] of $G$ such that $a \in H$.
That is:
: $K \le G: a \in K \implies H \subseteq K$ | From [[Powers of Element form Subgroup]], $H = \set {a^n: n \in \Z}$ is a [[Definition:Subgroup|subgroup]] of $G$.
Let $K \le G: a \in K$.
Then $\forall n \in \Z: a^n \in K$.
Thus, $H \subseteq K$.
{{qed}} | Existence of Unique Subgroup Generated by Subset/Singleton Generator | https://proofwiki.org/wiki/Existence_of_Unique_Subgroup_Generated_by_Subset/Singleton_Generator | https://proofwiki.org/wiki/Existence_of_Unique_Subgroup_Generated_by_Subset/Singleton_Generator | [
"Generated Subgroups"
] | [
"Definition:Unique",
"Definition:Subgroup"
] | [
"Powers of Element form Subgroup",
"Definition:Subgroup"
] |
proofwiki-6130 | Intersection of Subsemigroups/General Result | Let $\mathbb H$ be a set of subsemigroups of $\struct {S, \circ}$, where $\mathbb H \ne \O$.
Then the intersection $\bigcap \mathbb H$ of the members of $\mathbb H$ is the largest subsemigroup of $\struct {S, \circ}$ contained in each member of $\mathbb H$. | Let $T = \bigcap \mathbb S$.
Then:
{{begin-eqn}}
{{eqn | l = a, b
| o = \in
| r = T
| c =
}}
{{eqn | ll= \leadsto
| q = \forall K \in \mathbb S
| l = a, b
| o = \in
| r = K
| c = {{Defof|Set Intersection}}
}}
{{eqn | ll= \leadsto
| q = \forall K \in \mathbb S
... | Let $\mathbb H$ be a [[Definition:Set|set]] of [[Definition:Subsemigroup|subsemigroups]] of $\struct {S, \circ}$, where $\mathbb H \ne \O$.
Then the [[Definition:Set Intersection|intersection]] $\bigcap \mathbb H$ of the members of $\mathbb H$ is the largest [[Definition:Subsemigroup|subsemigroup]] of $\struct {S, \c... | Let $T = \bigcap \mathbb S$.
Then:
{{begin-eqn}}
{{eqn | l = a, b
| o = \in
| r = T
| c =
}}
{{eqn | ll= \leadsto
| q = \forall K \in \mathbb S
| l = a, b
| o = \in
| r = K
| c = {{Defof|Set Intersection}}
}}
{{eqn | ll= \leadsto
| q = \forall K \in \mathbb S
... | Intersection of Subsemigroups/General Result/Proof 1 | https://proofwiki.org/wiki/Intersection_of_Subsemigroups/General_Result | https://proofwiki.org/wiki/Intersection_of_Subsemigroups/General_Result/Proof_1 | [
"Intersection of Subsemigroups"
] | [
"Definition:Set",
"Definition:Subsemigroup",
"Definition:Set Intersection",
"Definition:Subsemigroup"
] | [
"Definition:Subsemigroup",
"Subsemigroup Closure Test",
"Definition:Subsemigroup",
"Definition:Subsemigroup",
"Definition:Subsemigroup"
] |
proofwiki-6131 | Intersection of Subsemigroups/General Result | Let $\mathbb H$ be a set of subsemigroups of $\struct {S, \circ}$, where $\mathbb H \ne \O$.
Then the intersection $\bigcap \mathbb H$ of the members of $\mathbb H$ is the largest subsemigroup of $\struct {S, \circ}$ contained in each member of $\mathbb H$. | From Set of Subsemigroups forms Complete Lattice:
:$\struct {\mathbb S, \subseteq}$ is a complete lattice.
where for every set $\mathbb H$ of subsemigroups of $S$:
:the infimum of $\mathbb H$ necessarily admitted by $\mathbb H$ is $\ds \bigcap \mathbb H$.
Hence the result, by definition of infimum.
{{qed}} | Let $\mathbb H$ be a [[Definition:Set|set]] of [[Definition:Subsemigroup|subsemigroups]] of $\struct {S, \circ}$, where $\mathbb H \ne \O$.
Then the [[Definition:Set Intersection|intersection]] $\bigcap \mathbb H$ of the members of $\mathbb H$ is the largest [[Definition:Subsemigroup|subsemigroup]] of $\struct {S, \c... | From [[Set of Subsemigroups forms Complete Lattice]]:
:$\struct {\mathbb S, \subseteq}$ is a [[Definition:Complete Lattice|complete lattice]].
where for every [[Definition:Set|set]] $\mathbb H$ of [[Definition:Subsemigroup|subsemigroups]] of $S$:
:the [[Definition:Infimum of Set|infimum]] of $\mathbb H$ necessarily ad... | Intersection of Subsemigroups/General Result/Proof 2 | https://proofwiki.org/wiki/Intersection_of_Subsemigroups/General_Result | https://proofwiki.org/wiki/Intersection_of_Subsemigroups/General_Result/Proof_2 | [
"Intersection of Subsemigroups"
] | [
"Definition:Set",
"Definition:Subsemigroup",
"Definition:Set Intersection",
"Definition:Subsemigroup"
] | [
"Set of Subsemigroups forms Complete Lattice",
"Definition:Complete Lattice",
"Definition:Set",
"Definition:Subsemigroup",
"Definition:Infimum of Set",
"Definition:Infimum of Set"
] |
proofwiki-6132 | Join of Subgroups is Group Generated by Union | Let $G$ be a group.
Let $H$ and $K$ be subgroups of $G$.
Let $S$ be the set of words of $H \cup K$.
Then $S$ is a subgroup of $K$ such that:
:$S = \gen {H \cup K} = H \vee K$
where $H \vee K$ denotes the join of $H$ and $K$. | By definition, the set of words in $H \cup K$ is:
:$S = \map W {H \cup K} := \set {s_1 \circ s_2 \circ \cdots \circ s_n: n \in \N_{>0}: s_i \in H \cup K 1 \le i \le n}$
Let $h \in H$.
Then setting $n = 1$ in the above definition and letting $s_1 = h$ it follows that $H \subseteq S$.
Similarly it is seen that $K \subset... | Let $G$ be a [[Definition:Group|group]].
Let $H$ and $K$ be [[Definition:Subgroup|subgroups]] of $G$.
Let $S$ be the [[Definition:Word (Abstract Algebra)|set of words]] of $H \cup K$.
Then $S$ is a [[Definition:Subgroup|subgroup]] of $K$ such that:
:$S = \gen {H \cup K} = H \vee K$
where $H \vee K$ denotes the [[De... | By definition, the [[Definition:Word (Abstract Algebra)|set of words]] in $H \cup K$ is:
:$S = \map W {H \cup K} := \set {s_1 \circ s_2 \circ \cdots \circ s_n: n \in \N_{>0}: s_i \in H \cup K 1 \le i \le n}$
Let $h \in H$.
Then setting $n = 1$ in the above definition and letting $s_1 = h$ it follows that $H \subsete... | Join of Subgroups is Group Generated by Union | https://proofwiki.org/wiki/Join_of_Subgroups_is_Group_Generated_by_Union | https://proofwiki.org/wiki/Join_of_Subgroups_is_Group_Generated_by_Union | [
"Group Theory"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Word (Abstract Algebra)",
"Definition:Subgroup",
"Definition:Join of Subgroups"
] | [
"Definition:Word (Abstract Algebra)",
"Union is Smallest Superset",
"Set of Words Generates Group",
"Definition:Subgroup",
"Definition:Generator of Subgroup"
] |
proofwiki-6133 | Order of Subgroup Product/Corollary | :$\size {H \vee K} \ge \dfrac {\order H \order K} {\order {H \cap K} }$
or
:$\dfrac {\size {H \vee K} } {\order H} \ge \dfrac {\order K} {\order {H \cap K} }$
where $H \vee K$ denotes join and $\order H$ denotes the order of $H$. | From Order of Subgroup Product:
: $(1): \quad \size {H K} = \dfrac {\order H \order K} {\order {H \cap K} }$
From Subset Product is Subset of Generator, we have that:
: $H K \subseteq H \vee K$
where $H K$ is the subset product of $H$ and $K$.
Thus:
: $(2): \quad \size {H \vee K} \ge \size {H K}$
The result follows by ... | :$\size {H \vee K} \ge \dfrac {\order H \order K} {\order {H \cap K} }$
or
:$\dfrac {\size {H \vee K} } {\order H} \ge \dfrac {\order K} {\order {H \cap K} }$
where $H \vee K$ denotes [[Definition:Join of Subgroups|join]] and $\order H$ denotes the [[Definition:Order of Structure|order of $H$]]. | From [[Order of Subgroup Product]]:
: $(1): \quad \size {H K} = \dfrac {\order H \order K} {\order {H \cap K} }$
From [[Subset Product is Subset of Generator]], we have that:
: $H K \subseteq H \vee K$
where $H K$ is the [[Definition:Subset Product|subset product]] of $H$ and $K$.
Thus:
: $(2): \quad \size {H \vee K... | Order of Subgroup Product/Corollary | https://proofwiki.org/wiki/Order_of_Subgroup_Product/Corollary | https://proofwiki.org/wiki/Order_of_Subgroup_Product/Corollary | [
"Order of Subgroup Product"
] | [
"Definition:Join of Subgroups",
"Definition:Order of Structure"
] | [
"Order of Subgroup Product",
"Subset Product is Subset of Generator",
"Definition:Subset Product"
] |
proofwiki-6134 | Abelian Group of Order Twice Odd has Exactly One Order 2 Element | Let $G$ be an abelian group whose identity element is $e$.
Let the order of $G$ be $2 n$ such that $n$ is odd.
Then there exists exactly one $g \in G$ with $g \ne e$ such that $g = g^{-1}$. | By Even Order Group has Order 2 Element, $G$ has an element $x$ of order $2$.
{{AimForCont}} $y$ is another element of order $2$.
Then $x y = y x$ is another element of order $2$.
The subset $H = \set {g \in G: g^2 = e} = \set {e, x, y, x y}$ of $G$ forms a subgroup of $G$.
Thus $\order H = 4$.
But as $n$ is odd, it fo... | Let $G$ be an [[Definition:Abelian Group|abelian group]] whose [[Definition:Identity Element|identity element]] is $e$.
Let the [[Definition:Order of Structure|order]] of $G$ be $2 n$ such that $n$ is [[Definition:Odd Integer|odd]].
Then there exists [[Definition:Unique|exactly one]] $g \in G$ with $g \ne e$ such th... | By [[Even Order Group has Order 2 Element]], $G$ has an [[Definition:Element|element]] $x$ of [[Definition:Order of Group Element|order]] $2$.
{{AimForCont}} $y$ is another [[Definition:Element|element]] of [[Definition:Order of Group Element|order]] $2$.
Then $x y = y x$ is another [[Definition:Element|element]] of ... | Abelian Group of Order Twice Odd has Exactly One Order 2 Element/Proof 2 | https://proofwiki.org/wiki/Abelian_Group_of_Order_Twice_Odd_has_Exactly_One_Order_2_Element | https://proofwiki.org/wiki/Abelian_Group_of_Order_Twice_Odd_has_Exactly_One_Order_2_Element/Proof_2 | [
"Abelian Groups",
"Abelian Group of Order Twice Odd has Exactly One Order 2 Element"
] | [
"Definition:Abelian Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Order of Structure",
"Definition:Odd Integer",
"Definition:Unique"
] | [
"Even Order Group has Order 2 Element",
"Definition:Element",
"Definition:Order of Group Element",
"Definition:Element",
"Definition:Order of Group Element",
"Definition:Element",
"Definition:Order of Group Element",
"Definition:Subset",
"Definition:Subgroup",
"Definition:Odd Integer",
"Definiti... |
proofwiki-6135 | Homomorphic Image of Group Element is Coset | Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a group homomorphism.
Let $\map \ker \phi$ be the kernel of $\phi$.
Let $h \in H$.
Then $\Preimg h$ is either the empty set or a coset of $\map \ker \phi$. | There are two possibilities for any $h \in H$.
:$(1): \quad \Preimg h = \O$
:$(2): \quad \Preimg h \ne \O$
If $(1)$, then the first disjunct of the result is satisfied.
Now suppose $(2)$ holds.
Let $e_G$ and $e_H$ be the identity elements of $G$ and $H$ respectively.
Let $K = \map \ker \phi$.
Let $x, y \in G$ such that... | Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a [[Definition:Group Homomorphism|group homomorphism]].
Let $\map \ker \phi$ be the [[Definition:Kernel of Group Homomorphism|kernel]] of $\phi$.
Let $h \in H$.
Then $\Preimg h$ is either the [[Definition:Empty Set|empty set]] or a [[Definition:Coset|coset]] of $... | There are two possibilities for any $h \in H$.
:$(1): \quad \Preimg h = \O$
:$(2): \quad \Preimg h \ne \O$
If $(1)$, then the first [[Definition:Disjunction|disjunct]] of the result is satisfied.
Now suppose $(2)$ holds.
Let $e_G$ and $e_H$ be the [[Definition:Identity Element|identity elements]] of $G$ and $H$ res... | Homomorphic Image of Group Element is Coset | https://proofwiki.org/wiki/Homomorphic_Image_of_Group_Element_is_Coset | https://proofwiki.org/wiki/Homomorphic_Image_of_Group_Element_is_Coset | [
"Group Homomorphisms",
"Cosets"
] | [
"Definition:Group Homomorphism",
"Definition:Kernel of Group Homomorphism",
"Definition:Empty Set",
"Definition:Coset"
] | [
"Definition:Disjunction",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Homomorphism to Group Preserves Inverses",
"Definition:Group Homomorphism",
"Element in Left Coset iff Product with Inverse in Subgroup",
"Definition:Subset",
"Kernel is Normal Subgroup of Domain",
"Definition:Grou... |
proofwiki-6136 | Characterization of Metacategory via Equations | Let $\mathbf C_0$ and $\mathbf C_1$ be collections of objects.
Let $\operatorname{Cdm}$ and $\operatorname{Dom}$ assign to every element of $\mathbf C_1$ an element of $\mathbf C_0$.
Let $\operatorname{id}$ assign to every element of $\mathbf C_0$ an element of $\mathbf C_1$.
Denote with $\mathbf C_2$ the collection of... | Let $\mathbf C_0$ and $\mathbf C_1$ be collections of objects.
Let $A$ in $C_O$ be arbitrary.
Let $f, g, h$ in $C_1$ be arbitrary.
By definition of identity morphism:
:$\Dom{\operatorname{id}_A} = \Cdm{\operatorname{id}_A} = A$
:$f \circ \operatorname{id}_A = f$
:$\operatorname{id}_A \circ g = g$
whenever $A$ is the do... | Let $\mathbf C_0$ and $\mathbf C_1$ be collections of [[Definition:Object|objects]].
Let $\operatorname{Cdm}$ and $\operatorname{Dom}$ assign to every element of $\mathbf C_1$ an element of $\mathbf C_0$.
Let $\operatorname{id}$ assign to every element of $\mathbf C_0$ an element of $\mathbf C_1$.
Denote with $\mat... | Let $\mathbf C_0$ and $\mathbf C_1$ be collections of [[Definition:Object|objects]].
Let $A$ in $C_O$ be arbitrary.
Let $f, g, h$ in $C_1$ be arbitrary.
By definition of [[Definition:Identity Morphism|identity morphism]]:
:$\Dom{\operatorname{id}_A} = \Cdm{\operatorname{id}_A} = A$
:$f \circ \operatorname{id}_A = f$... | Characterization of Metacategory via Equations | https://proofwiki.org/wiki/Characterization_of_Metacategory_via_Equations | https://proofwiki.org/wiki/Characterization_of_Metacategory_via_Equations | [
"Category Theory"
] | [
"Definition:Object",
"Definition:Metacategory",
"Definition:Object (Category Theory)",
"Definition:Morphism",
"Definition:Domain (Category Theory)",
"Definition:Codomain (Category Theory)",
"Definition:Morphism",
"Definition:Identity Morphism",
"Definition:Composable Morphisms",
"Definition:Compos... | [
"Definition:Object",
"Definition:Identity Morphism",
"Definition:Domain (Category Theory)",
"Definition:Codomain (Category Theory)",
"Definition:Identity Morphism",
"Definition:Identity Morphism",
"Definition:Identity Morphism",
"Definition:Identity Morphism"
] |
proofwiki-6137 | Kernel of Group Homomorphism Corresponds with Normal Subgroup of Domain | Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups.
Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a group homomorphism.
Let $\map \ker \phi$ be the kernel of $\phi$.
Then there exists $N \lhd G$, a normal subgroup of $G$ such that:
:$N = \map \ker \phi$
Conversely, let $N \lhd G$ be normal subgroup of $G$.
... | The first statement is Kernel is Normal Subgroup of Domain:
:The kernel of $\phi$ is a normal subgroup of the domain of $\phi$:
::$\map \ker \phi \lhd \Dom \phi$
The second statement is Quotient Group Epimorphism is Epimorphism:
:The mapping $\phi: G \to G / N$, defined as:
::$\phi: G \to G / N: \map \phi x = x N$
:is ... | Let $\struct {G, \circ}$ and $\struct {H, *}$ be [[Definition:Group|groups]].
Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a [[Definition:Group Homomorphism|group homomorphism]].
Let $\map \ker \phi$ be the [[Definition:Kernel of Group Homomorphism|kernel]] of $\phi$.
Then there exists $N \lhd G$, a [[Defin... | The first statement is [[Kernel is Normal Subgroup of Domain]]:
:The [[Definition:Kernel of Group Homomorphism|kernel]] of $\phi$ is a [[Definition:Normal Subgroup|normal subgroup]] of the [[Definition:Domain of Mapping|domain]] of $\phi$:
::$\map \ker \phi \lhd \Dom \phi$
The second statement is [[Quotient Group Ep... | Kernel of Group Homomorphism Corresponds with Normal Subgroup of Domain | https://proofwiki.org/wiki/Kernel_of_Group_Homomorphism_Corresponds_with_Normal_Subgroup_of_Domain | https://proofwiki.org/wiki/Kernel_of_Group_Homomorphism_Corresponds_with_Normal_Subgroup_of_Domain | [
"Kernels of Group Homomorphisms",
"Normal Subgroups",
"Group Homomorphisms"
] | [
"Definition:Group",
"Definition:Group Homomorphism",
"Definition:Kernel of Group Homomorphism",
"Definition:Normal Subgroup",
"Definition:Normal Subgroup",
"Definition:Group Homomorphism",
"Definition:Kernel of Group Homomorphism"
] | [
"Kernel is Normal Subgroup of Domain",
"Definition:Kernel of Group Homomorphism",
"Definition:Normal Subgroup",
"Definition:Domain (Set Theory)/Mapping",
"Quotient Epimorphism is Epimorphism/Group",
"Definition:Group Epimorphism",
"Definition:Kernel of Group Homomorphism"
] |
proofwiki-6138 | Inner Automorphism Maps Subgroup to Itself iff Normal | Let $G$ be a group.
For $x \in G$, let $\kappa_x$ denote the inner automorphism of $x$ in $G$.
Let $H$ be a subgroup of $G$.
Then:
:$\forall x \in G: \kappa_x \sqbrk H = H$
{{iff}}:
:$H$ is a normal subgroup of $G$. | === Sufficient Condition ===
{{:Inner Automorphism Maps Subgroup to Itself iff Normal/Sufficient Condition}}{{qed|lemma}} | Let $G$ be a [[Definition:Group|group]].
For $x \in G$, let $\kappa_x$ denote the [[Definition:Inner Automorphism|inner automorphism]] of $x$ in $G$.
Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
Then:
:$\forall x \in G: \kappa_x \sqbrk H = H$
{{iff}}:
:$H$ is a [[Definition:Normal Subgroup|normal subgroup]... | === [[Inner Automorphism Maps Subgroup to Itself iff Normal/Sufficient Condition|Sufficient Condition]] ===
{{:Inner Automorphism Maps Subgroup to Itself iff Normal/Sufficient Condition}}{{qed|lemma}} | Inner Automorphism Maps Subgroup to Itself iff Normal | https://proofwiki.org/wiki/Inner_Automorphism_Maps_Subgroup_to_Itself_iff_Normal | https://proofwiki.org/wiki/Inner_Automorphism_Maps_Subgroup_to_Itself_iff_Normal | [
"Normal Subgroups",
"Inner Automorphisms",
"Inner Automorphism Maps Subgroup to Itself iff Normal"
] | [
"Definition:Group",
"Definition:Inner Automorphism",
"Definition:Subgroup",
"Definition:Normal Subgroup"
] | [
"Inner Automorphism Maps Subgroup to Itself iff Normal/Sufficient Condition"
] |
proofwiki-6139 | Group Epimorphism Preserves Normal Subgroups | Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups.
Let $\phi: G \to H$ be a group epimorphism.
Let $N \lhd G$, where $\lhd$ denotes that $N$ is a normal subgroup of $G$.
Then $\phi \sqbrk N \lhd H$.
That is, the image under $\phi$ of a normal subgroup is itself normal. | Let $N' := \phi \sqbrk N$.
From Group Homomorphism Preserves Subgroups, $N'$ is a subgroup of $H$.
It remains to show that $N'$ is normal in $H$.
Let $h \in H$ be arbitrary.
Let $n' \in N'$ be arbitrary.
Because $\phi$ is an epimorphism, it is by definition surjective.
Therefore:
:$\exists n \in N: \map \phi n = n'$
:$... | Let $\struct {G, \circ}$ and $\struct {H, *}$ be [[Definition:Group|groups]].
Let $\phi: G \to H$ be a [[Definition:Group Epimorphism|group epimorphism]].
Let $N \lhd G$, where $\lhd$ denotes that $N$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
Then $\phi \sqbrk N \lhd H$.
That is, the [[Definitio... | Let $N' := \phi \sqbrk N$.
From [[Group Homomorphism Preserves Subgroups]], $N'$ is a [[Definition:Subgroup|subgroup]] of $H$.
It remains to show that $N'$ is [[Definition:Normal Subgroup|normal]] in $H$.
Let $h \in H$ be arbitrary.
Let $n' \in N'$ be arbitrary.
Because $\phi$ is an [[Definition:Group Epimorphis... | Group Epimorphism Preserves Normal Subgroups | https://proofwiki.org/wiki/Group_Epimorphism_Preserves_Normal_Subgroups | https://proofwiki.org/wiki/Group_Epimorphism_Preserves_Normal_Subgroups | [
"Group Epimorphisms",
"Normal Subgroups"
] | [
"Definition:Group",
"Definition:Group Epimorphism",
"Definition:Normal Subgroup",
"Definition:Image (Set Theory)/Mapping/Subset",
"Definition:Normal Subgroup",
"Definition:Normal Subgroup"
] | [
"Group Homomorphism Preserves Subgroups",
"Definition:Subgroup",
"Definition:Normal Subgroup",
"Definition:Group Epimorphism",
"Definition:Surjection",
"Definition:Image (Set Theory)/Mapping/Subset",
"Definition:Morphism Property",
"Group Homomorphism Preserves Inverses",
"Definition:Normal Subgroup... |
proofwiki-6140 | Group Epimorphism Induces Bijection between Subgroups | Let $G_1$ and $G_2$ be groups whose identities are $e_{G_1}$ and $e_{G_2}$ respectively.
Let $\phi: G_1 \to G_2$ be a group epimorphism.
Let $K := \map \ker \phi$ be the kernel of $\phi$.
Let $\mathbb H_1 = \set {H \subseteq G_1: H \le G_1, K \subseteq H}$ be the set of subgroups of $G_1$ which contain $K$.
Let $\mathb... | Let $Q$ be the mapping defined as:
:$\forall H \le \mathbb H_1: \map Q H = \set {\map \phi h: h \in H}$
Let $H$ be a subgroup of $G_1$ such that $K \subseteq H$.
From Group Homomorphism Preserves Subgroups, $\phi \sqbrk H$ is a subgroup of $G_2$.
This establishes that $Q$ is actually a mapping.
Let $N \lhd G_1$.
From G... | Let $G_1$ and $G_2$ be [[Definition:Group|groups]] whose [[Definition:Identity Element|identities]] are $e_{G_1}$ and $e_{G_2}$ respectively.
Let $\phi: G_1 \to G_2$ be a [[Definition:Group Epimorphism|group epimorphism]].
Let $K := \map \ker \phi$ be the [[Definition:Kernel of Group Homomorphism|kernel]] of $\phi$.
... | Let $Q$ be the [[Definition:Mapping|mapping]] defined as:
:$\forall H \le \mathbb H_1: \map Q H = \set {\map \phi h: h \in H}$
Let $H$ be a [[Definition:Subgroup|subgroup]] of $G_1$ such that $K \subseteq H$.
From [[Group Homomorphism Preserves Subgroups]], $\phi \sqbrk H$ is a [[Definition:Subgroup|subgroup]] of $G_... | Group Epimorphism Induces Bijection between Subgroups | https://proofwiki.org/wiki/Group_Epimorphism_Induces_Bijection_between_Subgroups | https://proofwiki.org/wiki/Group_Epimorphism_Induces_Bijection_between_Subgroups | [
"Group Epimorphisms"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Group Epimorphism",
"Definition:Kernel of Group Homomorphism",
"Definition:Subgroup",
"Definition:Subgroup",
"Definition:Bijection",
"Definition:Normal Subgroup",
"Definition:Normal Subgroup",
"Definition:... | [
"Definition:Mapping",
"Definition:Subgroup",
"Group Homomorphism Preserves Subgroups",
"Definition:Subgroup",
"Definition:Mapping",
"Group Epimorphism Preserves Normal Subgroups",
"Definition:Normal Subgroup",
"Definition:Bijection",
"Definition:Subgroup",
"Definition:Subgroup",
"Definition:Bije... |
proofwiki-6141 | Group Epimorphism Induces Bijection between Subgroups/Corollary | Let $H \le G$ denote that $H$ is a subgroup of $G$.
Then:
:$\forall H \le G, K \subseteq H: \phi \sqbrk H \cong H / K$
where $H / K$ denotes the quotient group of $H$ by $K$. | Let $H$ be a subgroup of $G$ such that $K \subseteq H$.
Consider the restriction of $\phi$ to $H$.
By Group Homomorphism Preserves Subgroups, $\phi_{\restriction_H} \sqbrk H$ is a group.
From Group Epimorphism Induces Bijection between Subgroups it follows that the First Isomorphism Theorem can be applied.
Hence the re... | Let $H \le G$ denote that $H$ is a [[Definition:Subgroup|subgroup]] of $G$.
Then:
:$\forall H \le G, K \subseteq H: \phi \sqbrk H \cong H / K$
where $H / K$ denotes the [[Definition:Quotient Group|quotient group]] of $H$ by $K$. | Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$ such that $K \subseteq H$.
Consider the [[Definition:Restriction of Mapping|restriction]] of $\phi$ to $H$.
By [[Group Homomorphism Preserves Subgroups]], $\phi_{\restriction_H} \sqbrk H$ is a [[Definition:Group|group]].
From [[Group Epimorphism Induces Bijection ... | Group Epimorphism Induces Bijection between Subgroups/Corollary | https://proofwiki.org/wiki/Group_Epimorphism_Induces_Bijection_between_Subgroups/Corollary | https://proofwiki.org/wiki/Group_Epimorphism_Induces_Bijection_between_Subgroups/Corollary | [
"Group Epimorphisms"
] | [
"Definition:Subgroup",
"Definition:Quotient Group"
] | [
"Definition:Subgroup",
"Definition:Restriction/Mapping",
"Group Homomorphism Preserves Subgroups",
"Definition:Group",
"Group Epimorphism Induces Bijection between Subgroups",
"First Isomorphism Theorem"
] |
proofwiki-6142 | Homomorphic Image of Quotient Group under Epimorphism | Let $G_1$ and $G_2$ be groups whose identities are $e_{G_1}$ and $e_{G_2}$ respectively.
Let $\phi: G_1 \to G_2$ be a group epimorphism.
Let $K := \map \ker \phi$ be the kernel of $\phi$.
Let $N$ be a normal subgroup of $G_1$ such that $K \subseteq N$.
Then:
:$\dfrac {G_1} N \cong \dfrac {G_2} {\map \phi N}$
where $\df... | From Group Epimorphism Preserves Normal Subgroups, $\map \phi N$ is normal in $G_2$.
Let $N' := \map \phi N$.
From Quotient Group Epimorphism is Epimorphism, we construct the (group) epimorphism $q: G_2 \to \dfrac {G_2} {N'}$.
Now consider the composite mapping $q \circ \phi: G \to \dfrac {G_2} {N'}$ defined as:
:$\for... | Let $G_1$ and $G_2$ be [[Definition:Group|groups]] whose [[Definition:Identity Element|identities]] are $e_{G_1}$ and $e_{G_2}$ respectively.
Let $\phi: G_1 \to G_2$ be a [[Definition:Group Epimorphism|group epimorphism]].
Let $K := \map \ker \phi$ be the [[Definition:Kernel of Group Homomorphism|kernel]] of $\phi$.
... | From [[Group Epimorphism Preserves Normal Subgroups]], $\map \phi N$ is [[Definition:Normal Subgroup|normal]] in $G_2$.
Let $N' := \map \phi N$.
From [[Quotient Group Epimorphism is Epimorphism]], we construct the [[Definition:Group Epimorphism|(group) epimorphism]] $q: G_2 \to \dfrac {G_2} {N'}$.
Now consider the [... | Homomorphic Image of Quotient Group under Epimorphism | https://proofwiki.org/wiki/Homomorphic_Image_of_Quotient_Group_under_Epimorphism | https://proofwiki.org/wiki/Homomorphic_Image_of_Quotient_Group_under_Epimorphism | [
"Group Epimorphisms",
"Quotient Groups"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Group Epimorphism",
"Definition:Kernel of Group Homomorphism",
"Definition:Normal Subgroup",
"Definition:Quotient Group"
] | [
"Group Epimorphism Preserves Normal Subgroups",
"Definition:Normal Subgroup",
"Quotient Epimorphism is Epimorphism/Group",
"Definition:Group Epimorphism",
"Definition:Composition of Mappings",
"Composite of Group Epimorphisms is Epimorphism",
"Definition:Group Epimorphism",
"Definition:Identity (Abstr... |
proofwiki-6143 | Cauchy's Group Theorem | Let $G$ be a finite group whose identity is $e$.
Let $p$ be a prime number which divides the order of $G$.
Then $G$ has a subgroup of order $p$. | === Lemma ===
Let $G$ be a finite abelian group whose identity is $e$.
Let $p$ be a prime number which divides the order of $G$.
Then $G$ has a subgroup of order $p$.
=== Proof of Lemma ===
Let $\order G$ be a prime number.
Then from Prime Group is Cyclic $G$ has a generator $\gen g$ such that $\order g = p$.
Now suppo... | Let $G$ be a [[Definition:Finite Group|finite group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $p$ be a [[Definition:Prime Number|prime number]] which [[Definition:Divisor of Integer|divides]] the [[Definition:Order of Group|order]] of $G$.
Then $G$ has a [[Definition:Subgroup|subgroup]] of [[Defin... | === Lemma ===
Let $G$ be a [[Definition:Finite Group|finite]] [[Definition:Abelian Group|abelian group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $p$ be a [[Definition:Prime Number|prime number]] which [[Definition:Divisor of Integer|divides]] the [[Definition:Order of Group|order]] of $G$.
Then $G... | Cauchy's Group Theorem/Proof 1 | https://proofwiki.org/wiki/Cauchy's_Group_Theorem | https://proofwiki.org/wiki/Cauchy's_Group_Theorem/Proof_1 | [
"Cauchy's Group Theorem",
"Abelian Groups",
"Subgroups",
"Prime Groups"
] | [
"Definition:Finite Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Prime Number",
"Definition:Divisor (Algebra)/Integer",
"Definition:Order of Structure",
"Definition:Subgroup",
"Definition:Order of Structure"
] | [
"Definition:Finite Group",
"Definition:Abelian Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Prime Number",
"Definition:Divisor (Algebra)/Integer",
"Definition:Order of Structure",
"Definition:Subgroup",
"Definition:Order of Structure",
"Definition:Prime Number",
... |
proofwiki-6144 | Cauchy's Group Theorem | Let $G$ be a finite group whose identity is $e$.
Let $p$ be a prime number which divides the order of $G$.
Then $G$ has a subgroup of order $p$. | This result follows as a special case of Group has Subgroups of All Prime Power Factors.
{{qed}} | Let $G$ be a [[Definition:Finite Group|finite group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $p$ be a [[Definition:Prime Number|prime number]] which [[Definition:Divisor of Integer|divides]] the [[Definition:Order of Group|order]] of $G$.
Then $G$ has a [[Definition:Subgroup|subgroup]] of [[Defin... | This result follows as a special case of [[Group has Subgroups of All Prime Power Factors]].
{{qed}} | Cauchy's Group Theorem/Proof 2 | https://proofwiki.org/wiki/Cauchy's_Group_Theorem | https://proofwiki.org/wiki/Cauchy's_Group_Theorem/Proof_2 | [
"Cauchy's Group Theorem",
"Abelian Groups",
"Subgroups",
"Prime Groups"
] | [
"Definition:Finite Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Prime Number",
"Definition:Divisor (Algebra)/Integer",
"Definition:Order of Structure",
"Definition:Subgroup",
"Definition:Order of Structure"
] | [
"Group has Subgroups of All Prime Power Factors"
] |
proofwiki-6145 | Morphisms-Only Metacategory Induces Metacategory | Let $\mathbf C$ be a morphisms-only metacategory.
Then $\mathbf C$ induces a metacategory $\mathbf C'$, as follows (phrased to fit with Characterization of Metacategory via Equations):
:Define $\mathbf C'_1$ to be the collection $\mathbf C_1$ of morphisms of $\mathbf C$.
:Define $\mathbf C'_0$ to be the image of the op... | It remains to verify that these definitions satisfy the premises for applying Characterization of Metacategory via Equations.
The first of these is the assertion:
:$\operatorname{dom} \operatorname{id}_A = A$
Here, $A \in \mathbf C'_0$, so by above definition, there is some $x \in \mathbf C_1$ with $A = \operatorname{d... | Let $\mathbf C$ be a [[Definition:Morphisms-Only Metacategory|morphisms-only metacategory]].
Then $\mathbf C$ induces a [[Definition:Metacategory|metacategory]] $\mathbf C'$, as follows (phrased to fit with [[Characterization of Metacategory via Equations]]):
:Define $\mathbf C'_1$ to be the collection $\mathbf C_1$... | It remains to verify that these definitions satisfy the premises for applying [[Characterization of Metacategory via Equations]].
The first of these is the assertion:
:$\operatorname{dom} \operatorname{id}_A = A$
Here, $A \in \mathbf C'_0$, so by above definition, there is some $x \in \mathbf C_1$ with $A = \operat... | Morphisms-Only Metacategory Induces Metacategory | https://proofwiki.org/wiki/Morphisms-Only_Metacategory_Induces_Metacategory | https://proofwiki.org/wiki/Morphisms-Only_Metacategory_Induces_Metacategory | [
"Category Theory"
] | [
"Definition:Morphisms-Only Metacategory",
"Definition:Metacategory",
"Characterization of Metacategory via Equations",
"Definition:Morphism",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Identity Mapping"
] | [
"Characterization of Metacategory via Equations",
"Definition:Morphisms-Only Metacategory",
"Definition:Associative Operation",
"Definition:Metacategory"
] |
proofwiki-6146 | Metacategory Induces Morphisms-Only Metacategory | Let $\mathbf C$ be a metacategory.
Then the collection of morphisms $\mathbf C_1$ of $\mathbf C$ is a morphisms-only metacategory. | In order to check that $\mathbf C_1$ is a morphisms-only metacategory, we need to interpret the symbols $\operatorname{dom}$, $\operatorname{cod}$ and $R_\circ$.
Set $\operatorname{dom} x := \operatorname{id}_{\operatorname{dom} x}$, and $\operatorname{cod} x := \operatorname{id}_{\operatorname{cod} x}$ with the symbol... | Let $\mathbf C$ be a [[Definition:Metacategory|metacategory]].
Then the collection of [[Definition:Morphism (Category Theory)|morphisms]] $\mathbf C_1$ of $\mathbf C$ is a [[Definition:Morphisms-Only Metacategory|morphisms-only metacategory]]. | In order to check that $\mathbf C_1$ is a [[Definition:Morphisms-Only Metacategory|morphisms-only metacategory]], we need to interpret the symbols $\operatorname{dom}$, $\operatorname{cod}$ and $R_\circ$.
Set $\operatorname{dom} x := \operatorname{id}_{\operatorname{dom} x}$, and $\operatorname{cod} x := \operatorname... | Metacategory Induces Morphisms-Only Metacategory | https://proofwiki.org/wiki/Metacategory_Induces_Morphisms-Only_Metacategory | https://proofwiki.org/wiki/Metacategory_Induces_Morphisms-Only_Metacategory | [
"Category Theory"
] | [
"Definition:Metacategory",
"Definition:Morphism",
"Definition:Morphisms-Only Metacategory"
] | [
"Definition:Morphisms-Only Metacategory",
"Definition:Morphisms-Only Metacategory",
"Definition:Universal Quantifier",
"Definition:Conditional/Antecedent",
"Definition:Composition of Morphisms",
"Definition:Composable Morphisms",
"Definition:Domain (Category Theory)",
"Definition:Codomain (Category Th... |
proofwiki-6147 | Connected Subspace of Linearly Ordered Space | Let $\struct {S, \preceq, \tau}$ be a linearly ordered space.
Let $Y \subseteq S$.
Then $Y$ is connected in $\struct {S, \tau}$ {{iff}} both of the following hold:
:$(1): \quad Y$ is order-convex in $S$
:$(2): \quad \struct {Y, \preceq \restriction_Y}$ is a linear continuum, where $\restriction$ denotes restriction. | === Necessary Conditions ===
Let $Y$ be connected in $\struct {S, \tau}$.
{{AimForCont}} $Y$ is not order-convex in $S$.
Then there exist $a, b, c \in S$ such that:
:$a \prec b \prec c$
:$a, c \in Y$ but $b \notin Y$
Recall that:
:$b^\prec$ denotes the (strict) lower closure of $b$: $b^\prec = \set {u \in S: u \prec b}... | Let $\struct {S, \preceq, \tau}$ be a [[Definition:Linearly Ordered Space|linearly ordered space]].
Let $Y \subseteq S$.
Then $Y$ is [[Definition:Connected Set (Topology)|connected]] in $\struct {S, \tau}$ {{iff}} both of the following hold:
:$(1): \quad Y$ is [[Definition:Order-Convex Set|order-convex]] in $S$
:$(... | === Necessary Conditions ===
Let $Y$ be [[Definition:Connected Set (Topology)|connected]] in $\struct {S, \tau}$.
{{AimForCont}} $Y$ is not [[Definition:Order-Convex Set|order-convex]] in $S$.
Then there exist $a, b, c \in S$ such that:
:$a \prec b \prec c$
:$a, c \in Y$ but $b \notin Y$
Recall that:
:$b^\prec$ den... | Connected Subspace of Linearly Ordered Space | https://proofwiki.org/wiki/Connected_Subspace_of_Linearly_Ordered_Space | https://proofwiki.org/wiki/Connected_Subspace_of_Linearly_Ordered_Space | [
"Linearly Ordered Spaces",
"Examples of Connected Topological Spaces"
] | [
"Definition:Linearly Ordered Space",
"Definition:Connected Set (Topology)",
"Definition:Order-Convex Set",
"Definition:Linear Continuum",
"Definition:Restriction of Ordering"
] | [
"Definition:Connected Set (Topology)",
"Definition:Order-Convex Set",
"Definition:Strict Lower Closure/Element",
"Definition:Strict Upper Closure/Element",
"Definition:Separated Sets",
"Definition:Disconnected (Topology)/Set",
"Proof by Contradiction",
"Definition:Order-Convex Set",
"Definition:Orde... |
proofwiki-6148 | Abnormal Subgroup is Self-Normalizing Subgroup | Let $G$ be a group.
Let $H$ be an abnormal subgroup of $G$.
Then $H$ is a self-normalizing subgroup of $G$. | We have:
* Abnormal Subgroup is Weakly Abnormal Subgroup
* Weakly Abnormal Subgroup is Self-Normalizing Subgroup
Hence the result.
{{qed}}
Category:Abnormal Subgroups
Category:Self-Normalizing Subgroups
83csah10cquwgpz3rxfviowrnan1c9w | Let $G$ be a [[Definition:Group|group]].
Let $H$ be an [[Definition:Abnormal Subgroup|abnormal subgroup]] of $G$.
Then $H$ is a [[Definition:Self-Normalizing Subgroup|self-normalizing subgroup]] of $G$. | We have:
* [[Abnormal Subgroup is Weakly Abnormal Subgroup]]
* [[Weakly Abnormal Subgroup is Self-Normalizing Subgroup]]
Hence the result.
{{qed}}
[[Category:Abnormal Subgroups]]
[[Category:Self-Normalizing Subgroups]]
83csah10cquwgpz3rxfviowrnan1c9w | Abnormal Subgroup is Self-Normalizing Subgroup | https://proofwiki.org/wiki/Abnormal_Subgroup_is_Self-Normalizing_Subgroup | https://proofwiki.org/wiki/Abnormal_Subgroup_is_Self-Normalizing_Subgroup | [
"Abnormal Subgroups",
"Self-Normalizing Subgroups"
] | [
"Definition:Group",
"Definition:Abnormal Subgroup",
"Definition:Self-Normalizing Subgroup"
] | [
"Abnormal Subgroup is Weakly Abnormal Subgroup",
"Weakly Abnormal Subgroup is Self-Normalizing Subgroup",
"Category:Abnormal Subgroups",
"Category:Self-Normalizing Subgroups"
] |
proofwiki-6149 | Duality Principle (Category Theory)/Conceptual Duality | Let $\Sigma$ be a statement about metacategories, be it in natural language or otherwise.
Suppose that $\Sigma$ holds for all metacategories.
Then so does its dual statement $\Sigma^*$. | From Dual Category of Dual Category, any metacategory $\mathbf C$ may be regarded as a dual category.
Thus it is sufficient to verify $\Sigma^*$ holds in all dual categories.
In any dual category $\mathbf C^{\mathrm {op}}$, we see that:
:$\operatorname{dom} f^{\mathrm {op}} = \paren {\operatorname{cod} f}^{\mathrm {op}... | Let $\Sigma$ be a [[Definition:Statement|statement]] about [[Definition:Metacategory|metacategories]], be it in [[Definition:Natural Language|natural language]] or otherwise.
Suppose that $\Sigma$ holds for all [[Definition:Metacategory|metacategories]].
Then so does its [[Definition:Dual Statement (Category Theory)... | From [[Dual Category of Dual Category]], any [[Definition:Metacategory|metacategory]] $\mathbf C$ may be regarded as a [[Definition:Dual Category|dual category]].
Thus it is sufficient to verify $\Sigma^*$ holds in all [[Definition:Dual Category|dual categories]].
In any [[Definition:Dual Category|dual category]] $\... | Duality Principle (Category Theory)/Conceptual Duality | https://proofwiki.org/wiki/Duality_Principle_(Category_Theory)/Conceptual_Duality | https://proofwiki.org/wiki/Duality_Principle_(Category_Theory)/Conceptual_Duality | [
"Duality Principle (Category Theory)"
] | [
"Definition:Statement",
"Definition:Metacategory",
"Definition:Natural Language",
"Definition:Metacategory",
"Definition:Dual Statement (Category Theory)"
] | [
"Dual Category of Dual Category",
"Definition:Metacategory",
"Definition:Dual Category",
"Definition:Dual Category",
"Definition:Dual Category"
] |
proofwiki-6150 | Product of Complex Number with Conjugate | Let $z = a + i b \in \C$ be a complex number.
Let $\overline z$ denote the complex conjugate of $z$.
Then:
:$z \overline z = a^2 + b^2 = \cmod z^2$
and thus is wholly real. | By the definition of a complex number, let $z = a + i b$ where $a$ and $b$ are real numbers.
Then:
{{begin-eqn}}
{{eqn | l = z \overline z
| r = \paren {a + i b} \paren {a - i b}
| c = {{Defof|Complex Conjugate}}
}}
{{eqn | r = a^2 + a \cdot i b + a \cdot \paren {-i b} + i \cdot \paren {-i} \cdot b^2
... | Let $z = a + i b \in \C$ be a [[Definition:Complex Number|complex number]].
Let $\overline z$ denote the [[Definition:Complex Conjugate|complex conjugate]] of $z$.
Then:
:$z \overline z = a^2 + b^2 = \cmod z^2$
and thus is [[Definition:Wholly Real|wholly real]]. | By the definition of a [[Definition:Complex Number|complex number]], let $z = a + i b$ where $a$ and $b$ are [[Definition:Real Number|real numbers]].
Then:
{{begin-eqn}}
{{eqn | l = z \overline z
| r = \paren {a + i b} \paren {a - i b}
| c = {{Defof|Complex Conjugate}}
}}
{{eqn | r = a^2 + a \cdot i b + a... | Product of Complex Number with Conjugate | https://proofwiki.org/wiki/Product_of_Complex_Number_with_Conjugate | https://proofwiki.org/wiki/Product_of_Complex_Number_with_Conjugate | [
"Complex Conjugates",
"Complex Modulus",
"Complex Multiplication"
] | [
"Definition:Complex Number",
"Definition:Complex Conjugate",
"Definition:Complex Number/Wholly Real"
] | [
"Definition:Complex Number",
"Definition:Real Number",
"Definition:Complex Number/Wholly Real"
] |
proofwiki-6151 | Compact Subspace of Linearly Ordered Space/Lemma 1 | Let $\struct {X, \preceq, \tau}$ be a linearly ordered space.
Let $Y \subseteq X$ be a non-empty subset of $X$.
Let $Y$ be a compact subspace of $\struct {X, \tau}$.
Then $\struct {Y, \preceq \restriction_Y}$ is a complete lattice, where $\restriction$ denotes restriction. | {{AimForCont}} $Y$ is not a complete lattice.
Then there is an $S \subseteq Y$ with no least upper bound in $Y$.
Let $U = \set {b \in Y: b \text { is an upper bound of } S}$.
Note that $U$ may be empty.
Let:
:$\UU = \set {u^\succeq: u \in U}$
:$\LL = \set {s^\preceq: s \in S}$.
where:
:$u^\succeq$ is the upper closure ... | Let $\struct {X, \preceq, \tau}$ be a [[Definition:Linearly Ordered Space|linearly ordered space]].
Let $Y \subseteq X$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Subset|subset]] of $X$.
Let $Y$ be a [[Definition:Compact Topological Subspace|compact subspace]] of $\struct {X, \tau}$.
Then $\struct {Y,... | {{AimForCont}} $Y$ is not a [[Definition:Complete Lattice|complete lattice]].
Then there is an $S \subseteq Y$ with no [[Definition:Supremum of Set|least upper bound]] in $Y$.
Let $U = \set {b \in Y: b \text { is an upper bound of } S}$.
Note that $U$ may be empty.
Let:
:$\UU = \set {u^\succeq: u \in U}$
:$\LL = \s... | Compact Subspace of Linearly Ordered Space/Lemma 1 | https://proofwiki.org/wiki/Compact_Subspace_of_Linearly_Ordered_Space/Lemma_1 | https://proofwiki.org/wiki/Compact_Subspace_of_Linearly_Ordered_Space/Lemma_1 | [
"Compact Subspace of Linearly Ordered Space"
] | [
"Definition:Linearly Ordered Space",
"Definition:Non-Empty Set",
"Definition:Subset",
"Definition:Compact Topological Space/Subspace",
"Definition:Complete Lattice",
"Definition:Restriction of Ordering"
] | [
"Definition:Complete Lattice",
"Definition:Supremum of Set",
"Definition:Upper Closure/Element",
"Definition:Lower Closure/Element",
"Definition:Cover of Set",
"Definition:Upper Bound",
"Definition:Supremum of Set",
"Definition:Upper Bound",
"Definition:Upper Bound",
"Definition:Subcover/Finite",
... |
proofwiki-6152 | Inclusion Mapping on Subgroup is Homomorphism | Let $\struct {G, \circ}$ be a group.
Let $\struct {H, \circ_{\restriction H} }$ be a subgroup of $G$.
Let $i: H \to G$ be the inclusion mapping from $H$ to $G$.
Then $i$ is a group homomorphism. | Let $x, y \in H$.
From {{Group-axiom|0}}, $x \circ_{\restriction H} y \in H$.
Then:
{{begin-eqn}}
{{eqn | l = \map i {x \circ_{\restriction H} y}
| r = x \circ_{\restriction H} y
| c = {{Defof|Inclusion Mapping}}
}}
{{eqn | r = x \circ y
| c = {{Defof|Restriction of Operation}}
}}
{{eqn | r = \map i x... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $\struct {H, \circ_{\restriction H} }$ be a [[Definition:Subgroup|subgroup]] of $G$.
Let $i: H \to G$ be the [[Definition:Inclusion Mapping|inclusion mapping]] from $H$ to $G$.
Then $i$ is a [[Definition:Group Homomorphism|group homomorphism]]. | Let $x, y \in H$.
From {{Group-axiom|0}}, $x \circ_{\restriction H} y \in H$.
Then:
{{begin-eqn}}
{{eqn | l = \map i {x \circ_{\restriction H} y}
| r = x \circ_{\restriction H} y
| c = {{Defof|Inclusion Mapping}}
}}
{{eqn | r = x \circ y
| c = {{Defof|Restriction of Operation}}
}}
{{eqn | r = \map ... | Inclusion Mapping on Subgroup is Homomorphism | https://proofwiki.org/wiki/Inclusion_Mapping_on_Subgroup_is_Homomorphism | https://proofwiki.org/wiki/Inclusion_Mapping_on_Subgroup_is_Homomorphism | [
"Group Homomorphisms",
"Inclusion Mappings"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Inclusion Mapping",
"Definition:Group Homomorphism"
] | [
"Definition:Group Homomorphism"
] |
proofwiki-6153 | Mapping to Identity is Unique Constant Homomorphism | Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups whose identities are $e_G$ and $e_H$ respectively.
Then there exists a unique constant mapping from $G$ to $H$ which is a homomorphism:
:$\phi_{e_H}: G \to H: \forall g \in G: \map {\phi_{e_H} } g = e_H$ | Let $h \in H$ such that $\phi_h: G \to H$ is a (group) homomorphism, where $\phi_h$ is defined as:
:$\forall g \in G: \map {\phi_h} g = h$
Then from Group Homomorphism Preserves Identity:
:$\map {\phi_h} {e_G} = e_H$
and so $h = e_H$.
Hence the result by definition of constant mapping.
It remains to prove that such a c... | Let $\struct {G, \circ}$ and $\struct {H, *}$ be [[Definition:Group|groups]] whose [[Definition:Identity Element|identities]] are $e_G$ and $e_H$ respectively.
Then there exists a [[Definition:Unique|unique]] [[Definition:Constant Mapping|constant mapping]] from $G$ to $H$ which is a [[Definition:Group Homomorphism|h... | Let $h \in H$ such that $\phi_h: G \to H$ is a [[Definition:Group Homomorphism|(group) homomorphism]], where $\phi_h$ is defined as:
:$\forall g \in G: \map {\phi_h} g = h$
Then from [[Group Homomorphism Preserves Identity]]:
:$\map {\phi_h} {e_G} = e_H$
and so $h = e_H$.
Hence the result by definition of [[Definiti... | Mapping to Identity is Unique Constant Homomorphism | https://proofwiki.org/wiki/Mapping_to_Identity_is_Unique_Constant_Homomorphism | https://proofwiki.org/wiki/Mapping_to_Identity_is_Unique_Constant_Homomorphism | [
"Group Homomorphisms",
"Constant Mappings"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Unique",
"Definition:Constant Mapping",
"Definition:Group Homomorphism"
] | [
"Definition:Group Homomorphism",
"Group Homomorphism Preserves Identity",
"Definition:Constant Mapping",
"Definition:Constant Mapping",
"Definition:Group Homomorphism",
"Definition:Morphism Property",
"Definition:Group Homomorphism"
] |
proofwiki-6154 | Subgroup is Superset of Conjugate iff Normal | Let $\struct {G, \circ}$ be a group.
Let $N$ be a subgroup of $G$.
Then $N$ is normal in $G$ (by definition 1) {{iff}}:
:$\forall g \in G: g \circ N \circ g^{-1} \subseteq N$
:$\forall g \in G: g^{-1} \circ N \circ g \subseteq N$ | By definition, a subgroup is normal in $G$ {{iff}}:
:$\forall g \in G: g \circ N = N \circ g$
First note that:
:$(1): \quad \paren {\forall g \in G: g \circ N \circ g^{-1} \subseteq N} \iff \paren {\forall g \in G: g^{-1} \circ N \circ g \subseteq N}$
which is shown by, for example, setting $h := g^{-1}$ and substituti... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $N$ be a [[Definition:Subgroup|subgroup]] of $G$.
Then $N$ is [[Definition:Normal Subgroup/Definition 1|normal in $G$ (by definition 1)]] {{iff}}:
:$\forall g \in G: g \circ N \circ g^{-1} \subseteq N$
:$\forall g \in G: g^{-1} \circ N \circ g \subseteq N... | By definition, a [[Definition:Subgroup|subgroup]] is [[Definition:Normal Subgroup/Definition 1|normal in $G$]] {{iff}}:
:$\forall g \in G: g \circ N = N \circ g$
First note that:
:$(1): \quad \paren {\forall g \in G: g \circ N \circ g^{-1} \subseteq N} \iff \paren {\forall g \in G: g^{-1} \circ N \circ g \subseteq N... | Subgroup is Superset of Conjugate iff Normal | https://proofwiki.org/wiki/Subgroup_is_Superset_of_Conjugate_iff_Normal | https://proofwiki.org/wiki/Subgroup_is_Superset_of_Conjugate_iff_Normal | [
"Conjugacy",
"Normal Subgroups"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Normal Subgroup/Definition 1"
] | [
"Definition:Subgroup",
"Definition:Normal Subgroup/Definition 1",
"Definition:Normal Subgroup",
"Subset Product within Semigroup is Associative/Corollary",
"Coset by Identity",
"Subset Product within Semigroup is Associative/Corollary",
"Coset by Identity",
"Definition:Subgroup",
"Subset Product wit... |
proofwiki-6155 | Set Union is Self-Distributive/General Result | Let $\family {\mathbb S_i}_{i \mathop \in I}$ be an $I$-indexed family of sets of sets.
Then:
:$\ds \bigcup_{i \mathop \in I} \bigcup \mathbb S_i = \bigcup \bigcup_{i \mathop \in I} \mathbb S_i$ | By the definition of set union, we have:
:$\ds \forall i \in I: \forall S \in \mathbb S_i: S \in \bigcup_{j \mathop \in I} \mathbb S_j$
Hence, by Set is Subset of Union: General Result:
:$\ds \forall i \in I: \forall S \in \mathbb S_i: S \subseteq \bigcup \bigcup_{j \mathop \in I} \mathbb S_j$
By Union is Smallest Supe... | Let $\family {\mathbb S_i}_{i \mathop \in I}$ be an [[Definition:Indexed Family of Sets|$I$-indexed family]] of [[Definition:Set of Sets|sets of sets]].
Then:
:$\ds \bigcup_{i \mathop \in I} \bigcup \mathbb S_i = \bigcup \bigcup_{i \mathop \in I} \mathbb S_i$ | By the definition of [[Definition:Union of Family|set union]], we have:
:$\ds \forall i \in I: \forall S \in \mathbb S_i: S \in \bigcup_{j \mathop \in I} \mathbb S_j$
Hence, by [[Set is Subset of Union/General Result|Set is Subset of Union: General Result]]:
:$\ds \forall i \in I: \forall S \in \mathbb S_i: S \subsete... | Set Union is Self-Distributive/General Result | https://proofwiki.org/wiki/Set_Union_is_Self-Distributive/General_Result | https://proofwiki.org/wiki/Set_Union_is_Self-Distributive/General_Result | [
"Set Union is Self-Distributive"
] | [
"Definition:Indexing Set/Family of Sets",
"Definition:Set of Sets"
] | [
"Definition:Set Union/Family of Sets",
"Set is Subset of Union/General Result",
"Union is Smallest Superset/General Result",
"Union is Smallest Superset/Family of Sets",
"Set is Subset of Union/General Result",
"Set is Subset of Union/Family of Sets",
"Subset Relation is Transitive",
"Definition:Set U... |
proofwiki-6156 | Subgroup is Subset of Conjugate iff Normal | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $N$ be a subgroup of $G$.
Then $N$ is normal in $G$ (by definition 1) {{iff}}:
:$\forall g \in G: N \subseteq g \circ N \circ g^{-1}$
:$\forall g \in G: N \subseteq g^{-1} \circ N \circ g$ | By definition, a subgroup is normal in $G$ {{iff}}:
:$\forall g \in G: g \circ N = N \circ g$
First note that:
:$(1): \quad \paren {\forall g \in G: N \subseteq g \circ N \circ g^{-1} } \iff \paren {\forall g \in G: N \subseteq g^{-1} \circ N \circ g}$
which is shown by, for example, setting $h := g^{-1}$ and substitut... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $N$ be a [[Definition:Subgroup|subgroup]] of $G$.
Then $N$ is [[Definition:Normal Subgroup/Definition 1|normal in $G$ (by definition 1)]] {{iff}}:
:$\forall g \in G: N \subseteq g \circ N \circ g^{-1}$... | By definition, a [[Definition:Subgroup|subgroup]] is [[Definition:Normal Subgroup/Definition 1|normal in $G$]] {{iff}}:
:$\forall g \in G: g \circ N = N \circ g$
First note that:
:$(1): \quad \paren {\forall g \in G: N \subseteq g \circ N \circ g^{-1} } \iff \paren {\forall g \in G: N \subseteq g^{-1} \circ N \circ ... | Subgroup is Subset of Conjugate iff Normal | https://proofwiki.org/wiki/Subgroup_is_Subset_of_Conjugate_iff_Normal | https://proofwiki.org/wiki/Subgroup_is_Subset_of_Conjugate_iff_Normal | [
"Conjugacy",
"Normal Subgroups"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Subgroup",
"Definition:Normal Subgroup/Definition 1"
] | [
"Definition:Subgroup",
"Definition:Normal Subgroup/Definition 1",
"Definition:Normal Subgroup",
"Subset Product within Semigroup is Associative/Corollary",
"Coset by Identity",
"Definition:Subgroup",
"Subset Relation is Compatible with Subset Product/Corollary 2",
"Subset Product within Semigroup is A... |
proofwiki-6157 | Existence and Uniqueness of Generated Topology | Let $X$ be a set.
Let $\SS \subseteq \powerset X$ be a subset of the power set of $X$.
Then there exists a unique topology $\map \tau \SS$ on $X$ such that:
:$(1): \quad \SS \subseteq \map \tau \SS$.
:$(2): \quad$ For any topology $\TT$ on $X$, the implication $\SS \subseteq \TT \implies \map \tau \SS \subseteq \TT$ ho... | === Existence ===
Define:
:$\mathbb E = \leftset {\TT \subseteq \powerset X: \SS \subseteq \TT}$ and $\TT$ is a topology on $\rightset X$
Since $\powerset X$ is a topology on $X$, it follows that $\mathbb E$ is non-empty.
Hence, we can define:
:$\ds \map \tau \SS = \bigcap \mathbb E$
It follows that $\map \tau \SS$ is ... | Let $X$ be a [[Definition:Set|set]].
Let $\SS \subseteq \powerset X$ be a [[Definition:Subset|subset]] of the [[Definition:Power Set|power set]] of $X$.
Then [[Definition:Existential Quantifier|there exists]] a [[Definition:Unique|unique]] [[Definition:Topology|topology]] $\map \tau \SS$ on $X$ such that:
:$(1): \qu... | === Existence ===
Define:
:$\mathbb E = \leftset {\TT \subseteq \powerset X: \SS \subseteq \TT}$ [[Definition:Conjunction|and]] $\TT$ is a [[Definition:Topology|topology]] on $\rightset X$
Since [[Discrete Topology is Topology|$\powerset X$ is a topology on $X$]], it follows that $\mathbb E$ is [[Definition:Non-Empt... | Existence and Uniqueness of Generated Topology | https://proofwiki.org/wiki/Existence_and_Uniqueness_of_Generated_Topology | https://proofwiki.org/wiki/Existence_and_Uniqueness_of_Generated_Topology | [
"Topology"
] | [
"Definition:Set",
"Definition:Subset",
"Definition:Power Set",
"Definition:Existential Quantifier",
"Definition:Unique",
"Definition:Topology",
"Definition:Universal Quantifier",
"Definition:Topology",
"Definition:Conditional"
] | [
"Definition:Conjunction",
"Definition:Topology",
"Discrete Topology is Topology",
"Definition:Non-Empty Set",
"Intersection of Topologies is Topology",
"Intersection is Largest Subset/General Result",
"Intersection is Subset/General Result",
"Definition:Topology",
"Definition:Topology"
] |
proofwiki-6158 | Equivalent Conditions for Cover by Collection of Subsets | Let $X$ be a set.
Then the following conditions are equivalent for a subset $\CC \subseteq \powerset X$ of the power set of $X$:
:$(1): \quad \CC$ is a cover for $X$.
:$(2): \quad \ds X = \bigcup \CC$.
:$(3): \quad \ds \exists \SS \subseteq \CC: X = \bigcup \SS$. | === $(1)$ implies $(2)$ ===
By definition, $\CC$ covers $X$ {{iff}} $X \subseteq \ds \bigcup \CC$.
By Union of Subsets is Subset, we have that:
:$\ds \bigcup \CC \subseteq X$
since $\CC \subseteq \powerset X$.
By definition of set equality, it follows that $X = \ds \bigcup \CC$.
{{qed|lemma}} | Let $X$ be a [[Definition:Set|set]].
Then the following conditions are [[Definition:Logical Equivalence|equivalent]] for a [[Definition:Subset|subset]] $\CC \subseteq \powerset X$ of the [[Definition:Power Set|power set]] of $X$:
:$(1): \quad \CC$ is a [[Definition:Cover of Set|cover]] for $X$.
:$(2): \quad \ds X = \... | === $(1)$ implies $(2)$ ===
By definition, $\CC$ [[Definition:Cover of Set|covers]] $X$ {{iff}} $X \subseteq \ds \bigcup \CC$.
By [[Union of Subsets is Subset/Set of Sets|Union of Subsets is Subset]], we have that:
:$\ds \bigcup \CC \subseteq X$
since $\CC \subseteq \powerset X$.
By definition of [[Definition:Set ... | Equivalent Conditions for Cover by Collection of Subsets | https://proofwiki.org/wiki/Equivalent_Conditions_for_Cover_by_Collection_of_Subsets | https://proofwiki.org/wiki/Equivalent_Conditions_for_Cover_by_Collection_of_Subsets | [
"Covers"
] | [
"Definition:Set",
"Definition:Logical Equivalence",
"Definition:Subset",
"Definition:Power Set",
"Definition:Cover of Set"
] | [
"Definition:Cover of Set",
"Union of Subsets is Subset/Set of Sets",
"Definition:Set Equality/Definition 2",
"Definition:Cover of Set"
] |
proofwiki-6159 | Subgroup equals Conjugate iff Normal | :$\forall g \in G: g \circ N \circ g^{-1} = N$
:$\forall g \in G: g^{-1} \circ N \circ g = N$ | By definition, a subgroup is normal in $G$ {{iff}}:
:$\forall g \in G: g \circ N = N \circ g$
First note that:
:$(1): \quad \paren {\forall g \in G: g \circ N \circ g^{-1} = N} \iff \paren {\forall g \in G: g^{-1} \circ N \circ g = N}$
which is shown by, for example, setting $h := g^{-1}$ and substituting.
=== Necessar... | :$\forall g \in G: g \circ N \circ g^{-1} = N$
:$\forall g \in G: g^{-1} \circ N \circ g = N$ | By definition, a [[Definition:Subgroup|subgroup]] is [[Definition:Normal Subgroup|normal in $G$]] {{iff}}:
:$\forall g \in G: g \circ N = N \circ g$
First note that:
:$(1): \quad \paren {\forall g \in G: g \circ N \circ g^{-1} = N} \iff \paren {\forall g \in G: g^{-1} \circ N \circ g = N}$
which is shown by, for exa... | Subgroup equals Conjugate iff Normal/Proof 1 | https://proofwiki.org/wiki/Subgroup_equals_Conjugate_iff_Normal | https://proofwiki.org/wiki/Subgroup_equals_Conjugate_iff_Normal/Proof_1 | [
"Conjugacy",
"Normal Subgroups",
"Subgroup equals Conjugate iff Normal"
] | [] | [
"Definition:Subgroup",
"Definition:Normal Subgroup",
"Definition:Normal Subgroup",
"Subset Product within Semigroup is Associative/Corollary",
"Coset by Identity",
"Subset Product within Semigroup is Associative/Corollary",
"Coset by Identity",
"Definition:Subgroup",
"Subset Product within Semigroup... |
proofwiki-6160 | Subgroup equals Conjugate iff Normal | :$\forall g \in G: g \circ N \circ g^{-1} = N$
:$\forall g \in G: g^{-1} \circ N \circ g = N$ | From Subgroup is Superset of Conjugate iff Normal, $N$ is normal in $G$ {{iff}}:
:$\forall g \in G: N \supseteq g \circ N \circ g^{-1}$
:$\forall g \in G: N \supseteq g^{-1} \circ N \circ g$
From Subgroup is Subset of Conjugate iff Normal, $N$ is normal in $G$ {{iff}}:
:$\forall g \in G: N \subseteq g \circ N \circ g^{... | :$\forall g \in G: g \circ N \circ g^{-1} = N$
:$\forall g \in G: g^{-1} \circ N \circ g = N$ | From [[Subgroup is Superset of Conjugate iff Normal]], $N$ is [[Definition:Normal Subgroup|normal in $G$]] {{iff}}:
:$\forall g \in G: N \supseteq g \circ N \circ g^{-1}$
:$\forall g \in G: N \supseteq g^{-1} \circ N \circ g$
From [[Subgroup is Subset of Conjugate iff Normal]], $N$ is [[Definition:Normal Subgroup|norm... | Subgroup equals Conjugate iff Normal/Proof 2 | https://proofwiki.org/wiki/Subgroup_equals_Conjugate_iff_Normal | https://proofwiki.org/wiki/Subgroup_equals_Conjugate_iff_Normal/Proof_2 | [
"Conjugacy",
"Normal Subgroups",
"Subgroup equals Conjugate iff Normal"
] | [] | [
"Subgroup is Superset of Conjugate iff Normal",
"Definition:Normal Subgroup",
"Subgroup is Subset of Conjugate iff Normal",
"Definition:Normal Subgroup",
"Definition:Set Equality/Definition 2"
] |
proofwiki-6161 | Subgroup is Normal iff Contains Conjugate Elements | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $N$ be a subgroup of $G$.
Then $N$ is normal in $G$ {{iff}}:
{{begin-eqn}}
{{eqn | n = 1
| q = \forall g \in G
| l = n \in N
| o = \iff
| r = g \circ n \circ g^{-1} \in N
}}
{{eqn | n = 2
| q = \forall g \in G
| l = n \in... | By definition, a subgroup is normal in $G$ {{iff}}:
:$\forall g \in G: g \circ N = N \circ g$
=== Necessary Condition ===
Suppose that $g \circ N = N \circ g$, by definition 1 of normality in $G$.
Let $n \in N$.
Then:
{{begin-eqn}}
{{eqn | l = g \circ n
| o = \in
| r = N \circ g
| c = {{Defof|Coset}}
... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $N$ be a [[Definition:Subgroup|subgroup]] of $G$.
Then $N$ is [[Definition:Normal Subgroup/Definition 1|normal in $G$]] {{iff}}:
{{begin-eqn}}
{{eqn | n = 1
| q = \forall g \in G
| l = n \... | By definition, a [[Definition:Subgroup|subgroup]] is [[Definition:Normal Subgroup/Definition 1|normal in $G$]] {{iff}}:
:$\forall g \in G: g \circ N = N \circ g$
=== Necessary Condition ===
Suppose that $g \circ N = N \circ g$, by [[Definition:Normal Subgroup/Definition 1|definition 1 of normality in $G$]].
Let $n... | Subgroup is Normal iff Contains Conjugate Elements | https://proofwiki.org/wiki/Subgroup_is_Normal_iff_Contains_Conjugate_Elements | https://proofwiki.org/wiki/Subgroup_is_Normal_iff_Contains_Conjugate_Elements | [
"Conjugacy",
"Normal Subgroups"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Subgroup",
"Definition:Normal Subgroup/Definition 1"
] | [
"Definition:Subgroup",
"Definition:Normal Subgroup/Definition 1",
"Definition:Normal Subgroup/Definition 1",
"Division Laws for Groups",
"Division Laws for Groups",
"Division Laws for Groups",
"Definition:Set Equality/Definition 2"
] |
proofwiki-6162 | Subgroup is Normal iff Left Cosets are Right Cosets | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $N$ be a subgroup of $G$.
Then $N$ is normal in $G$ (by definition 1) {{iff}}:
:Every right coset of $N$ in $G$ is a left coset
or equivalently:
:The right coset space of $N$ in $G$ equals its left coset space. | === Necessary Condition ===
Let $N$ be a normal subgroup of $G$ by Definition 1.
Then the equality of the coset spaces follows directly from definition of normal subgroup and coset.
{{qed|lemma}}
=== Sufficient Condition ===
Suppose that every right coset of $N$ in $G$ is a left coset of $N$ in $G$.
Let $g \in G$.
Sinc... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $N$ be a [[Definition:Subgroup|subgroup]] of $G$.
Then $N$ is [[Definition:Normal Subgroup/Definition 1|normal in $G$ (by definition 1)]] {{iff}}:
:Every [[Definition:Right Coset|right coset]] of $N$ ... | === Necessary Condition ===
Let $N$ be a normal subgroup of $G$ by [[Definition:Normal Subgroup/Definition 1|Definition 1]].
Then the equality of the coset spaces follows directly from definition of [[Definition:Normal Subgroup|normal subgroup]] and [[Definition:Coset|coset]].
{{qed|lemma}}
=== Sufficient Condition... | Subgroup is Normal iff Left Cosets are Right Cosets | https://proofwiki.org/wiki/Subgroup_is_Normal_iff_Left_Cosets_are_Right_Cosets | https://proofwiki.org/wiki/Subgroup_is_Normal_iff_Left_Cosets_are_Right_Cosets | [
"Conjugacy",
"Normal Subgroups"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Subgroup",
"Definition:Normal Subgroup/Definition 1",
"Definition:Coset/Right Coset",
"Definition:Coset/Left Coset",
"Definition:Coset Space/Right Coset Space",
"Definition:Coset Space/Left Coset Space"
] | [
"Definition:Normal Subgroup/Definition 1",
"Definition:Normal Subgroup",
"Definition:Coset",
"Definition:Coset/Right Coset",
"Definition:Coset/Left Coset",
"Element of Group is in its own Coset",
"Element in Left Coset iff Product with Inverse in Subgroup",
"Coset by Identity",
"Subset Product withi... |
proofwiki-6163 | Symmetric Group has Non-Normal Subgroup | Let $S_n$ be the (full) symmetric group on $n$ elements, where $n \ge 3$.
Then $S_n$ contains at least one subgroup which is not normal. | Let $S_n$ act on the set $S$.
Let $e$ be the identity of $S_n$, by definition the identity mapping $I_S$ on $S$.
As $S$ has at least three elements, three can be arbitrary selected and called $a$, $b$ and $c$.
Let $\rho$ be a transposition of $S_n$, transposing elements $a$ and $b$.
$\rho$ can be described in cycle not... | Let $S_n$ be the [[Definition:Symmetric Group|(full) symmetric group on $n$ elements]], where $n \ge 3$.
Then $S_n$ contains at least one [[Definition:Subgroup|subgroup]] which is not [[Definition:Normal Subgroup|normal]]. | Let $S_n$ [[Definition:Group Action|act on]] the [[Definition:Set|set]] $S$.
Let $e$ be the [[Definition:Identity Element|identity]] of $S_n$, by definition the [[Definition:Identity Mapping|identity mapping]] $I_S$ on $S$.
As $S$ has at least three [[Definition:Element|elements]], three can be arbitrary selected an... | Symmetric Group has Non-Normal Subgroup | https://proofwiki.org/wiki/Symmetric_Group_has_Non-Normal_Subgroup | https://proofwiki.org/wiki/Symmetric_Group_has_Non-Normal_Subgroup | [
"Normal Subgroups",
"Symmetric Groups"
] | [
"Definition:Symmetric Group",
"Definition:Subgroup",
"Definition:Normal Subgroup"
] | [
"Definition:Group Action",
"Definition:Set",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Identity Mapping",
"Definition:Element",
"Definition:Transposition",
"Definition:Permutation on n Letters/Cycle Notation",
"Transposition is Self-Inverse",
"Definition:Subgroup",
"... |
proofwiki-6164 | Disjoint Union is Coproduct in Category of Sets | Let $\mathbf{Set}$ be the category of sets.
Let $S$ and $T$ be sets.
Then their disjoint union $S \sqcup T$ is a coproduct in $\mathbf{Set}$. | We have the implicit mappings $i_1: S \to S \sqcup T$ and $i_2: T \to S \sqcup T$ defined by:
:$i_1 \left({s}\right) = \left({s, 1}\right)$
:$i_2 \left({t}\right) = \left({t, 2}\right)$
Now given a set $V$ and mappings $f: S \to V$ and $g: T \to V$, there is to be a unique $\left[{f, g}\right]: S \sqcup T \to V$ such t... | Let $\mathbf{Set}$ be the [[Definition:Category of Sets|category of sets]].
Let $S$ and $T$ be [[Definition:Set|sets]].
Then their [[Definition:Disjoint Union (Set Theory)|disjoint union]] $S \sqcup T$ is a [[Definition:Coproduct|coproduct]] in $\mathbf{Set}$. | We have the implicit [[Definition:Mapping|mappings]] $i_1: S \to S \sqcup T$ and $i_2: T \to S \sqcup T$ defined by:
:$i_1 \left({s}\right) = \left({s, 1}\right)$
:$i_2 \left({t}\right) = \left({t, 2}\right)$
Now given a set $V$ and [[Definition:Mapping|mappings]] $f: S \to V$ and $g: T \to V$, there is to be a [[Def... | Disjoint Union is Coproduct in Category of Sets | https://proofwiki.org/wiki/Disjoint_Union_is_Coproduct_in_Category_of_Sets | https://proofwiki.org/wiki/Disjoint_Union_is_Coproduct_in_Category_of_Sets | [
"Category of Sets",
"Coproducts"
] | [
"Definition:Category of Sets",
"Definition:Set",
"Definition:Disjoint Union (Set Theory)",
"Definition:Coproduct"
] | [
"Definition:Mapping",
"Definition:Mapping",
"Definition:Unique",
"Definition:Coproduct"
] |
proofwiki-6165 | Inclusion Mapping on Subgroup is Monomorphism | Let $\struct {G, \circ}$ be a group.
Let $\struct {H, \circ {\restriction_H} }$ be a subgroup of $G$.
Let $i: H \to G$ be the inclusion mapping from $H$ to $G$.
Then $i$ is a group monomorphism. | We have:
* Inclusion Mapping on Subgroup is Homomorphism
* Inclusion Mapping is Injection
The result follows by definition of (group) monomorphism.
{{qed}} | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $\struct {H, \circ {\restriction_H} }$ be a [[Definition:Subgroup|subgroup]] of $G$.
Let $i: H \to G$ be the [[Definition:Inclusion Mapping|inclusion mapping]] from $H$ to $G$.
Then $i$ is a [[Definition:Group Monomorphism|group monomorphism]]. | We have:
* [[Inclusion Mapping on Subgroup is Homomorphism]]
* [[Inclusion Mapping is Injection]]
The result follows by definition of [[Definition:Group Monomorphism|(group) monomorphism]].
{{qed}} | Inclusion Mapping on Subgroup is Monomorphism | https://proofwiki.org/wiki/Inclusion_Mapping_on_Subgroup_is_Monomorphism | https://proofwiki.org/wiki/Inclusion_Mapping_on_Subgroup_is_Monomorphism | [
"Group Monomorphisms",
"Inclusion Mappings"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Inclusion Mapping",
"Definition:Group Monomorphism"
] | [
"Inclusion Mapping on Subgroup is Homomorphism",
"Inclusion Mapping is Injection",
"Definition:Group Monomorphism"
] |
proofwiki-6166 | Inverse of Group Isomorphism is Isomorphism | Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups.
Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a mapping.
Then $\phi$ is an isomorphism {{iff}} $\phi^{-1}: \struct {H, *} \to \struct {G, \circ}$ is also an isomorphism. | A specific instance of Inverse of Algebraic Structure Isomorphism is Isomorphism.
{{qed}} | Let $\struct {G, \circ}$ and $\struct {H, *}$ be [[Definition:Group|groups]].
Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a [[Definition:Mapping|mapping]].
Then $\phi$ is an [[Definition:Group Isomorphism|isomorphism]] {{iff}} $\phi^{-1}: \struct {H, *} \to \struct {G, \circ}$ is also an [[Definition:Group ... | A specific instance of [[Inverse of Algebraic Structure Isomorphism is Isomorphism]].
{{qed}} | Inverse of Group Isomorphism is Isomorphism/Proof 1 | https://proofwiki.org/wiki/Inverse_of_Group_Isomorphism_is_Isomorphism | https://proofwiki.org/wiki/Inverse_of_Group_Isomorphism_is_Isomorphism/Proof_1 | [
"Inverse of Group Isomorphism is Isomorphism",
"Group Isomorphisms"
] | [
"Definition:Group",
"Definition:Mapping",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism"
] | [
"Inverse of Algebraic Structure Isomorphism is Isomorphism"
] |
proofwiki-6167 | Inverse of Group Isomorphism is Isomorphism | Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups.
Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a mapping.
Then $\phi$ is an isomorphism {{iff}} $\phi^{-1}: \struct {H, *} \to \struct {G, \circ}$ is also an isomorphism. | === Necessary Condition ===
Let $\phi: G \to H$ be an isomorphism.
Then by definition $\phi$ is a bijection.
From Bijection iff Inverse is Bijection it follows that:
:$\exists \phi^{-1}: \struct {H, *} \to \struct {G, \circ}$
such that $\phi^{-1}$ is also a bijection.
Thus:
{{begin-eqn}}
{{eqn | q = \forall g \in G, h ... | Let $\struct {G, \circ}$ and $\struct {H, *}$ be [[Definition:Group|groups]].
Let $\phi: \struct {G, \circ} \to \struct {H, *}$ be a [[Definition:Mapping|mapping]].
Then $\phi$ is an [[Definition:Group Isomorphism|isomorphism]] {{iff}} $\phi^{-1}: \struct {H, *} \to \struct {G, \circ}$ is also an [[Definition:Group ... | === Necessary Condition ===
Let $\phi: G \to H$ be an [[Definition:Group Isomorphism|isomorphism]].
Then by definition $\phi$ is a [[Definition:Bijection|bijection]].
From [[Bijection iff Inverse is Bijection]] it follows that:
:$\exists \phi^{-1}: \struct {H, *} \to \struct {G, \circ}$
such that $\phi^{-1}$ is also... | Inverse of Group Isomorphism is Isomorphism/Proof 2 | https://proofwiki.org/wiki/Inverse_of_Group_Isomorphism_is_Isomorphism | https://proofwiki.org/wiki/Inverse_of_Group_Isomorphism_is_Isomorphism/Proof_2 | [
"Inverse of Group Isomorphism is Isomorphism",
"Group Isomorphisms"
] | [
"Definition:Group",
"Definition:Mapping",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism"
] | [
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Bijection",
"Inverse of Bijection is Bijection",
"Definition:Bijection",
"Inverse Element of Bijection",
"Inverse Element of Bijection",
"Definition:Group Homomorphism",
"Definition:Bijection",
"Definition:Isomorphism (Abstra... |
proofwiki-6168 | Topology as Magma of Sets | The concept of a topology is an instance of a magma of sets. | It will suffice to define partial mappings such that the axiom for a magma of sets crystallises into the axioms for a topology.
Let $X$ be any set, and let $\powerset X$ be its power set.
Define:
:$\phi_1: \powerset X \to \powerset X: \map {\phi_1} S := X$
:$\phi_2: \powerset X^2 \to \powerset X: \map {\phi_2} {S, T} :... | The concept of a [[Definition:Topology|topology]] is an instance of a [[Definition:Magma of Sets|magma of sets]]. | It will suffice to define [[Definition:Partial Mapping|partial mappings]] such that the axiom for a [[Definition:Magma of Sets|magma of sets]] crystallises into the axioms for a [[Definition:Topology|topology]].
Let $X$ be any [[Definition:Set|set]], and let $\powerset X$ be its [[Definition:Power Set|power set]].
D... | Topology as Magma of Sets | https://proofwiki.org/wiki/Topology_as_Magma_of_Sets | https://proofwiki.org/wiki/Topology_as_Magma_of_Sets | [
"Topology",
"Magmas of Sets"
] | [
"Definition:Topology",
"Definition:Magma of Sets"
] | [
"Definition:Many-to-One Relation",
"Definition:Magma of Sets",
"Definition:Topology",
"Definition:Set",
"Definition:Power Set",
"Definition:Indexing Set",
"Definition:Topology",
"Category:Topology",
"Category:Magmas of Sets"
] |
proofwiki-6169 | Composite of Group Homomorphisms is Homomorphism | Let:
:$\struct {G_1, \circ}$
:$\struct {G_2, *}$
:$\struct {G_3, \oplus}$
be groups.
Let:
:$\phi: \struct {G_1, \circ} \to \struct {G_2, *}$
:$\psi: \struct {G_2, *} \to \struct {G_3, \oplus}$
be homomorphisms.
Then the composite of $\phi$ and $\psi$ is also a homomorphism. | A specific instance of Composite of Homomorphisms on Algebraic Structure is Homomorphism.
{{qed}} | Let:
:$\struct {G_1, \circ}$
:$\struct {G_2, *}$
:$\struct {G_3, \oplus}$
be [[Definition:Group|groups]].
Let:
:$\phi: \struct {G_1, \circ} \to \struct {G_2, *}$
:$\psi: \struct {G_2, *} \to \struct {G_3, \oplus}$
be [[Definition:Group Homomorphism|homomorphisms]].
Then the [[Definition:Composition of Mappings|compo... | A specific instance of [[Composite of Homomorphisms on Algebraic Structure is Homomorphism]].
{{qed}} | Composite of Group Homomorphisms is Homomorphism/Proof 1 | https://proofwiki.org/wiki/Composite_of_Group_Homomorphisms_is_Homomorphism | https://proofwiki.org/wiki/Composite_of_Group_Homomorphisms_is_Homomorphism/Proof_1 | [
"Group Homomorphisms",
"Composite of Group Homomorphisms is Homomorphism"
] | [
"Definition:Group",
"Definition:Group Homomorphism",
"Definition:Composition of Mappings",
"Definition:Group Homomorphism"
] | [
"Composite of Homomorphisms is Homomorphism/Algebraic Structure"
] |
proofwiki-6170 | Composite of Group Homomorphisms is Homomorphism | Let:
:$\struct {G_1, \circ}$
:$\struct {G_2, *}$
:$\struct {G_3, \oplus}$
be groups.
Let:
:$\phi: \struct {G_1, \circ} \to \struct {G_2, *}$
:$\psi: \struct {G_2, *} \to \struct {G_3, \oplus}$
be homomorphisms.
Then the composite of $\phi$ and $\psi$ is also a homomorphism. | So as to alleviate possible confusion over notation, let the composite of $\phi$ and $\psi$ be denoted $\psi \bullet \phi$ instead of the more usual $\psi \circ \phi$.
Then what we are trying to prove is denoted:
:$\paren {\psi \bullet \phi}: \struct {G_1, \circ} \to \struct {G_3, \oplus}$ is a homomorphism.
To prove t... | Let:
:$\struct {G_1, \circ}$
:$\struct {G_2, *}$
:$\struct {G_3, \oplus}$
be [[Definition:Group|groups]].
Let:
:$\phi: \struct {G_1, \circ} \to \struct {G_2, *}$
:$\psi: \struct {G_2, *} \to \struct {G_3, \oplus}$
be [[Definition:Group Homomorphism|homomorphisms]].
Then the [[Definition:Composition of Mappings|compo... | So as to alleviate possible confusion over notation, let the [[Definition:Composition of Mappings|composite]] of $\phi$ and $\psi$ be denoted $\psi \bullet \phi$ instead of the more usual $\psi \circ \phi$.
Then what we are trying to prove is denoted:
:$\paren {\psi \bullet \phi}: \struct {G_1, \circ} \to \struct {G_... | Composite of Group Homomorphisms is Homomorphism/Proof 2 | https://proofwiki.org/wiki/Composite_of_Group_Homomorphisms_is_Homomorphism | https://proofwiki.org/wiki/Composite_of_Group_Homomorphisms_is_Homomorphism/Proof_2 | [
"Group Homomorphisms",
"Composite of Group Homomorphisms is Homomorphism"
] | [
"Definition:Group",
"Definition:Group Homomorphism",
"Definition:Composition of Mappings",
"Definition:Group Homomorphism"
] | [
"Definition:Composition of Mappings",
"Definition:Group Homomorphism",
"Definition:Morphism Property",
"Definition:Morphism Property",
"Definition:Group Homomorphism"
] |
proofwiki-6171 | Composite of Group Monomorphisms is Monomorphism | Let:
:$\struct {G_1, \circ}$
:$\struct {G_2, *}$
:$\struct {G_3, \oplus}$
be groups.
Let:
:$\phi: \struct {G_1, \circ} \to \struct {G_2, *}$
:$\psi: \struct {G_2, *} \to \struct {G_3, \oplus}$
be monomorphisms.
Then the composite of $\phi$ and $\psi$ is also a monomorphism. | A monomorphism is a homomorphism which is also an injection.
From Composite of Group Homomorphisms is Homomorphism, $\psi \circ \phi$ is a homomorphism.
From Composite of Injections is Injection, $\psi \circ \phi$ is an injection.
{{qed}} | Let:
:$\struct {G_1, \circ}$
:$\struct {G_2, *}$
:$\struct {G_3, \oplus}$
be [[Definition:Group|groups]].
Let:
:$\phi: \struct {G_1, \circ} \to \struct {G_2, *}$
:$\psi: \struct {G_2, *} \to \struct {G_3, \oplus}$
be [[Definition:Group Monomorphism|monomorphisms]].
Then the [[Definition:Composition of Mappings|compo... | A [[Definition:Group Monomorphism|monomorphism]] is a [[Definition:Group Homomorphism|homomorphism]] which is also an [[Definition:Injection|injection]].
From [[Composite of Group Homomorphisms is Homomorphism]], $\psi \circ \phi$ is a [[Definition:Group Homomorphism|homomorphism]].
From [[Composite of Injections is... | Composite of Group Monomorphisms is Monomorphism | https://proofwiki.org/wiki/Composite_of_Group_Monomorphisms_is_Monomorphism | https://proofwiki.org/wiki/Composite_of_Group_Monomorphisms_is_Monomorphism | [
"Group Monomorphisms"
] | [
"Definition:Group",
"Definition:Group Monomorphism",
"Definition:Composition of Mappings",
"Definition:Group Monomorphism"
] | [
"Definition:Group Monomorphism",
"Definition:Group Homomorphism",
"Definition:Injection",
"Composite of Group Homomorphisms is Homomorphism",
"Definition:Group Homomorphism",
"Composite of Injections is Injection",
"Definition:Injection"
] |
proofwiki-6172 | Composite of Group Epimorphisms is Epimorphism | Let:
:$\struct {G_1, \odot}$
:$\struct {G_2, *}$
:$\struct {G_3, \oplus}$
be groups.
Let:
:$\phi: \struct {G_1, \odot} \to \struct {G_2, *}$
:$\psi: \struct {G_2, *} \to \struct {G_3, \oplus}$
be (group) epimorphisms.
Then the composite of $\phi$ and $\psi$ is also a (group) epimorphism. | A group epimorphism is a group homomorphism which is also a surection.
From Composite of Group Homomorphisms is Homomorphism, $\psi \circ \phi$ is a group homomorphism.
From Composite of Surjections is Surjection, $\psi \circ \phi$ is a surection.
{{qed}} | Let:
:$\struct {G_1, \odot}$
:$\struct {G_2, *}$
:$\struct {G_3, \oplus}$
be [[Definition:Group|groups]].
Let:
:$\phi: \struct {G_1, \odot} \to \struct {G_2, *}$
:$\psi: \struct {G_2, *} \to \struct {G_3, \oplus}$
be [[Definition:Group Epimorphism|(group) epimorphisms]].
Then the [[Definition:Composition of Mappings... | A [[Definition:Group Epimorphism|group epimorphism]] is a [[Definition:Group Homomorphism|group homomorphism]] which is also a [[Definition:Surjection|surection]].
From [[Composite of Group Homomorphisms is Homomorphism]], $\psi \circ \phi$ is a [[Definition:Group Homomorphism|group homomorphism]].
From [[Composite ... | Composite of Group Epimorphisms is Epimorphism | https://proofwiki.org/wiki/Composite_of_Group_Epimorphisms_is_Epimorphism | https://proofwiki.org/wiki/Composite_of_Group_Epimorphisms_is_Epimorphism | [
"Group Epimorphisms"
] | [
"Definition:Group",
"Definition:Group Epimorphism",
"Definition:Composition of Mappings",
"Definition:Group Epimorphism"
] | [
"Definition:Group Epimorphism",
"Definition:Group Homomorphism",
"Definition:Surjection",
"Composite of Group Homomorphisms is Homomorphism",
"Definition:Group Homomorphism",
"Composite of Surjections is Surjection",
"Definition:Surjection"
] |
proofwiki-6173 | Composite of Group Isomorphisms is Isomorphism | Let $\struct {G_1, \circ}$, $\struct {G_2, *}$ and $\struct {G_3, \oplus}$ be groups.
Let $\phi: \struct {G_1, \circ} \to \struct {G_2, *}$ and $\psi: \struct {G_2, *} \to \struct {G_3, \oplus}$ be group isomorphisms.
Then the composite of $\psi$ with $\phi$ is also a group isomorphism. | A group isomorphism is a group homomorphism which is also a bijection.
From Composite of Group Homomorphisms is Homomorphism, $\psi \circ \phi$ is a group homomorphism.
From Composite of Bijections is Bijection, $\psi \circ \phi$ is a bijection.
{{qed}} | Let $\struct {G_1, \circ}$, $\struct {G_2, *}$ and $\struct {G_3, \oplus}$ be [[Definition:Group|groups]].
Let $\phi: \struct {G_1, \circ} \to \struct {G_2, *}$ and $\psi: \struct {G_2, *} \to \struct {G_3, \oplus}$ be [[Definition:Group Isomorphism|group isomorphisms]].
Then the [[Definition:Composition of Mappings... | A [[Definition:Group Isomorphism|group isomorphism]] is a [[Definition:Group Homomorphism|group homomorphism]] which is also a [[Definition:Bijection|bijection]].
From [[Composite of Group Homomorphisms is Homomorphism]], $\psi \circ \phi$ is a [[Definition:Group Homomorphism|group homomorphism]].
From [[Composite o... | Composite of Group Isomorphisms is Isomorphism | https://proofwiki.org/wiki/Composite_of_Group_Isomorphisms_is_Isomorphism | https://proofwiki.org/wiki/Composite_of_Group_Isomorphisms_is_Isomorphism | [
"Group Isomorphisms"
] | [
"Definition:Group",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Composition of Mappings",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism"
] | [
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Group Homomorphism",
"Definition:Bijection",
"Composite of Group Homomorphisms is Homomorphism",
"Definition:Group Homomorphism",
"Composite of Bijections is Bijection",
"Definition:Bijection"
] |
proofwiki-6174 | Sum of Deviations from Mean | Let $S = \set {x_1, x_2, \ldots, x_n}$ be a set of real numbers.
Let $\overline x$ denote the arithmetic mean of $S$.
Then:
:$\ds \sum_{i \mathop = 1}^n \paren {x_i - \overline x} = 0$ | For brevity, let us write $\ds \sum$ for $\ds \sum_{i \mathop = 1}^n$.
Then:
{{begin-eqn}}
{{eqn | l = \sum \paren {x_i - \overline x}
| r = x_1 - \overline x + x_2 - \overline x + \cdots + x_n - \overline x
| c = {{Defof|Summation}}
}}
{{eqn | r = x_1 - \sum \frac {x_i} n + x_2 - \sum \frac {x_i} n + \cdot... | Let $S = \set {x_1, x_2, \ldots, x_n}$ be a [[Definition:Set|set]] of [[Definition:Real Number|real numbers]].
Let $\overline x$ denote the [[Definition:Arithmetic Mean|arithmetic mean]] of $S$.
Then:
:$\ds \sum_{i \mathop = 1}^n \paren {x_i - \overline x} = 0$ | For brevity, let us write $\ds \sum$ for $\ds \sum_{i \mathop = 1}^n$.
Then:
{{begin-eqn}}
{{eqn | l = \sum \paren {x_i - \overline x}
| r = x_1 - \overline x + x_2 - \overline x + \cdots + x_n - \overline x
| c = {{Defof|Summation}}
}}
{{eqn | r = x_1 - \sum \frac {x_i} n + x_2 - \sum \frac {x_i} n + \cd... | Sum of Deviations from Mean | https://proofwiki.org/wiki/Sum_of_Deviations_from_Mean | https://proofwiki.org/wiki/Sum_of_Deviations_from_Mean | [
"Arithmetic Mean"
] | [
"Definition:Set",
"Definition:Real Number",
"Definition:Arithmetic Mean"
] | [] |
proofwiki-6175 | Identity of Group Direct Product | Let $\struct {G \times H, \circ}$ be the group direct product of the two groups $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$.
Let $e_G$ be the identity for $\struct {G, \circ_1}$.
Let $e_H$ be the identity for $\struct {H, \circ_2}$.
Then $\tuple {e_G, e_H}$ is the identity for $\struct {G \times H, \circ}$. | {{begin-eqn}}
{{eqn | l = \tuple {g, h} \circ \tuple {e_G, e_H}
| r = \tuple {g \circ_1 e_G, h \circ_2 e_H} = \tuple {g, h}
| c =
}}
{{eqn | l = \tuple {e_G, e_H} \circ \tuple {g, h}
| r = \tuple {e_G \circ_1 g, e_H \circ_2 h} = \tuple {g, h}
| c =
}}
{{end-eqn}}
So the identity is $\tuple {e_... | Let $\struct {G \times H, \circ}$ be the [[Definition:Group Direct Product|group direct product]] of the two [[Definition:Group|groups]] $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$.
Let $e_G$ be the [[Definition:Identity Element|identity]] for $\struct {G, \circ_1}$.
Let $e_H$ be the [[Definition:Identity Eleme... | {{begin-eqn}}
{{eqn | l = \tuple {g, h} \circ \tuple {e_G, e_H}
| r = \tuple {g \circ_1 e_G, h \circ_2 e_H} = \tuple {g, h}
| c =
}}
{{eqn | l = \tuple {e_G, e_H} \circ \tuple {g, h}
| r = \tuple {e_G \circ_1 g, e_H \circ_2 h} = \tuple {g, h}
| c =
}}
{{end-eqn}}
So the [[Definition:Identity... | Identity of Group Direct Product/Proof 1 | https://proofwiki.org/wiki/Identity_of_Group_Direct_Product | https://proofwiki.org/wiki/Identity_of_Group_Direct_Product/Proof_1 | [
"Group Direct Products",
"Identity of Group Direct Product"
] | [
"Definition:Group Direct Product",
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] |
proofwiki-6176 | Identity of Group Direct Product | Let $\struct {G \times H, \circ}$ be the group direct product of the two groups $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$.
Let $e_G$ be the identity for $\struct {G, \circ_1}$.
Let $e_H$ be the identity for $\struct {H, \circ_2}$.
Then $\tuple {e_G, e_H}$ is the identity for $\struct {G \times H, \circ}$. | A specific instance of External Direct Product Identity, where the algebraic structures in question are groups.
{{qed}} | Let $\struct {G \times H, \circ}$ be the [[Definition:Group Direct Product|group direct product]] of the two [[Definition:Group|groups]] $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$.
Let $e_G$ be the [[Definition:Identity Element|identity]] for $\struct {G, \circ_1}$.
Let $e_H$ be the [[Definition:Identity Eleme... | A specific instance of [[External Direct Product Identity]], where the [[Definition:Algebraic Structure with One Operation|algebraic structures]] in question are [[Definition:Group|groups]].
{{qed}} | Identity of Group Direct Product/Proof 2 | https://proofwiki.org/wiki/Identity_of_Group_Direct_Product | https://proofwiki.org/wiki/Identity_of_Group_Direct_Product/Proof_2 | [
"Group Direct Products",
"Identity of Group Direct Product"
] | [
"Definition:Group Direct Product",
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"External Direct Product Identity",
"Definition:Algebraic Structure/One Operation",
"Definition:Group"
] |
proofwiki-6177 | Inverses in Group Direct Product | Let $\struct {G \times H, \circ}$ be the group direct product of the two groups $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$.
Let $g^{-1}$ be an inverse of $g \in \struct {G, \circ_1}$.
Let $h^{-1}$ be an inverse of $h \in \struct {H, \circ_2}$.
Then $\tuple {g^{-1}, h^{-1} }$ is the inverse of $\tuple {g, h} \in ... | Let $e_G$ be the identity for $\struct {G, \circ_1}$
Let $e_H$ be the identity for $\struct {H, \circ_2}$.
Then:
{{begin-eqn}}
{{eqn | l = \tuple {g, h} \circ \tuple {g^{-1}, h^{-1} }
| r = \tuple {g \circ_1 g^{-1}, h \circ_2 h^{-1} }
| c =
}}
{{eqn | r = \tuple {e_G, e_H}
| c =
}}
{{eqn | r = \tupl... | Let $\struct {G \times H, \circ}$ be the [[Definition:Group Direct Product|group direct product]] of the two [[Definition:Group|groups]] $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$.
Let $g^{-1}$ be an [[Definition:Inverse Element|inverse]] of $g \in \struct {G, \circ_1}$.
Let $h^{-1}$ be an [[Definition:Inverse... | Let $e_G$ be the [[Definition:Identity Element|identity]] for $\struct {G, \circ_1}$
Let $e_H$ be the [[Definition:Identity Element|identity]] for $\struct {H, \circ_2}$.
Then:
{{begin-eqn}}
{{eqn | l = \tuple {g, h} \circ \tuple {g^{-1}, h^{-1} }
| r = \tuple {g \circ_1 g^{-1}, h \circ_2 h^{-1} }
| c =... | Inverses in Group Direct Product/Proof 1 | https://proofwiki.org/wiki/Inverses_in_Group_Direct_Product | https://proofwiki.org/wiki/Inverses_in_Group_Direct_Product/Proof_1 | [
"Group Direct Products",
"Inverses in Group Direct Product"
] | [
"Definition:Group Direct Product",
"Definition:Group",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Inverse (Abstract Algebra)/Inverse"
] | [
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] |
proofwiki-6178 | Inverses in Group Direct Product | Let $\struct {G \times H, \circ}$ be the group direct product of the two groups $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$.
Let $g^{-1}$ be an inverse of $g \in \struct {G, \circ_1}$.
Let $h^{-1}$ be an inverse of $h \in \struct {H, \circ_2}$.
Then $\tuple {g^{-1}, h^{-1} }$ is the inverse of $\tuple {g, h} \in ... | A specific instance of External Direct Product Inverses, where the algebraic structures in question are groups.
{{qed}} | Let $\struct {G \times H, \circ}$ be the [[Definition:Group Direct Product|group direct product]] of the two [[Definition:Group|groups]] $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$.
Let $g^{-1}$ be an [[Definition:Inverse Element|inverse]] of $g \in \struct {G, \circ_1}$.
Let $h^{-1}$ be an [[Definition:Inverse... | A specific instance of [[External Direct Product Inverses]], where the [[Definition:Algebraic Structure with One Operation|algebraic structures]] in question are [[Definition:Group|groups]].
{{qed}} | Inverses in Group Direct Product/Proof 2 | https://proofwiki.org/wiki/Inverses_in_Group_Direct_Product | https://proofwiki.org/wiki/Inverses_in_Group_Direct_Product/Proof_2 | [
"Group Direct Products",
"Inverses in Group Direct Product"
] | [
"Definition:Group Direct Product",
"Definition:Group",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Inverse (Abstract Algebra)/Inverse"
] | [
"External Direct Product Inverses",
"Definition:Algebraic Structure/One Operation",
"Definition:Group"
] |
proofwiki-6179 | Associativity of Operation in Group Direct Product | Let $\struct {G \times H, \circ}$ be the group direct product of the two groups $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$.
Then the operation $\circ$ in $\struct {G \times T, \circ}$ is associative. | {{begin-eqn}}
{{eqn | l = \paren {\tuple {g_1, h_1} \circ \tuple {g_2, h_2} } \circ \tuple {g_3, h_3}
| r = \tuple {\paren {g_1 \circ_1 g_2} \circ_1 g_3, \paren {h_1 \circ_2 h_2} \circ_2 h_3}
| c =
}}
{{eqn | r = \tuple {g_1 \circ_1 g_2 \circ_1 g_3, h_1 \circ_2 h_2 \circ_2 h_3}
| c =
}}
{{eqn | r = ... | Let $\struct {G \times H, \circ}$ be the [[Definition:Group Direct Product|group direct product]] of the two [[Definition:Group|groups]] $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$.
Then the operation $\circ$ in $\struct {G \times T, \circ}$ is [[Definition:Associative Operation|associative]]. | {{begin-eqn}}
{{eqn | l = \paren {\tuple {g_1, h_1} \circ \tuple {g_2, h_2} } \circ \tuple {g_3, h_3}
| r = \tuple {\paren {g_1 \circ_1 g_2} \circ_1 g_3, \paren {h_1 \circ_2 h_2} \circ_2 h_3}
| c =
}}
{{eqn | r = \tuple {g_1 \circ_1 g_2 \circ_1 g_3, h_1 \circ_2 h_2 \circ_2 h_3}
| c =
}}
{{eqn | r = ... | Associativity of Operation in Group Direct Product/Proof 1 | https://proofwiki.org/wiki/Associativity_of_Operation_in_Group_Direct_Product | https://proofwiki.org/wiki/Associativity_of_Operation_in_Group_Direct_Product/Proof_1 | [
"Group Direct Products",
"Associativity of Operation in Group Direct Product"
] | [
"Definition:Group Direct Product",
"Definition:Group",
"Definition:Associative Operation"
] | [
"Definition:Associative Operation"
] |
proofwiki-6180 | Associativity of Operation in Group Direct Product | Let $\struct {G \times H, \circ}$ be the group direct product of the two groups $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$.
Then the operation $\circ$ in $\struct {G \times T, \circ}$ is associative. | By definition of group, both $\circ_1$ and $\circ_2$ are associative operations.
The result follows from External Direct Product Associativity, where the algebraic structures in question are groups.
{{qed}} | Let $\struct {G \times H, \circ}$ be the [[Definition:Group Direct Product|group direct product]] of the two [[Definition:Group|groups]] $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$.
Then the operation $\circ$ in $\struct {G \times T, \circ}$ is [[Definition:Associative Operation|associative]]. | By definition of [[Definition:Group|group]], both $\circ_1$ and $\circ_2$ are [[Definition:Associative Operation|associative operations]].
The result follows from [[External Direct Product Associativity]], where the [[Definition:Algebraic Structure with One Operation|algebraic structures]] in question are [[Definition... | Associativity of Operation in Group Direct Product/Proof 2 | https://proofwiki.org/wiki/Associativity_of_Operation_in_Group_Direct_Product | https://proofwiki.org/wiki/Associativity_of_Operation_in_Group_Direct_Product/Proof_2 | [
"Group Direct Products",
"Associativity of Operation in Group Direct Product"
] | [
"Definition:Group Direct Product",
"Definition:Group",
"Definition:Associative Operation"
] | [
"Definition:Group",
"Definition:Associative Operation",
"External Direct Product Associativity",
"Definition:Algebraic Structure/One Operation",
"Definition:Group"
] |
proofwiki-6181 | Anticommutativity of External Direct Product | Let $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ be algebraic structures where $S$ and $T$ both have at least two distinct elements.
Let $\struct {S \times T, \circ}$ be their external direct product.
Then $\struct {S \times T, \circ}$ is anticommutative {{iff}} at least one of $\struct {S, \circ_1}$ and $\struct ... | === Sufficient Condition ===
Suppose $\struct {S \times T, \circ}$ is anticommutative and neither $\struct {S, \circ_1}$ nor $\struct {T, \circ_2}$ is.
Then for some distinct $a, b \in S$:
:$a \circ_1 b = b \circ_1 a$
Similarly, for some distinct $c, d \in T$:
:$c \circ_2 d = d \circ_2 c$
Then we compute by definition ... | Let $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ be [[Definition:Algebraic Structure with One Operation|algebraic structures]] where $S$ and $T$ both have at least two [[Definition:Distinct Elements|distinct elements]].
Let $\struct {S \times T, \circ}$ be their [[Definition:External Direct Product|external direc... | === Sufficient Condition ===
Suppose $\struct {S \times T, \circ}$ is [[Definition:Anticommutative Structure with One Operation|anticommutative]] and neither $\struct {S, \circ_1}$ nor $\struct {T, \circ_2}$ is.
Then for some distinct $a, b \in S$:
:$a \circ_1 b = b \circ_1 a$
Similarly, for some distinct $c, d \in... | Anticommutativity of External Direct Product | https://proofwiki.org/wiki/Anticommutativity_of_External_Direct_Product | https://proofwiki.org/wiki/Anticommutativity_of_External_Direct_Product | [
"Anticommutativity",
"External Direct Products"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Distinct/Plural",
"Definition:External Direct Product",
"Definition:Anticommutative/Structure with One Operation"
] | [
"Definition:Anticommutative/Structure with One Operation",
"Definition:External Direct Product",
"Definition:Anticommutative/Structure with One Operation",
"Definition:Anticommutative/Structure with One Operation",
"Definition:Anticommutative/Structure with One Operation",
"Definition:Anticommutative/Stru... |
proofwiki-6182 | Projection on Group Direct Product is Epimorphism | Let $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$ be groups.
Let $\struct {G, \circ}$ be the group direct product of $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$.
Then:
:$\pr_1$ is an epimorphism from $\struct {G, \circ}$ to $\struct {G_1, \circ_1}$
:$\pr_2$ is an epimorphism from $\struct {G, \circ}$ ... | From Projection is Surjection, $\pr_1$ and $\pr_2$ are surjections.
We now need to show they are homomorphisms.
Let $g, h \in \struct {G, \circ}$ where $g = \tuple {g_1, g_2}$ and $h = \tuple {h_1, h_2}$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\pr_1} {g \circ h}
| r = \map {\pr_1} {\tuple {g_1, g_2} \circ \tuple {... | Let $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$ be [[Definition:Group|groups]].
Let $\struct {G, \circ}$ be the [[Definition:Group Direct Product|group direct product]] of $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$.
Then:
:$\pr_1$ is an [[Definition:Group Epimorphism|epimorphism]] from $\struct ... | From [[Projection is Surjection]], $\pr_1$ and $\pr_2$ are [[Definition:Surjection|surjections]].
We now need to show they are [[Definition:Group Homomorphism|homomorphisms]].
Let $g, h \in \struct {G, \circ}$ where $g = \tuple {g_1, g_2}$ and $h = \tuple {h_1, h_2}$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\pr_1} {g... | Projection on Group Direct Product is Epimorphism/Proof 1 | https://proofwiki.org/wiki/Projection_on_Group_Direct_Product_is_Epimorphism | https://proofwiki.org/wiki/Projection_on_Group_Direct_Product_is_Epimorphism/Proof_1 | [
"Group Epimorphisms",
"Group Direct Products",
"Projection on Group Direct Product is Epimorphism"
] | [
"Definition:Group",
"Definition:Group Direct Product",
"Definition:Group Epimorphism",
"Definition:Group Epimorphism",
"Definition:Projection (Mapping Theory)/First Projection",
"Definition:Projection (Mapping Theory)/Second Projection"
] | [
"Projection is Surjection",
"Definition:Surjection",
"Definition:Group Homomorphism",
"Definition:Morphism Property"
] |
proofwiki-6183 | Projection on Group Direct Product is Epimorphism | Let $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$ be groups.
Let $\struct {G, \circ}$ be the group direct product of $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$.
Then:
:$\pr_1$ is an epimorphism from $\struct {G, \circ}$ to $\struct {G_1, \circ_1}$
:$\pr_2$ is an epimorphism from $\struct {G, \circ}$ ... | A specific instance of Projection is Epimorphism, where the algebraic structures in question are groups.
{{qed}} | Let $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$ be [[Definition:Group|groups]].
Let $\struct {G, \circ}$ be the [[Definition:Group Direct Product|group direct product]] of $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$.
Then:
:$\pr_1$ is an [[Definition:Group Epimorphism|epimorphism]] from $\struct ... | A specific instance of [[Projection is Epimorphism]], where the [[Definition:Algebraic Structure with One Operation|algebraic structures]] in question are [[Definition:Group|groups]].
{{qed}} | Projection on Group Direct Product is Epimorphism/Proof 2 | https://proofwiki.org/wiki/Projection_on_Group_Direct_Product_is_Epimorphism | https://proofwiki.org/wiki/Projection_on_Group_Direct_Product_is_Epimorphism/Proof_2 | [
"Group Epimorphisms",
"Group Direct Products",
"Projection on Group Direct Product is Epimorphism"
] | [
"Definition:Group",
"Definition:Group Direct Product",
"Definition:Group Epimorphism",
"Definition:Group Epimorphism",
"Definition:Projection (Mapping Theory)/First Projection",
"Definition:Projection (Mapping Theory)/Second Projection"
] | [
"Projection is Epimorphism",
"Definition:Algebraic Structure/One Operation",
"Definition:Group"
] |
proofwiki-6184 | Canonical Injection on Group Direct Product is Monomorphism | Let $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$ be groups with identities $e_1, e_2$ respectively.
Let $\struct {G_1 \times G_2, \circ}$ be the group direct product of $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$
Then the canonical injections:
:$\inj_1: \struct {G_1, \circ_1} \to \struct {G_1, \circ_... | From Canonical Injection is Injection we have that the canonical injections are in fact injective.
It remains to prove the morphism property.
Let $x, y \in G_1$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\inj_1} {x \circ_1 y}
| r = \tuple {x \circ_1 y, e_2}
| c = Definition of $\inj_1$
}}
{{eqn | r = \tuple {x ... | Let $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$ be [[Definition:Group|groups]] with [[Definition:Identity Element|identities]] $e_1, e_2$ respectively.
Let $\struct {G_1 \times G_2, \circ}$ be the [[Definition:Group Direct Product|group direct product]] of $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}... | From [[Canonical Injection is Injection]] we have that the [[Definition:Canonical Injection (Abstract Algebra)|canonical injections]] are in fact [[Definition:Injection|injective]].
It remains to prove the [[Definition:Morphism Property|morphism property]].
Let $x, y \in G_1$.
Then:
{{begin-eqn}}
{{eqn | l = \map ... | Canonical Injection on Group Direct Product is Monomorphism/Proof 1 | https://proofwiki.org/wiki/Canonical_Injection_on_Group_Direct_Product_is_Monomorphism | https://proofwiki.org/wiki/Canonical_Injection_on_Group_Direct_Product_is_Monomorphism/Proof_1 | [
"Group Monomorphisms",
"Group Direct Products",
"Canonical Injection on Group Direct Product is Monomorphism"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Group Direct Product",
"Definition:Canonical Injection (Abstract Algebra)",
"Definition:Group Monomorphism"
] | [
"Canonical Injection is Injection",
"Definition:Canonical Injection (Abstract Algebra)",
"Definition:Injection",
"Definition:Morphism Property",
"Definition:Morphism Property",
"Definition:Injection",
"Definition:Group Homomorphism",
"Definition:Group Monomorphism"
] |
proofwiki-6185 | Canonical Injection on Group Direct Product is Monomorphism | Let $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$ be groups with identities $e_1, e_2$ respectively.
Let $\struct {G_1 \times G_2, \circ}$ be the group direct product of $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$
Then the canonical injections:
:$\inj_1: \struct {G_1, \circ_1} \to \struct {G_1, \circ_... | A specific instance of Canonical Injection is Monomorphism, where the algebraic structures in question are groups.
{{qed}} | Let $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$ be [[Definition:Group|groups]] with [[Definition:Identity Element|identities]] $e_1, e_2$ respectively.
Let $\struct {G_1 \times G_2, \circ}$ be the [[Definition:Group Direct Product|group direct product]] of $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}... | A specific instance of [[Canonical Injection is Monomorphism]], where the [[Definition:Algebraic Structure with One Operation|algebraic structures]] in question are [[Definition:Group|groups]].
{{qed}} | Canonical Injection on Group Direct Product is Monomorphism/Proof 2 | https://proofwiki.org/wiki/Canonical_Injection_on_Group_Direct_Product_is_Monomorphism | https://proofwiki.org/wiki/Canonical_Injection_on_Group_Direct_Product_is_Monomorphism/Proof_2 | [
"Group Monomorphisms",
"Group Direct Products",
"Canonical Injection on Group Direct Product is Monomorphism"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Group Direct Product",
"Definition:Canonical Injection (Abstract Algebra)",
"Definition:Group Monomorphism"
] | [
"Canonical Injection is Monomorphism",
"Definition:Algebraic Structure/One Operation",
"Definition:Group"
] |
proofwiki-6186 | Canonical Injection is Right Inverse of Projection | Let $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$ be groups with identity elements $e_1$ and $e_2$ respectively.
Let $\struct {G_1 \times G_2, \circ}$ be the group direct product of $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$
Let:
:$\pr_1: \struct {G_1 \times G_2, \circ} \to \struct {G_1, \circ_1}$ be... | This is a specific instance of External Direct Product of Projection with Canonical Injection, where the algebraic structures in question are groups.
{{qed}} | Let $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$ be [[Definition:Group|groups]] with [[Definition:Identity Element|identity elements]] $e_1$ and $e_2$ respectively.
Let $\struct {G_1 \times G_2, \circ}$ be the [[Definition:Group Direct Product|group direct product]] of $\struct {G_1, \circ_1}$ and $\struct {G... | This is a specific instance of [[External Direct Product of Projection with Canonical Injection]], where the [[Definition:Algebraic Structure|algebraic structures]] in question are [[Definition:Group|groups]].
{{qed}} | Canonical Injection is Right Inverse of Projection | https://proofwiki.org/wiki/Canonical_Injection_is_Right_Inverse_of_Projection | https://proofwiki.org/wiki/Canonical_Injection_is_Right_Inverse_of_Projection | [
"Group Homomorphisms",
"Group Direct Products"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Group Direct Product",
"Definition:Projection (Mapping Theory)/First Projection",
"Definition:Projection (Mapping Theory)/Second Projection",
"Definition:Canonical Injection (Abstract Algebra)",
"Definition:Cano... | [
"External Direct Product of Projection with Canonical Injection",
"Definition:Algebraic Structure",
"Definition:Group"
] |
proofwiki-6187 | Image of Canonical Injection is Kernel of Projection | Let $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$ be groups with identity elements $e_1$ and $e_2$ respectively.
Let $\struct {G_1 \times G_2, \circ}$ be the group direct product of $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$
Let:
:$\pr_1: \struct {G_1 \times G_2, \circ} \to \struct {G_1, \circ_1}$ be... | The canonical injection $\inj_1: \struct {G_1, \circ_1} \to \struct {G_1 \times G_2, \circ}$ is defined as:
:$\forall x \in G_1: \map {\inj_1} x = \tuple {x, e_2}$
Thus:
:$\Img {\inj_1} = \set {\tuple {x, e_2}: x \in G_1}$
The second projection $\pr_2: \struct {G_1 \times G_2, \circ} \to \struct {G_2, \circ_2}$ is defi... | Let $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$ be [[Definition:Group|groups]] with [[Definition:Identity Element|identity elements]] $e_1$ and $e_2$ respectively.
Let $\struct {G_1 \times G_2, \circ}$ be the [[Definition:Group Direct Product|group direct product]] of $\struct {G_1, \circ_1}$ and $\struct {G... | The [[Definition:Canonical Injection (Abstract Algebra)|canonical injection]] $\inj_1: \struct {G_1, \circ_1} \to \struct {G_1 \times G_2, \circ}$ is defined as:
:$\forall x \in G_1: \map {\inj_1} x = \tuple {x, e_2}$
Thus:
:$\Img {\inj_1} = \set {\tuple {x, e_2}: x \in G_1}$
The [[Definition:Second Projection|secon... | Image of Canonical Injection is Kernel of Projection | https://proofwiki.org/wiki/Image_of_Canonical_Injection_is_Kernel_of_Projection | https://proofwiki.org/wiki/Image_of_Canonical_Injection_is_Kernel_of_Projection | [
"Group Homomorphisms",
"Group Direct Products"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Group Direct Product",
"Definition:Projection (Mapping Theory)/First Projection",
"Definition:Projection (Mapping Theory)/Second Projection",
"Definition:Canonical Injection (Abstract Algebra)",
"Definition:Cano... | [
"Definition:Canonical Injection (Abstract Algebra)",
"Definition:Projection (Mapping Theory)/Second Projection",
"Definition:Kernel of Group Homomorphism"
] |
proofwiki-6188 | Image of Canonical Injection is Normal Subgroup | Let $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$ be groups with identity elements $e_1$ and $e_2$ respectively.
Let $\struct {G_1 \times G_2, \circ}$ be the group direct product of $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$.
Let:
:$\inj_1: \struct {G_1, \circ_1} \to \struct {G_1 \times G_2, \circ}$ ... | From Image of Canonical Injection is Kernel of Projection:
:$\Img {\inj_1} = \map \ker {\pr_2}$
:$\Img {\inj_2} = \map \ker {\pr_1}$
That is:
:the image of the (first) canonical injection is the kernel of the second projection
:the image of the (second) canonical injection is the kernel of the first projection.
The dom... | Let $\struct {G_1, \circ_1}$ and $\struct {G_2, \circ_2}$ be [[Definition:Group|groups]] with [[Definition:Identity Element|identity elements]] $e_1$ and $e_2$ respectively.
Let $\struct {G_1 \times G_2, \circ}$ be the [[Definition:Group Direct Product|group direct product]] of $\struct {G_1, \circ_1}$ and $\struct {G... | From [[Image of Canonical Injection is Kernel of Projection]]:
:$\Img {\inj_1} = \map \ker {\pr_2}$
:$\Img {\inj_2} = \map \ker {\pr_1}$
That is:
:the [[Definition:Image of Mapping|image]] of the [[Definition:Canonical Injection (Abstract Algebra)|(first) canonical injection]] is the [[Definition:Kernel of Group Homo... | Image of Canonical Injection is Normal Subgroup | https://proofwiki.org/wiki/Image_of_Canonical_Injection_is_Normal_Subgroup | https://proofwiki.org/wiki/Image_of_Canonical_Injection_is_Normal_Subgroup | [
"Normal Subgroups",
"Group Homomorphisms",
"Group Direct Products"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Group Direct Product",
"Definition:Canonical Injection (Abstract Algebra)",
"Definition:Canonical Injection (Abstract Algebra)",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Canonical Injection (... | [
"Image of Canonical Injection is Kernel of Projection",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Canonical Injection (Abstract Algebra)",
"Definition:Kernel of Group Homomorphism",
"Definition:Projection (Mapping Theory)/Second Projection",
"Definition:Image (Set Theory)/Mapping/Mappin... |
proofwiki-6189 | Direct Product of Group Homomorphisms is Homomorphism | Let $\struct {G, \circ}, \struct {H_1, *_1}$ and $\struct {H_2, *_2}$ be groups.
Let $\struct {H_1 \times H_2, *}$ be the group direct product of $H_1$ and $H_2$.
Let $f_1: G \to H_1$ and $f_2: G \to H_2$ be group homomorphisms.
Let $f_1 \times f_2: g \to H_1 \times H_2$ be the direct product of $f_1$ and $f_2$.
Then $... | The direct product of $f_1$ and $f_2$ $f_1 \times f_2: g \to H_1 \times H_2$ is defined as:
:$\forall g \in G: \map {\paren {f_1 \times f_2} } g = \tuple {\map {f_1} g, \map {f_2} g}$
From External Direct Product of Groups is Group, the group direct product $H_1 \times H_2$ is a group.
It remains to be shown that $f_1 ... | Let $\struct {G, \circ}, \struct {H_1, *_1}$ and $\struct {H_2, *_2}$ be [[Definition:Group|groups]].
Let $\struct {H_1 \times H_2, *}$ be the [[Definition:Group Direct Product|group direct product]] of $H_1$ and $H_2$.
Let $f_1: G \to H_1$ and $f_2: G \to H_2$ be [[Definition:Group Homomorphism|group homomorphisms]]... | The [[Definition:Direct Product of Group Homomorphisms|direct product of $f_1$ and $f_2$]] $f_1 \times f_2: g \to H_1 \times H_2$ is defined as:
:$\forall g \in G: \map {\paren {f_1 \times f_2} } g = \tuple {\map {f_1} g, \map {f_2} g}$
From [[External Direct Product of Groups is Group]], the [[Definition:Group Direc... | Direct Product of Group Homomorphisms is Homomorphism | https://proofwiki.org/wiki/Direct_Product_of_Group_Homomorphisms_is_Homomorphism | https://proofwiki.org/wiki/Direct_Product_of_Group_Homomorphisms_is_Homomorphism | [
"Group Homomorphisms",
"Group Direct Products"
] | [
"Definition:Group",
"Definition:Group Direct Product",
"Definition:Group Homomorphism",
"Definition:Direct Product of Group Homomorphisms",
"Definition:Group Homomorphism"
] | [
"Definition:Direct Product of Group Homomorphisms",
"External Direct Product of Groups is Group",
"Definition:Group Direct Product",
"Definition:Group",
"Definition:Morphism Property",
"Definition:Group Homomorphism"
] |
proofwiki-6190 | Projections on Direct Product of Group Homomorphisms | Let $G, H_1$ and $H_2$ be groups.
Let $H_1 \times H_2$ be the group direct product of $H_1$ and $H_2$.
Let $f_1: G \to H_1$ and $f_2: G \to H_2$ be group homomorphisms.
Let $f_1 \times f_2: g \to H_1 \times H_2$ be the direct product of $f_1$ and $f_2$.
Let:
:$\pr_1: H_1 \times H_2 \to H_1$ be the first projection from... | The proof of $(2)$ follows exactly the same lines as the proof of $(1)$, so the proof of $(1)$ only will be shown here. | Let $G, H_1$ and $H_2$ be [[Definition:Group|groups]].
Let $H_1 \times H_2$ be the [[Definition:Group Direct Product|group direct product]] of $H_1$ and $H_2$.
Let $f_1: G \to H_1$ and $f_2: G \to H_2$ be [[Definition:Group Homomorphism|group homomorphisms]].
Let $f_1 \times f_2: g \to H_1 \times H_2$ be the [[Defi... | The proof of $(2)$ follows exactly the same lines as the proof of $(1)$, so the proof of $(1)$ only will be shown here. | Projections on Direct Product of Group Homomorphisms | https://proofwiki.org/wiki/Projections_on_Direct_Product_of_Group_Homomorphisms | https://proofwiki.org/wiki/Projections_on_Direct_Product_of_Group_Homomorphisms | [
"Group Homomorphisms",
"Group Direct Products"
] | [
"Definition:Group",
"Definition:Group Direct Product",
"Definition:Group Homomorphism",
"Definition:Direct Product of Group Homomorphisms",
"Definition:Projection (Mapping Theory)/First Projection",
"Definition:Projection (Mapping Theory)/Second Projection",
"Definition:Composition of Mappings"
] | [] |
proofwiki-6191 | Gaussian Integers form Subgroup of Complex Numbers under Addition | The set of Gaussian integers $\Z \sqbrk i$, under the operation of complex addition, forms a subgroup of the set of additive group of complex numbers $\struct {\C, +}$. | We will use the One-Step Subgroup Test.
This is valid, as the Gaussian integers are a subset of the complex numbers.
We note that $\Z \sqbrk i$ is not empty, as (for example) $0 + 0 i \in \Z \sqbrk i$.
Let $a + b i, c + d i \in \Z \sqbrk i$.
Then we have $-\paren {c + d i} = -c - d i$, and so:
{{begin-eqn}}
{{eqn | l =... | The set of [[Definition:Gaussian Integer|Gaussian integers]] $\Z \sqbrk i$, under the operation of [[Definition:Complex Addition|complex addition]], forms a [[Definition:Subgroup|subgroup]] of the set of [[Definition:Additive Group of Complex Numbers|additive group of complex numbers]] $\struct {\C, +}$. | We will use the [[One-Step Subgroup Test]].
This is valid, as the [[Definition:Gaussian Integer|Gaussian integers]] are a [[Definition:Subset|subset]] of the [[Definition:Complex Number|complex numbers]].
We note that $\Z \sqbrk i$ is not [[Definition:Empty Set|empty]], as (for example) $0 + 0 i \in \Z \sqbrk i$.
... | Gaussian Integers form Subgroup of Complex Numbers under Addition | https://proofwiki.org/wiki/Gaussian_Integers_form_Subgroup_of_Complex_Numbers_under_Addition | https://proofwiki.org/wiki/Gaussian_Integers_form_Subgroup_of_Complex_Numbers_under_Addition | [
"Subgroups",
"Gaussian Integers"
] | [
"Definition:Gaussian Integer",
"Definition:Addition/Complex Numbers",
"Definition:Subgroup",
"Definition:Additive Group of Complex Numbers"
] | [
"One-Step Subgroup Test",
"Definition:Gaussian Integer",
"Definition:Subset",
"Definition:Complex Number",
"Definition:Empty Set",
"Integers form Integral Domain",
"Definition:Integral Domain",
"Definition:Ring (Abstract Algebra)",
"One-Step Subgroup Test",
"Definition:Subgroup"
] |
proofwiki-6192 | Pointwise Addition on Real-Valued Functions is Associative | Let $f, g, h: S \to \R$ be real-valued functions.
Let $f + g: S \to \R$ denote the pointwise sum of $f$ and $g$.
Then:
:$\paren {f + g} + h = f + \paren {g + h}$ | {{begin-eqn}}
{{eqn | q = \forall x \in S
| l = \map {\paren {\paren {f + g} + h} } x
| r = \paren {\map f x + \map g x} + \map h x
| c = {{Defof|Pointwise Addition of Real-Valued Functions}}
}}
{{eqn | r = \map f x + \paren {\map g x + \map h x}
| c = Real Addition is Associative
}}
{{eqn | r =... | Let $f, g, h: S \to \R$ be [[Definition:Real-Valued Function|real-valued functions]].
Let $f + g: S \to \R$ denote the [[Definition:Pointwise Addition of Real-Valued Functions|pointwise sum of $f$ and $g$]].
Then:
:$\paren {f + g} + h = f + \paren {g + h}$ | {{begin-eqn}}
{{eqn | q = \forall x \in S
| l = \map {\paren {\paren {f + g} + h} } x
| r = \paren {\map f x + \map g x} + \map h x
| c = {{Defof|Pointwise Addition of Real-Valued Functions}}
}}
{{eqn | r = \map f x + \paren {\map g x + \map h x}
| c = [[Real Addition is Associative]]
}}
{{eqn |... | Pointwise Addition on Real-Valued Functions is Associative | https://proofwiki.org/wiki/Pointwise_Addition_on_Real-Valued_Functions_is_Associative | https://proofwiki.org/wiki/Pointwise_Addition_on_Real-Valued_Functions_is_Associative | [
"Pointwise Addition is Associative",
"Real Addition"
] | [
"Definition:Real-Valued Function",
"Definition:Pointwise Addition of Real-Valued Functions"
] | [
"Real Addition is Associative",
"Category:Pointwise Addition is Associative",
"Category:Real Addition"
] |
proofwiki-6193 | Coproduct of Free Monoids | Let $\mathbf {Mon}$ be the category of monoids.
Let $\map M A$ and $\map M B$ be free monoids on sets $A$ and $B$, respectively.
Let $A \sqcup B$ be the disjoint union of $A$ and $B$.
Then the free monoid $\map M {A \sqcup B}$ on $A \sqcup B$ is the coproduct of $\map M A$ and $\map M B$ in $\mathbf {Mon}$. | By Coproduct is Unique, it suffices to verify that $\map M {A \sqcup B}$ is a coproduct for $\map M A$ and $\map M B$.
By the universal mapping property of $\map M A$, $\map M B$ and $\map M {A \sqcup B}$, we have the following commutative diagram:
$\quad\quad \begin {xy}
<-5em,0em>*+{\map M A} = "MA",
<0em,0em... | Let $\mathbf {Mon}$ be the [[Definition:Category of Monoids|category of monoids]].
Let $\map M A$ and $\map M B$ be [[Definition:Free Monoid|free monoids]] on [[Definition:Set|sets]] $A$ and $B$, respectively.
Let $A \sqcup B$ be the [[Definition:Disjoint Union (Set Theory)|disjoint union]] of $A$ and $B$.
Then the... | By [[Coproduct is Unique]], it suffices to verify that $\map M {A \sqcup B}$ is a [[Definition:Coproduct|coproduct]] for $\map M A$ and $\map M B$.
By the [[Definition:Coproduct UMP|universal mapping property]] of $\map M A$, $\map M B$ and $\map M {A \sqcup B}$, we have the following [[Definition:Commutative Diagram... | Coproduct of Free Monoids | https://proofwiki.org/wiki/Coproduct_of_Free_Monoids | https://proofwiki.org/wiki/Coproduct_of_Free_Monoids | [
"Category of Monoids",
"Coproducts"
] | [
"Definition:Category of Monoids",
"Definition:Free Monoid",
"Definition:Set",
"Definition:Disjoint Union (Set Theory)",
"Definition:Free Monoid",
"Definition:Coproduct"
] | [
"Coproduct is Unique",
"Definition:Coproduct",
"Definition:Coproduct UMP",
"Definition:Commutative Diagram",
"Definition:Free Monoid",
"Definition:Coproduct",
"Definition:Commutative Diagram",
"Definition:Monoid",
"Definition:Monoid Homomorphism",
"Definition:Monoid Homomorphism",
"Definition:Co... |
proofwiki-6194 | Pointwise Addition is Associative | Let $S$ be a non-empty set.
Let $\mathbb F$ be one of the standard number sets: $\Z, \Q, \R$ or $\C$.
Let $f, g, h: S \to \mathbb F$ be functions.
Let $f + g: S \to \mathbb F$ denote the pointwise sum of $f$ and $g$.
Then:
:$\paren {f + g} + h = f + \paren {g + h}$
That is, pointwise addition is associative. | {{begin-eqn}}
{{eqn | q = \forall x \in S
| l = \map {\paren {\paren {f + g} + h} } x
| r = \paren {\map f x + \map g x} + \map h x
| c = {{Defof|Pointwise Addition}}
}}
{{eqn | r = \map f x + \paren {\map g x + \map h x}
| c = Associative Law of Addition
}}
{{eqn | r = \map {\paren {f + \paren ... | Let $S$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Set|set]].
Let $\mathbb F$ be one of the [[Definition:Number|standard number sets]]: $\Z, \Q, \R$ or $\C$.
Let $f, g, h: S \to \mathbb F$ be [[Definition:Function|functions]].
Let $f + g: S \to \mathbb F$ denote the [[Definition:Pointwise Addition|poin... | {{begin-eqn}}
{{eqn | q = \forall x \in S
| l = \map {\paren {\paren {f + g} + h} } x
| r = \paren {\map f x + \map g x} + \map h x
| c = {{Defof|Pointwise Addition}}
}}
{{eqn | r = \map f x + \paren {\map g x + \map h x}
| c = [[Associative Law of Addition]]
}}
{{eqn | r = \map {\paren {f + \pa... | Pointwise Addition is Associative | https://proofwiki.org/wiki/Pointwise_Addition_is_Associative | https://proofwiki.org/wiki/Pointwise_Addition_is_Associative | [
"Pointwise Addition",
"Examples of Associative Operations",
"Pointwise Addition is Associative"
] | [
"Definition:Non-Empty Set",
"Definition:Set",
"Definition:Number",
"Definition:Function",
"Definition:Pointwise Addition",
"Definition:Pointwise Addition",
"Definition:Associative Operation"
] | [
"Associative Law of Addition"
] |
proofwiki-6195 | Pointwise Addition on Complex-Valued Functions is Associative | Let $f, g, h: S \to \C$ be complex-valued functions.
Let $f + g: S \to \C$ denote the pointwise sum of $f$ and $g$.
Then:
:$\paren {f + g} + h = f + \paren {g + h}$ | {{begin-eqn}}
{{eqn | q = \forall x \in S
| l = \map {\paren {\paren {f + g} + h} } x
| r = \paren {\map f x + \map g x} + \map h x
| c = {{Defof|Pointwise Addition of Complex-Valued Functions}}
}}
{{eqn | r = \map f x + \paren {\map g x + \map h x}
| c = Complex Addition is Associative
}}
{{eqn... | Let $f, g, h: S \to \C$ be [[Definition:Complex-Valued Function|complex-valued functions]].
Let $f + g: S \to \C$ denote the [[Definition:Pointwise Addition of Complex-Valued Functions|pointwise sum of $f$ and $g$]].
Then:
:$\paren {f + g} + h = f + \paren {g + h}$ | {{begin-eqn}}
{{eqn | q = \forall x \in S
| l = \map {\paren {\paren {f + g} + h} } x
| r = \paren {\map f x + \map g x} + \map h x
| c = {{Defof|Pointwise Addition of Complex-Valued Functions}}
}}
{{eqn | r = \map f x + \paren {\map g x + \map h x}
| c = [[Complex Addition is Associative]]
}}
{... | Pointwise Addition on Complex-Valued Functions is Associative | https://proofwiki.org/wiki/Pointwise_Addition_on_Complex-Valued_Functions_is_Associative | https://proofwiki.org/wiki/Pointwise_Addition_on_Complex-Valued_Functions_is_Associative | [
"Pointwise Addition is Associative",
"Complex Addition"
] | [
"Definition:Complex-Valued Function",
"Definition:Pointwise Addition of Complex-Valued Functions"
] | [
"Complex Addition is Associative",
"Category:Pointwise Addition is Associative",
"Category:Complex Addition"
] |
proofwiki-6196 | Pointwise Addition is Commutative | Let $S$ be a non-empty set.
Let $\mathbb F$ be one of the standard number sets: $\Z, \Q, \R$ or $\C$.
Let $f, g: S \to \mathbb F$ be functions.
Let $f + g: S \to \mathbb F$ denote the pointwise sum of $f$ and $g$.
Then:
:$f + g = g + f$
</onlyinclude>
That is, pointwise addition is commutative. | {{begin-eqn}}
{{eqn | q = \forall x \in S
| l = \map {\paren {f + g} } x
| r = \map f x + \map g x
| c = {{Defof|Pointwise Addition}}
}}
{{eqn | r = \map g x + \map f x
| c = Commutative Law of Addition
}}
{{eqn | r = \map {\paren {g + f} } x
| c = {{Defof|Pointwise Addition}}
}}
{{end-eqn... | Let $S$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Set|set]].
Let $\mathbb F$ be one of the [[Definition:Number|standard number sets]]: $\Z, \Q, \R$ or $\C$.
Let $f, g: S \to \mathbb F$ be [[Definition:Function|functions]].
Let $f + g: S \to \mathbb F$ denote the [[Definition:Pointwise Addition|pointwi... | {{begin-eqn}}
{{eqn | q = \forall x \in S
| l = \map {\paren {f + g} } x
| r = \map f x + \map g x
| c = {{Defof|Pointwise Addition}}
}}
{{eqn | r = \map g x + \map f x
| c = [[Commutative Law of Addition]]
}}
{{eqn | r = \map {\paren {g + f} } x
| c = {{Defof|Pointwise Addition}}
}}
{{end... | Pointwise Addition is Commutative | https://proofwiki.org/wiki/Pointwise_Addition_is_Commutative | https://proofwiki.org/wiki/Pointwise_Addition_is_Commutative | [
"Pointwise Addition",
"Examples of Commutative Operations",
"Pointwise Addition is Commutative"
] | [
"Definition:Non-Empty Set",
"Definition:Set",
"Definition:Number",
"Definition:Function",
"Definition:Pointwise Addition",
"Definition:Pointwise Addition",
"Definition:Commutative/Operation"
] | [
"Commutative Law of Addition"
] |
proofwiki-6197 | Pointwise Multiplication is Associative | Let $S$ be a non-empty set.
Let $\mathbb F$ be one of the standard number sets: $\Z, \Q, \R$ or $\C$.
Let $f, g, h: S \to \mathbb F$ be functions.
Let $f \times g: S \to \mathbb F$ denote the pointwise product of $f$ and $g$.
Then:
:$\paren {f \times g} \times h = f \times \paren {g \times h}$
That is, pointwise multip... | {{begin-eqn}}
{{eqn | q = \forall x \in S
| l = \map {\paren {\paren {f \times g} \times h} } x
| r = \paren {\map f x \times \map g x} \times \map h x
| c = {{Defof|Pointwise Multiplication}}
}}
{{eqn | r = \map f x \times \paren {\map g x \times \map h x}
| c = Associative Law of Multiplicatio... | Let $S$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Set|set]].
Let $\mathbb F$ be one of the [[Definition:Number|standard number sets]]: $\Z, \Q, \R$ or $\C$.
Let $f, g, h: S \to \mathbb F$ be [[Definition:Function|functions]].
Let $f \times g: S \to \mathbb F$ denote the [[Definition:Pointwise Multipli... | {{begin-eqn}}
{{eqn | q = \forall x \in S
| l = \map {\paren {\paren {f \times g} \times h} } x
| r = \paren {\map f x \times \map g x} \times \map h x
| c = {{Defof|Pointwise Multiplication}}
}}
{{eqn | r = \map f x \times \paren {\map g x \times \map h x}
| c = [[Associative Law of Multiplicat... | Pointwise Multiplication is Associative | https://proofwiki.org/wiki/Pointwise_Multiplication_is_Associative | https://proofwiki.org/wiki/Pointwise_Multiplication_is_Associative | [
"Pointwise Multiplication",
"Examples of Associative Operations",
"Pointwise Multiplication is Associative"
] | [
"Definition:Non-Empty Set",
"Definition:Set",
"Definition:Number",
"Definition:Function",
"Definition:Pointwise Multiplication",
"Definition:Pointwise Multiplication",
"Definition:Associative Operation"
] | [
"Associative Law of Multiplication"
] |
proofwiki-6198 | Pointwise Multiplication is Commutative | Let $S$ be a non-empty set.
Let $\mathbb F$ be one of the standard number sets: $\Z, \Q, \R$ or $\C$.
Let $f, g, h: S \to \mathbb F$ be functions.
Let $f \times g: S \to \mathbb F$ denote the pointwise product of $f$ and $g$.
Then:
:$f \times g = g \times f$
That is, pointwise multiplication is commutative. | {{begin-eqn}}
{{eqn | q = \forall x \in S
| l = \map {\paren {f \times g} } x
| r = \map f x \times \map g x
| c = {{Defof|Pointwise Multiplication}}
}}
{{eqn | r = \map g x \times \map f x
| c = Commutative Law of Multiplication
}}
{{eqn | r = \map {\paren {g \times f} } x
| c = {{Defof|P... | Let $S$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Set|set]].
Let $\mathbb F$ be one of the [[Definition:Number|standard number sets]]: $\Z, \Q, \R$ or $\C$.
Let $f, g, h: S \to \mathbb F$ be [[Definition:Function|functions]].
Let $f \times g: S \to \mathbb F$ denote the [[Definition:Pointwise Multipli... | {{begin-eqn}}
{{eqn | q = \forall x \in S
| l = \map {\paren {f \times g} } x
| r = \map f x \times \map g x
| c = {{Defof|Pointwise Multiplication}}
}}
{{eqn | r = \map g x \times \map f x
| c = [[Commutative Law of Multiplication]]
}}
{{eqn | r = \map {\paren {g \times f} } x
| c = {{Def... | Pointwise Multiplication is Commutative | https://proofwiki.org/wiki/Pointwise_Multiplication_is_Commutative | https://proofwiki.org/wiki/Pointwise_Multiplication_is_Commutative | [
"Pointwise Multiplication",
"Examples of Commutative Operations",
"Pointwise Multiplication is Commutative"
] | [
"Definition:Non-Empty Set",
"Definition:Set",
"Definition:Number",
"Definition:Function",
"Definition:Pointwise Multiplication",
"Definition:Pointwise Multiplication",
"Definition:Commutative/Operation"
] | [
"Commutative Law of Multiplication"
] |
proofwiki-6199 | Pointwise Addition on Integer-Valued Functions is Associative | Let $f, g, h: S \to \Z$ be integer-valued functions.
Let $f + g: S \to \Z$ denote the pointwise sum of $f$ and $g$.
Then:
:$\paren {f + g} + h = f + \paren {g + h}$ | {{begin-eqn}}
{{eqn | q = \forall x \in S
| l = \map {\paren {\paren {f + g} + h} } x
| r = \paren {\map f x + \map g x} + \map h x
| c = {{Defof|Pointwise Addition of Integer-Valued Functions}}
}}
{{eqn | r = \map f x + \paren {\map g x + \map h x}
| c = Integer Addition is Associative
}}
{{eqn... | Let $f, g, h: S \to \Z$ be [[Definition:Integer-Valued Function|integer-valued functions]].
Let $f + g: S \to \Z$ denote the [[Definition:Pointwise Addition of Integer-Valued Functions|pointwise sum of $f$ and $g$]].
Then:
:$\paren {f + g} + h = f + \paren {g + h}$ | {{begin-eqn}}
{{eqn | q = \forall x \in S
| l = \map {\paren {\paren {f + g} + h} } x
| r = \paren {\map f x + \map g x} + \map h x
| c = {{Defof|Pointwise Addition of Integer-Valued Functions}}
}}
{{eqn | r = \map f x + \paren {\map g x + \map h x}
| c = [[Integer Addition is Associative]]
}}
{... | Pointwise Addition on Integer-Valued Functions is Associative | https://proofwiki.org/wiki/Pointwise_Addition_on_Integer-Valued_Functions_is_Associative | https://proofwiki.org/wiki/Pointwise_Addition_on_Integer-Valued_Functions_is_Associative | [
"Pointwise Addition is Associative",
"Integer Addition"
] | [
"Definition:Integer-Valued Function",
"Definition:Pointwise Addition of Integer-Valued Functions"
] | [
"Integer Addition is Associative",
"Category:Pointwise Addition is Associative",
"Category:Integer Addition"
] |
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