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-500 | Right Operation is Idempotent | The right operation is idempotent:
:$\forall x: x \rightarrow x = x$ | Immediate from the definition of the right operation. | The [[Definition:Right Operation|right operation]] is [[Definition:Idempotent Operation|idempotent]]:
:$\forall x: x \rightarrow x = x$ | Immediate from the definition of the [[Definition:Right Operation|right operation]]. | Right Operation is Idempotent | https://proofwiki.org/wiki/Right_Operation_is_Idempotent | https://proofwiki.org/wiki/Right_Operation_is_Idempotent | [
"Right Operation",
"Examples of Idempotence"
] | [
"Definition:Right Operation",
"Definition:Idempotence/Operation"
] | [
"Definition:Right Operation"
] |
proofwiki-501 | Left Operation is Anticommutative | The left operation is anticommutative:
:$\forall x, y: x \leftarrow y = y \leftarrow x \iff x = y$ | Immediate from the definition of the left operation.
{{qed}} | The [[Definition:Left Operation|left operation]] is [[Definition:Anticommutative|anticommutative]]:
:$\forall x, y: x \leftarrow y = y \leftarrow x \iff x = y$ | Immediate from the definition of the [[Definition:Left Operation|left operation]].
{{qed}} | Left Operation is Anticommutative | https://proofwiki.org/wiki/Left_Operation_is_Anticommutative | https://proofwiki.org/wiki/Left_Operation_is_Anticommutative | [
"Left Operation",
"Examples of Anticommutativity"
] | [
"Definition:Left Operation",
"Definition:Anticommutative"
] | [
"Definition:Left Operation"
] |
proofwiki-502 | Right Operation is Anticommutative | The right operation is anticommutative:
:$\forall x, y: x \rightarrow y = y \rightarrow x \iff x = y$ | Immediate from the definition of the right operation.
{{qed}} | The [[Definition:Right Operation|right operation]] is [[Definition:Anticommutative|anticommutative]]:
:$\forall x, y: x \rightarrow y = y \rightarrow x \iff x = y$ | Immediate from the definition of the [[Definition:Right Operation|right operation]].
{{qed}} | Right Operation is Anticommutative | https://proofwiki.org/wiki/Right_Operation_is_Anticommutative | https://proofwiki.org/wiki/Right_Operation_is_Anticommutative | [
"Right Operation",
"Examples of Anticommutativity"
] | [
"Definition:Right Operation",
"Definition:Anticommutative"
] | [
"Definition:Right Operation"
] |
proofwiki-503 | Left Operation is Associative | The left operation is associative:
:$\forall x, y, z: \paren {x \gets y} \gets z = x \gets \paren {y \gets z}$ | {{begin-eqn}}
{{eqn | l = x \gets \paren {y \gets z}
| r = x \gets y
| c = {{Defof|Left Operation}}
}}
{{eqn | r = x
| c = {{Defof|Left Operation}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \paren {x \gets y} \gets z
| r = x \gets z
| c = {{Defof|Left Operation}}
}}
{{eqn | r = x
| c ... | The [[Definition:Left Operation|left operation]] is [[Definition:Associative Operation|associative]]:
:$\forall x, y, z: \paren {x \gets y} \gets z = x \gets \paren {y \gets z}$ | {{begin-eqn}}
{{eqn | l = x \gets \paren {y \gets z}
| r = x \gets y
| c = {{Defof|Left Operation}}
}}
{{eqn | r = x
| c = {{Defof|Left Operation}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \paren {x \gets y} \gets z
| r = x \gets z
| c = {{Defof|Left Operation}}
}}
{{eqn | r = x
| ... | Left Operation is Associative | https://proofwiki.org/wiki/Left_Operation_is_Associative | https://proofwiki.org/wiki/Left_Operation_is_Associative | [
"Left Operation",
"Examples of Associative Operations"
] | [
"Definition:Left Operation",
"Definition:Associative Operation"
] | [] |
proofwiki-504 | Right Operation is Associative | The right operation is associative:
:$\forall x, y, z: \paren {x \to y} \to z = x \to \paren {y \to z}$ | {{begin-eqn}}
{{eqn | l = \paren {x \to y} \to z
| r = y \to z
| c = {{Defof|Right Operation}}
}}
{{eqn | r = z
| c = {{Defof|Right Operation}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = x \to \paren {y \to z}
| r = x \to z
| c = {{Defof|Right Operation}}
}}
{{eqn | r = z
| c = {{Defof... | The [[Definition:Right Operation|right operation]] is [[Definition:Associative Operation|associative]]:
:$\forall x, y, z: \paren {x \to y} \to z = x \to \paren {y \to z}$ | {{begin-eqn}}
{{eqn | l = \paren {x \to y} \to z
| r = y \to z
| c = {{Defof|Right Operation}}
}}
{{eqn | r = z
| c = {{Defof|Right Operation}}
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = x \to \paren {y \to z}
| r = x \to z
| c = {{Defof|Right Operation}}
}}
{{eqn | r = z
| c = {{Def... | Right Operation is Associative | https://proofwiki.org/wiki/Right_Operation_is_Associative | https://proofwiki.org/wiki/Right_Operation_is_Associative | [
"Right Operation",
"Examples of Associative Operations"
] | [
"Definition:Right Operation",
"Definition:Associative Operation"
] | [] |
proofwiki-505 | Min Operation is Idempotent | The min operation operation is idempotent:
:$\map \min {x, x} = x$ | Follows immediately from the definition of min operation:
:$\map \min {a, b} = \begin {cases} a & : a \le b \\ b & : b \le a \end {cases}$
Setting $x = a = b$ returns the result.
{{Qed}} | The [[Definition:Min Operation|min operation]] operation is [[Definition:Idempotent Operation|idempotent]]:
:$\map \min {x, x} = x$ | Follows immediately from the definition of [[Definition:Min Operation|min operation]]:
:$\map \min {a, b} = \begin {cases} a & : a \le b \\ b & : b \le a \end {cases}$
Setting $x = a = b$ returns the result.
{{Qed}} | Min Operation is Idempotent | https://proofwiki.org/wiki/Min_Operation_is_Idempotent | https://proofwiki.org/wiki/Min_Operation_is_Idempotent | [
"Min Operation",
"Examples of Idempotence"
] | [
"Definition:Min Operation",
"Definition:Idempotence/Operation"
] | [
"Definition:Min Operation"
] |
proofwiki-506 | Max and Min Operations are Distributive over Each Other | The Max and Min operations are distributive over each other:
:$\map \max {x, \map \min {y, z} } = \map \min {\map \max {x, y}, \map \max {x, z} }$
:$\map \max {\map \min {x, y}, z} = \map \min {\map \max {x, z}, \map \max {y, z} }$
:$\map \min {x, \map \max {y, z} } = \map \max {\map \min {x, y}, \map \min {x, z} }$
:$... | To simplify our notation, let $\map \max {x, y}$ be (temporarily) denoted $x \overline \wedge y$, and let $\map \min {x, y}$ be (temporarily) denoted $x \underline \vee y$.
Note that, once we have proved:
:$x \overline \wedge \paren {y \underline \vee z} = \paren {x \overline \wedge y} \underline \vee \paren {x \overli... | The [[Definition:Max Operation|Max]] and [[Definition:Min Operation|Min]] operations are [[Definition:Distributive Operation|distributive]] over each other:
:$\map \max {x, \map \min {y, z} } = \map \min {\map \max {x, y}, \map \max {x, z} }$
:$\map \max {\map \min {x, y}, z} = \map \min {\map \max {x, z}, \map \max ... | To simplify our notation, let $\map \max {x, y}$ be (temporarily) denoted $x \overline \wedge y$, and let $\map \min {x, y}$ be (temporarily) denoted $x \underline \vee y$.
Note that, once we have proved:
:$x \overline \wedge \paren {y \underline \vee z} = \paren {x \overline \wedge y} \underline \vee \paren {x \over... | Max and Min Operations are Distributive over Each Other | https://proofwiki.org/wiki/Max_and_Min_Operations_are_Distributive_over_Each_Other | https://proofwiki.org/wiki/Max_and_Min_Operations_are_Distributive_over_Each_Other | [
"Max and Min Operations",
"Examples of Distributive Operations"
] | [
"Definition:Max Operation",
"Definition:Min Operation",
"Definition:Distributive Operation"
] | [
"Max Operation is Commutative",
"Min Operation is Commutative",
"Category:Max and Min Operations",
"Category:Examples of Distributive Operations"
] |
proofwiki-507 | Equality of Algebraic Structures | Two algebraic structures $\struct {S, \circ}$ and $\struct {T, *}$ are equal {{iff}}:
:$S = T$
:$\forall a, b \in S: a \circ b = a * b$ | This follows from set equality and Equality of Ordered Pairs.
Category:Algebraic Structures
Category:Equality
gjfvo3rl6m5v407sjg2xuqj8qs3g51r | Two [[Definition:Algebraic Structure with One Operation|algebraic structures]] $\struct {S, \circ}$ and $\struct {T, *}$ are [[Definition:Equality|equal]] {{iff}}:
:$S = T$
:$\forall a, b \in S: a \circ b = a * b$ | This follows from [[Definition:Set Equality|set equality]] and [[Equality of Ordered Pairs]].
[[Category:Algebraic Structures]]
[[Category:Equality]]
gjfvo3rl6m5v407sjg2xuqj8qs3g51r | Equality of Algebraic Structures | https://proofwiki.org/wiki/Equality_of_Algebraic_Structures | https://proofwiki.org/wiki/Equality_of_Algebraic_Structures | [
"Algebraic Structures",
"Equality"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Equals"
] | [
"Definition:Set Equality",
"Equality of Ordered Pairs",
"Category:Algebraic Structures",
"Category:Equality"
] |
proofwiki-508 | Restriction of Associative Operation is Associative | Let $\struct {S, \circ}$ be an algebraic structure.
Let $T \subseteq S$.
Let $\circ_T$ denote the restriction of $\circ$ to $T$.
Let the operation $\circ$ be associative on $\struct {S, \circ}$.
Then $\circ_T$ is associative on $\struct {T, \circ_T}$. | {{begin-eqn}}
{{eqn | l = T
| o = \subseteq
| r = S
| c =
}}
{{eqn | ll= \leadsto
| q = \forall a, b, c \in T
| l = a, b, c
| o = \in
| r = S
| c = {{Defof|Subset}}
}}
{{eqn | ll= \leadsto
| l = a \mathop {\circ_T} \paren {b \mathop {\circ_T} c}
| r = a \circ... | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]].
Let $T \subseteq S$.
Let $\circ_T$ denote the [[Definition:Restriction of Operation|restriction]] of $\circ$ to $T$.
Let the [[Definition:Binary Operation|operation]] $\circ$ be [[Definition:Associative Operation|associative]] on ... | {{begin-eqn}}
{{eqn | l = T
| o = \subseteq
| r = S
| c =
}}
{{eqn | ll= \leadsto
| q = \forall a, b, c \in T
| l = a, b, c
| o = \in
| r = S
| c = {{Defof|Subset}}
}}
{{eqn | ll= \leadsto
| l = a \mathop {\circ_T} \paren {b \mathop {\circ_T} c}
| r = a \circ... | Restriction of Associative Operation is Associative | https://proofwiki.org/wiki/Restriction_of_Associative_Operation_is_Associative | https://proofwiki.org/wiki/Restriction_of_Associative_Operation_is_Associative | [
"Associativity",
"Semigroups"
] | [
"Definition:Algebraic Structure",
"Definition:Restriction/Operation",
"Definition:Operation/Binary Operation",
"Definition:Associative Operation",
"Definition:Associative Operation"
] | [] |
proofwiki-509 | Restriction of Commutative Operation is Commutative | Let $\struct {S, \circ}$ be an algebraic structure.
Let $T \subseteq S$.
Let the operation $\circ$ be commutative on $\struct {S, \circ}$.
Then the restriction $\circ {\restriction_T}$ of $\circ$ to $T$ is also commutative. | {{begin-eqn}}
{{eqn | l = T
| o = \subseteq
| r = S
| c =
}}
{{eqn | ll= \leadsto
| q = \forall a, b \in T
| l = a, b
| o = \in
| r = S
| c = {{Defof|Subset}}
}}
{{eqn | ll= \leadsto
| l = a \mathop {\circ {\restriction_T} } b
| r = a \circ b
| c =
}}
... | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]].
Let $T \subseteq S$.
Let the [[Definition:Binary Operation|operation]] $\circ$ be [[Definition:Commutative Operation|commutative]] on $\struct {S, \circ}$.
Then the [[Definition:Restriction of Operation|restric... | {{begin-eqn}}
{{eqn | l = T
| o = \subseteq
| r = S
| c =
}}
{{eqn | ll= \leadsto
| q = \forall a, b \in T
| l = a, b
| o = \in
| r = S
| c = {{Defof|Subset}}
}}
{{eqn | ll= \leadsto
| l = a \mathop {\circ {\restriction_T} } b
| r = a \circ b
| c =
}}
... | Restriction of Commutative Operation is Commutative | https://proofwiki.org/wiki/Restriction_of_Commutative_Operation_is_Commutative | https://proofwiki.org/wiki/Restriction_of_Commutative_Operation_is_Commutative | [
"Commutativity"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Operation/Binary Operation",
"Definition:Commutative/Operation",
"Definition:Restriction/Operation",
"Definition:Commutative/Operation"
] | [
"Definition:Commutative/Operation"
] |
proofwiki-510 | Idempotent Magma Element forms Singleton Submagma | Let $\struct {S, \circ}$ be a magma.
Let $x \in S$ be an idempotent element of $\struct {S, \circ}$.
Then $\struct {\set x, \circ}$ is a submagma of $\struct {S, \circ}$. | By Singleton of Element is Subset:
:$x \in S \iff \set x \subseteq S$
By the definition of idempotence:
:$x \circ x = x \in \set x$
Thus $\set x$ is a subset of $S$ which is closed under $\circ$.
By the definition of submagma, the result follows.
{{qed}} | Let $\struct {S, \circ}$ be a [[Definition:Magma|magma]].
Let $x \in S$ be an [[Definition:Idempotent Element|idempotent]] element of $\struct {S, \circ}$.
Then $\struct {\set x, \circ}$ is a [[Definition:Submagma|submagma]] of $\struct {S, \circ}$. | By [[Singleton of Element is Subset]]:
:$x \in S \iff \set x \subseteq S$
By the definition of [[Definition:Idempotent Element|idempotence]]:
:$x \circ x = x \in \set x$
Thus $\set x$ is a [[Definition:Subset|subset]] of $S$ which is [[Definition:Closed Algebraic Structure|closed]] under $\circ$.
By the definition o... | Idempotent Magma Element forms Singleton Submagma | https://proofwiki.org/wiki/Idempotent_Magma_Element_forms_Singleton_Submagma | https://proofwiki.org/wiki/Idempotent_Magma_Element_forms_Singleton_Submagma | [
"Magmas"
] | [
"Definition:Magma",
"Definition:Idempotence/Element",
"Definition:Submagma"
] | [
"Singleton of Element is Subset",
"Definition:Idempotence/Element",
"Definition:Subset",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Submagma"
] |
proofwiki-511 | Magma Subset Product with Self | Let $\struct {S, \circ}$ be a magma.
Let $T \subseteq S$.
Then $\struct {T, \circ}$ is a magma {{iff}} $T \circ T \subseteq T$, where $T \circ T$ is the subset product of $T$ with itself. | By definition:
:$T \circ T = \set {x = a \circ b: a, b \in T}$ | Let $\struct {S, \circ}$ be a [[Definition:Magma|magma]].
Let $T \subseteq S$.
Then $\struct {T, \circ}$ is a [[Definition:Magma|magma]] {{iff}} $T \circ T \subseteq T$, where $T \circ T$ is the [[Definition:Subset Product|subset product]] of $T$ with itself. | By definition:
:$T \circ T = \set {x = a \circ b: a, b \in T}$ | Magma Subset Product with Self | https://proofwiki.org/wiki/Magma_Subset_Product_with_Self | https://proofwiki.org/wiki/Magma_Subset_Product_with_Self | [
"Subset Products",
"Magmas"
] | [
"Definition:Magma",
"Definition:Magma",
"Definition:Subset Product"
] | [] |
proofwiki-512 | Subset Product within Semigroup is Associative | Let $\struct {S, \circ}$ be a semigroup.
Then the operation $\circ_\PP$ induced on the power set of $S$ is also associative. | Let $X, Y, Z \in \powerset S$.
Then:
{{begin-eqn}}
{{eqn | l = X \circ_\PP \paren {Y \circ_\PP Z}
| r = \set {x \circ \paren {y \circ z}: x \in X, y \in Y, z \in Z}
| c = {{Defof|Subset Product}}
}}
{{eqn | r = \set {\paren {x \circ y} \circ z: x \in X, y \in Y, z \in Z}
| c = {{Semigroup-axiom|1}}
}}... | Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]].
Then the [[Definition:Operation Induced on Power Set|operation $\circ_\PP$ induced]] on the [[Definition:Power Set|power set]] of $S$ is also [[Definition:Associative Operation|associative]]. | Let $X, Y, Z \in \powerset S$.
Then:
{{begin-eqn}}
{{eqn | l = X \circ_\PP \paren {Y \circ_\PP Z}
| r = \set {x \circ \paren {y \circ z}: x \in X, y \in Y, z \in Z}
| c = {{Defof|Subset Product}}
}}
{{eqn | r = \set {\paren {x \circ y} \circ z: x \in X, y \in Y, z \in Z}
| c = {{Semigroup-axiom|1}}... | Subset Product within Semigroup is Associative | https://proofwiki.org/wiki/Subset_Product_within_Semigroup_is_Associative | https://proofwiki.org/wiki/Subset_Product_within_Semigroup_is_Associative | [
"Subset Products",
"Associativity",
"Semigroups",
"Subset Product within Semigroup is Associative"
] | [
"Definition:Semigroup",
"Definition:Subset Product",
"Definition:Power Set",
"Definition:Associative Operation"
] | [
"Definition:Associative Operation"
] |
proofwiki-513 | Subset Product within Commutative Structure is Commutative | Let $\struct {S, \circ}$ be a magma.
If $\circ$ is commutative, then the operation $\circ_\PP$ induced on the power set of $S$ is also commutative. | Let $\struct {S, \circ}$ be a magma in which $\circ$ is commutative.
Let $X, Y \in \powerset S$.
Then:
{{begin-eqn}}
{{eqn | l = X \circ_\PP Y
| r = \set {x \circ y: x \in X, y \in Y}
| c = Definition of $\circ_\PP$
}}
{{eqn | r = \set {y \circ x: x \in X, y \in Y}
| c = {{Defof|Commutative Operation}... | Let $\struct {S, \circ}$ be a [[Definition:Magma|magma]].
If $\circ$ is [[Definition:Commutative Operation|commutative]], then the [[Definition:Operation Induced on Power Set|operation $\circ_\PP$ induced]] on the [[Definition:Power Set|power set]] of $S$ is also [[Definition:Commutative Operation|commutative]]. | Let $\struct {S, \circ}$ be a [[Definition:Magma|magma]] in which $\circ$ is [[Definition:Commutative Operation|commutative]].
Let $X, Y \in \powerset S$.
Then:
{{begin-eqn}}
{{eqn | l = X \circ_\PP Y
| r = \set {x \circ y: x \in X, y \in Y}
| c = Definition of $\circ_\PP$
}}
{{eqn | r = \set {y \circ x... | Subset Product within Commutative Structure is Commutative | https://proofwiki.org/wiki/Subset_Product_within_Commutative_Structure_is_Commutative | https://proofwiki.org/wiki/Subset_Product_within_Commutative_Structure_is_Commutative | [
"Subset Products"
] | [
"Definition:Magma",
"Definition:Commutative/Operation",
"Definition:Subset Product",
"Definition:Power Set",
"Definition:Commutative/Operation"
] | [
"Definition:Magma",
"Definition:Commutative/Operation",
"Definition:Commutative/Operation"
] |
proofwiki-514 | Subset Relation is Compatible with Subset Product | Let $\struct {S, \circ}$ be a magma.
Let $\struct {\powerset S, \circ_\PP}$ be the power structure of $\struct {S, \circ}$.
Then the subset relation on $S$ is compatible with $\circ_\PP$.
That is:
{{begin-eqn}}
{{eqn | q = \forall X, Y, Z \in \powerset S
| l = X \subseteq Y
| o = \implies
| r = \paren... | Let $x \in X, z \in Z$.
Then:
:$x \circ z \in X \circ Z$ and $z \circ x \in Z \circ X$
Now:
:$Y \circ Z = \set {y \circ z: y \in Y, z \in Z}$
:$Z \circ Y = \set {z \circ y: y \in Y, z \in Z}$
But by the definition of a subset:
:$x \in X \implies x \in Y$
Thus:
:$x \circ z \in Y \circ Z$ and $z \circ x \in Z \circ Y$
an... | Let $\struct {S, \circ}$ be a [[Definition:Magma|magma]].
Let $\struct {\powerset S, \circ_\PP}$ be the [[Definition:Power Structure|power structure]] of $\struct {S, \circ}$.
Then the [[Definition:Subset|subset relation]] on $S$ is [[Definition:Relation Compatible with Operation|compatible]] with $\circ_\PP$.
Tha... | Let $x \in X, z \in Z$.
Then:
:$x \circ z \in X \circ Z$ and $z \circ x \in Z \circ X$
Now:
:$Y \circ Z = \set {y \circ z: y \in Y, z \in Z}$
:$Z \circ Y = \set {z \circ y: y \in Y, z \in Z}$
But by the definition of a [[Definition:Subset|subset]]:
:$x \in X \implies x \in Y$
Thus:
:$x \circ z \in Y \circ Z$ and $z... | Subset Relation is Compatible with Subset Product | https://proofwiki.org/wiki/Subset_Relation_is_Compatible_with_Subset_Product | https://proofwiki.org/wiki/Subset_Relation_is_Compatible_with_Subset_Product | [
"Power Structures",
"Subset Products",
"Compatible Relations"
] | [
"Definition:Magma",
"Definition:Power Structure",
"Definition:Subset",
"Definition:Relation Compatible with Operation"
] | [
"Definition:Subset"
] |
proofwiki-515 | Cancellable Element is Cancellable in Subset | Let $\struct {S, \circ}$ be an algebraic structure.
Let $\struct {T, \circ} \subseteq \struct {S, \circ}$.
Let $x \in T$ be cancellable in $S$.
Then $x$ is also cancellable in $T$. | Let $x \in T$ be cancellable in $S$.
Then by definition, $x \in T$ is left cancellable in $S$.
It follows from Left Cancellable Element is Left Cancellable in Subset that $x \in T$ is left cancellable in $T$.
Again by definition, $x \in T$ is right cancellable in $S$.
It follows from Right Cancellable Element is Right ... | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]].
Let $\struct {T, \circ} \subseteq \struct {S, \circ}$.
Let $x \in T$ be [[Definition:Cancellable Element|cancellable]] in $S$.
Then $x$ is also [[Definition:Cancellable Element|cancellable]] in $T$. | Let $x \in T$ be [[Definition:Cancellable Element|cancellable]] in $S$.
Then by definition, $x \in T$ is [[Definition:Left Cancellable Element|left cancellable]] in $S$.
It follows from [[Left Cancellable Element is Left Cancellable in Subset]] that $x \in T$ is [[Definition:Left Cancellable Element|left cancellable]... | Cancellable Element is Cancellable in Subset | https://proofwiki.org/wiki/Cancellable_Element_is_Cancellable_in_Subset | https://proofwiki.org/wiki/Cancellable_Element_is_Cancellable_in_Subset | [
"Cancellability"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Cancellable Element",
"Definition:Cancellable Element"
] | [
"Definition:Cancellable Element",
"Definition:Cancellable Element/Left Cancellable",
"Left Cancellable Element is Left Cancellable in Subset",
"Definition:Cancellable Element/Left Cancellable",
"Definition:Cancellable Element/Right Cancellable",
"Right Cancellable Element is Right Cancellable in Subset",
... |
proofwiki-516 | Subsemigroup Closure Test | To show that an algebraic structure $\struct {T, \circ_T}$ is a subsemigroup of a semigroup $\struct {S, \circ}$, we need to show only that:
:$(1): \quad T \subseteq S$
:$(2): \quad \circ$ is a closed operation in $T$.
where $\circ_T$ denotes the restriction of $\circ$ to $T$. | From Restriction of Associative Operation is Associative, if $\circ$ is associative on $\struct {S, \circ}$, then it will also be associative on $\struct {T, \circ_T}$.
Thus we do not need to check for associativity in $\struct {T, \circ_T}$, as that has been inherited from its extension $\struct {S, \circ}$.
So, once ... | To show that an [[Definition:Algebraic Structure with One Operation|algebraic structure]] $\struct {T, \circ_T}$ is a [[Definition:Subsemigroup|subsemigroup]] of a [[Definition:Semigroup|semigroup]] $\struct {S, \circ}$, we need to show only that:
:$(1): \quad T \subseteq S$
:$(2): \quad \circ$ is a [[Definition:Clos... | From [[Restriction of Associative Operation is Associative]], if $\circ$ is [[Definition:Associative Operation|associative]] on $\struct {S, \circ}$, then it will also be [[Definition:Associative Operation|associative]] on $\struct {T, \circ_T}$.
Thus we do not need to check for [[Definition:Associative Operation|asso... | Subsemigroup Closure Test | https://proofwiki.org/wiki/Subsemigroup_Closure_Test | https://proofwiki.org/wiki/Subsemigroup_Closure_Test | [
"Subsemigroups"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Subsemigroup",
"Definition:Semigroup",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Restriction/Operation"
] | [
"Restriction of Associative Operation is Associative",
"Definition:Associative Operation",
"Definition:Associative Operation",
"Definition:Associative Operation",
"Definition:Extension of Operation",
"Definition:Closure (Abstract Algebra)/Algebraic Structure"
] |
proofwiki-517 | Cancellable iff Regular Representations Injective | Let $\struct {S, \circ}$ be an algebraic structure.
Then $a \in S$ is cancellable {{iff}}:
:the left regular representation $\map {\lambda_a} x$ is injective
and:
:the right regular representation $\map {\rho_a} x$ is injective. | === Left Cancellable ===
{{:Left Cancellable iff Left Regular Representation Injective}} | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]].
Then $a \in S$ is [[Definition:Cancellable Element|cancellable]] {{iff}}:
:the [[Definition:Left Regular Representation|left regular representation]] $\map {\lambda_a} x$ is [[Definition:Injection|injective]]
and... | === [[Left Cancellable iff Left Regular Representation Injective|Left Cancellable]] ===
{{:Left Cancellable iff Left Regular Representation Injective}} | Cancellable iff Regular Representations Injective | https://proofwiki.org/wiki/Cancellable_iff_Regular_Representations_Injective | https://proofwiki.org/wiki/Cancellable_iff_Regular_Representations_Injective | [
"Regular Representations",
"Cancellability"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Cancellable Element",
"Definition:Regular Representations/Left Regular Representation",
"Definition:Injection",
"Definition:Regular Representations/Right Regular Representation",
"Definition:Injection"
] | [
"Left Cancellable iff Left Regular Representation Injective"
] |
proofwiki-518 | More than one Left Identity then no Right Identity | Let $\struct {S, \circ}$ be an algebraic structure.
If $\struct {S, \circ}$ has more than one left identity, then it has no right identity. | Let $\struct {S, \circ}$ be an algebraic structure with more than one left identity.
Take any two of these, and call them $e_{L_1}$ and $e_{L_2}$, where $e_{L_1} \ne e_{L_2}$.
Suppose $\struct {S, \circ}$ has a right identity.
Call this right identity $e_R$.
Then, by the behaviour of $e_R$, $e_{L_1}$ and $e_{L_2}$:
:$e... | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]].
If $\struct {S, \circ}$ has more than one [[Definition:Left Identity|left identity]], then it has no [[Definition:Right Identity|right identity]]. | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]] with more than one [[Definition:Left Identity|left identity]].
Take any two of these, and call them $e_{L_1}$ and $e_{L_2}$, where $e_{L_1} \ne e_{L_2}$.
Suppose $\struct {S, \circ}$ has a [[Definition:Right Ident... | More than one Left Identity then no Right Identity | https://proofwiki.org/wiki/More_than_one_Left_Identity_then_no_Right_Identity | https://proofwiki.org/wiki/More_than_one_Left_Identity_then_no_Right_Identity | [
"Identity Elements"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Identity (Abstract Algebra)/Left Identity",
"Definition:Identity (Abstract Algebra)/Right Identity"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Identity (Abstract Algebra)/Left Identity",
"Definition:Identity (Abstract Algebra)/Right Identity",
"Definition:Identity (Abstract Algebra)/Right Identity",
"Definition:Algebraic Structure/One Operation",
"Definition:Identity (Abstract Algebra)/... |
proofwiki-519 | Element under Left Operation is Right Identity | Let $\struct {S, \gets}$ be an algebraic structure in which the operation $\gets$ is the left operation.
Then all of the elements of $\struct {S, \gets}$ are right identities. | From Structure under Left Operation is Semigroup, $\struct {S, \gets}$ is a semigroup.
From the definition of left operation:
:$\forall x, y \in S: x \gets y = x$
from which it is apparent that all elements of $S$ are right identities.
{{qed}} | Let $\struct {S, \gets}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]] in which the [[Definition:Binary Operation|operation]] $\gets$ is the [[Definition:Left Operation|left operation]].
Then all of the [[Definition:Element|elements]] of $\struct {S, \gets}$ are [[Definition:Right Id... | From [[Structure under Left Operation is Semigroup]], $\struct {S, \gets}$ is a [[Definition:Semigroup|semigroup]].
From the definition of [[Definition:Left Operation|left operation]]:
:$\forall x, y \in S: x \gets y = x$
from which it is apparent that all [[Definition:Element|elements]] of $S$ are [[Definition:Right... | Element under Left Operation is Right Identity | https://proofwiki.org/wiki/Element_under_Left_Operation_is_Right_Identity | https://proofwiki.org/wiki/Element_under_Left_Operation_is_Right_Identity | [
"Left Operation",
"Examples of Identity Elements"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Operation/Binary Operation",
"Definition:Left Operation",
"Definition:Element",
"Definition:Identity (Abstract Algebra)/Right Identity"
] | [
"Structure under Left Operation is Semigroup",
"Definition:Semigroup",
"Definition:Left Operation",
"Definition:Element",
"Definition:Identity (Abstract Algebra)/Right Identity"
] |
proofwiki-520 | Element under Right Operation is Left Identity | Let $\struct {S, \to}$ be an algebraic structure in which the operation $\to$ is the right operation.
Then all of the elements of $\struct {S, \to}$ left identities. | From Structure under Left Operation is Semigroup, $\struct {S, \to}$ is a semigroup.
From the definition of right operation:
:$\forall x, y \in S: x \to y = y$
from which it is apparent that all elements of $S$ are left identities.
{{qed}} | Let $\struct {S, \to}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]] in which the [[Definition:Binary Operation|operation]] $\to$ is the [[Definition:Right Operation|right operation]].
Then all of the [[Definition:Element|elements]] of $\struct {S, \to}$ [[Definition:Left Identity|le... | From [[Structure under Left Operation is Semigroup]], $\struct {S, \to}$ is a [[Definition:Semigroup|semigroup]].
From the definition of [[Definition:Right Operation|right operation]]:
:$\forall x, y \in S: x \to y = y$
from which it is apparent that all [[Definition:Element|elements]] of $S$ are [[Definition:Left Id... | Element under Right Operation is Left Identity | https://proofwiki.org/wiki/Element_under_Right_Operation_is_Left_Identity | https://proofwiki.org/wiki/Element_under_Right_Operation_is_Left_Identity | [
"Right Operation",
"Examples of Identity Elements"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Operation/Binary Operation",
"Definition:Right Operation",
"Definition:Element",
"Definition:Identity (Abstract Algebra)/Left Identity"
] | [
"Structure under Left Operation is Semigroup",
"Definition:Semigroup",
"Definition:Right Operation",
"Definition:Element",
"Definition:Identity (Abstract Algebra)/Left Identity"
] |
proofwiki-521 | Left and Right Identity are the Same | Let $\struct {S, \circ}$ be an algebraic structure.
Let $e_L \in S$ be a left identity, and $e_R \in S$ be a right identity.
Then:
:$e_L = e_R$
that is, both the left identity and right identity are the same, and are therefore an identity $e$.
Furthermore, $e$ is the only left identity and right identity for $\circ$. | Let $\struct {S, \circ}$ be an algebraic structure such that:
:$\exists e_L \in S: \forall x \in S: e_L \circ x = x$
:$\exists e_R \in S: \forall x \in S: x \circ e_R = x$
Then $e_L = e_L \circ e_R = e_R$ by both the above, hence the result.
The uniqueness of the left and right identity is a direct result of Identity i... | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]].
Let $e_L \in S$ be a [[Definition:Left Identity|left identity]], and $e_R \in S$ be a [[Definition:Right Identity|right identity]].
Then:
:$e_L = e_R$
that is, both the [[Definition:Left Identity|left identity]]... | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]] such that:
:$\exists e_L \in S: \forall x \in S: e_L \circ x = x$
:$\exists e_R \in S: \forall x \in S: x \circ e_R = x$
Then $e_L = e_L \circ e_R = e_R$ by both the above, hence the result.
The uniqueness of th... | Left and Right Identity are the Same | https://proofwiki.org/wiki/Left_and_Right_Identity_are_the_Same | https://proofwiki.org/wiki/Left_and_Right_Identity_are_the_Same | [
"Identity Elements"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Identity (Abstract Algebra)/Left Identity",
"Definition:Identity (Abstract Algebra)/Right Identity",
"Definition:Identity (Abstract Algebra)/Left Identity",
"Definition:Identity (Abstract Algebra)/Right Identity",
"Definition:Identity (Abstract A... | [
"Definition:Algebraic Structure/One Operation",
"Identity is Unique"
] |
proofwiki-522 | Identity Property in Semigroup | Let $\struct {S, \circ}$ be a semigroup.
Let $s \in S$ be such that:
:$\forall a \in S: \exists x, y \in S: s \circ x = a = y \circ s$
Then $\struct {S, \circ}$ has an identity. | Suppose that:
:$\forall a \in S: \exists x, y \in S: s \circ x = a = y \circ s$.
Since $s \in S$, it follows that:
:$\exists v, w \in S: s \circ v = s = w \circ s$.
Let $a \in S$.
Then:
:$\exists x, y \in S: s \circ x = a = y \circ s$.
Thus:
{{begin-eqn}}
{{eqn | l = a
| r = s \circ x
| c =
}}
{{eqn | ll= ... | Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]].
Let $s \in S$ be such that:
:$\forall a \in S: \exists x, y \in S: s \circ x = a = y \circ s$
Then $\struct {S, \circ}$ has an [[Definition:Identity Element|identity]]. | Suppose that:
:$\forall a \in S: \exists x, y \in S: s \circ x = a = y \circ s$.
Since $s \in S$, it follows that:
:$\exists v, w \in S: s \circ v = s = w \circ s$.
Let $a \in S$.
Then:
:$\exists x, y \in S: s \circ x = a = y \circ s$.
Thus:
{{begin-eqn}}
{{eqn | l = a
| r = s \circ x
| c =
}}
{{eq... | Identity Property in Semigroup | https://proofwiki.org/wiki/Identity_Property_in_Semigroup | https://proofwiki.org/wiki/Identity_Property_in_Semigroup | [
"Semigroups",
"Identity Elements"
] | [
"Definition:Semigroup",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] |
proofwiki-523 | Identity of Monoid is Cancellable | Let $\struct {S, \circ}$ be a monoid.
Let $e$ be an identity element of $\struct {S, \circ}$.
Then $e$ is a cancellable element of $\struct {S, \circ}$. | Let $x, y \in S$ such that $x \circ e = y \circ e$.
Then, by definition of identity element:
:$x = x \circ e = y \circ e = y$
Thus $x = y$.
{{qed}} | Let $\struct {S, \circ}$ be a [[Definition:Monoid|monoid]].
Let $e$ be an [[Definition:Identity Element|identity element]] of $\struct {S, \circ}$.
Then $e$ is a [[Definition:Cancellable Element|cancellable element]] of $\struct {S, \circ}$. | Let $x, y \in S$ such that $x \circ e = y \circ e$.
Then, by definition of [[Definition:Identity Element|identity element]]:
:$x = x \circ e = y \circ e = y$
Thus $x = y$.
{{qed}} | Identity of Monoid is Cancellable | https://proofwiki.org/wiki/Identity_of_Monoid_is_Cancellable | https://proofwiki.org/wiki/Identity_of_Monoid_is_Cancellable | [
"Cancellability",
"Monoids"
] | [
"Definition:Monoid",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Cancellable Element"
] | [
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] |
proofwiki-524 | Identity is only Idempotent Cancellable Element | Let $e_S$ be the identity of an algebraic structure $\struct {S, \circ}$.
Then $e_S$ is the only cancellable element of $\struct {S, \circ}$ that is idempotent. | By Identity Element is Idempotent, $e_S$ is idempotent.
Let $x$ be a cancellable idempotent element of $\struct {S, \circ}$.
{{begin-eqn}}
{{eqn | l = x \circ x
| r = x
| c = $x$ is idempotent
}}
{{eqn | r = x \circ e_S
| c = {{Defof|Identity Element}}
}}
{{end-eqn}}
So $x \circ x = x \circ e_S$.
But ... | Let $e_S$ be the [[Definition:Identity Element|identity]] of an [[Definition:Algebraic Structure with One Operation|algebraic structure]] $\struct {S, \circ}$.
Then $e_S$ is the only [[Definition:Cancellable Element|cancellable element]] of $\struct {S, \circ}$ that is [[Definition:Idempotent Element|idempotent]]. | By [[Identity Element is Idempotent]], $e_S$ is [[Definition:Idempotent Element|idempotent]].
Let $x$ be a [[Definition:Cancellable Element|cancellable]] [[Definition:Idempotent Element|idempotent element]] of $\struct {S, \circ}$.
{{begin-eqn}}
{{eqn | l = x \circ x
| r = x
| c = [[Definition:Idempotent ... | Identity is only Idempotent Cancellable Element | https://proofwiki.org/wiki/Identity_is_only_Idempotent_Cancellable_Element | https://proofwiki.org/wiki/Identity_is_only_Idempotent_Cancellable_Element | [
"Cancellability",
"Identities are Idempotent"
] | [
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Algebraic Structure/One Operation",
"Definition:Cancellable Element",
"Definition:Idempotence/Element"
] | [
"Identity Element is Idempotent",
"Definition:Idempotence/Element",
"Definition:Cancellable Element",
"Definition:Idempotence/Element",
"Definition:Idempotence/Element",
"Definition:Cancellable Element"
] |
proofwiki-525 | Set of all Self-Maps under Composition forms Monoid | Let $S$ be a set.
Let $S^S$ be the set of all mappings from $S$ to itself.
Let the operation $\circ$ represent composition of mappings.
Then the algebraic structure $\struct {S^S, \circ}$ is a monoid whose identity element is the identity mapping on $S$. | By Set of all Self-Maps under Composition forms Semigroup, $\struct {S^S, \circ}$ is a semigroup.
By Identity Mapping is Left Identity and Identity Mapping is Right Identity the identity mapping on $S$ is the identity element of $\struct {S^S, \circ}$.
Since $\struct {S^S, \circ}$ is a semigroup with an identity elemen... | Let $S$ be a [[Definition:Set|set]].
Let $S^S$ be the [[Definition:Set of All Mappings|set of all mappings]] from $S$ to itself.
Let the [[Definition:Binary Operation|operation]] $\circ$ represent [[Definition:Composition of Mappings|composition of mappings]].
Then the [[Definition:Algebraic Structure with One Oper... | By [[Set of all Self-Maps under Composition forms Semigroup]], $\struct {S^S, \circ}$ is a [[Definition:Semigroup|semigroup]].
By [[Identity Mapping is Left Identity]] and [[Identity Mapping is Right Identity]] the [[Definition:Identity Mapping|identity mapping]] on $S$ is the [[Definition:Identity Element|identity el... | Set of all Self-Maps under Composition forms Monoid | https://proofwiki.org/wiki/Set_of_all_Self-Maps_under_Composition_forms_Monoid | https://proofwiki.org/wiki/Set_of_all_Self-Maps_under_Composition_forms_Monoid | [
"Composite Mappings",
"Examples of Monoids",
"Self-Maps"
] | [
"Definition:Set",
"Definition:Set of All Mappings",
"Definition:Operation/Binary Operation",
"Definition:Composition of Mappings",
"Definition:Algebraic Structure/One Operation",
"Definition:Monoid",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Identity Mapping"
] | [
"Set of all Self-Maps under Composition forms Semigroup",
"Definition:Semigroup",
"Identity Mapping is Left Identity",
"Identity Mapping is Right Identity",
"Definition:Identity Mapping",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Semigroup",
"Definition:Identity (Abstrac... |
proofwiki-526 | Test for Submonoid | To show that $\struct {T, \circ}$ is a submonoid of a monoid $\struct {S, \circ}$, we need to show that:
:$(1): \quad T \subseteq S$
:$(2): \quad \struct {T, \circ}$ is a magma (that is, that it is closed)
:$(3): \quad \struct {T, \circ}$ has an identity. | From Subsemigroup Closure Test, $(1)$ and $(2)$ are sufficient to show that $\struct {T, \circ}$ is a subsemigroup of $\struct {S, \circ}$.
Demonstrating the presence of an identity is then sufficient to show that it is a monoid.
{{qed}}
Category:Monoids
s8ylqenh1apfl8zb5btm7g43qjgonsg | To show that $\struct {T, \circ}$ is a [[Definition:Submonoid|submonoid]] of a [[Definition:Monoid|monoid]] $\struct {S, \circ}$, we need to show that:
:$(1): \quad T \subseteq S$
:$(2): \quad \struct {T, \circ}$ is a [[Definition:Magma|magma]] (that is, that it is [[Definition:Closed Algebraic Structure|closed]])
:$(... | From [[Subsemigroup Closure Test]], $(1)$ and $(2)$ are sufficient to show that $\struct {T, \circ}$ is a [[Definition:Subsemigroup|subsemigroup]] of $\struct {S, \circ}$.
Demonstrating the presence of an [[Definition:Identity Element|identity]] is then sufficient to show that it is a [[Definition:Monoid|monoid]].
{{q... | Test for Submonoid | https://proofwiki.org/wiki/Test_for_Submonoid | https://proofwiki.org/wiki/Test_for_Submonoid | [
"Monoids"
] | [
"Definition:Submonoid",
"Definition:Monoid",
"Definition:Magma",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Subsemigroup Closure Test",
"Definition:Subsemigroup",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Monoid",
"Category:Monoids"
] |
proofwiki-527 | Identity of Cancellable Monoid is Identity of Submonoid | Let $\struct {S, \circ}$ be a monoid, all of whose elements are cancellable.
Let $\struct {T, \circ}$ be a submonoid of $\struct {S, \circ}$.
Then the identity of $\struct {T, \circ}$ is the same element as the identity of $\struct {S, \circ}$. | By Identity of Monoid is Unique:
: there is only one identity element of $\struct {S, \circ}$
and:
: there is only one identity element of $\struct {T, \circ}$.
Let $e_S$ be the identity of $\struct {S, \circ}$, and $e_T$ the identity of $\struct {T, \circ}$.
From Identity is only Idempotent Cancellable Element, $e_S$ ... | Let $\struct {S, \circ}$ be a [[Definition:Monoid|monoid]], all of whose [[Definition:Element|elements]] are [[Definition:Cancellable Element|cancellable]].
Let $\struct {T, \circ}$ be a [[Definition:Submonoid|submonoid]] of $\struct {S, \circ}$.
Then the [[Definition:Identity Element|identity]] of $\struct {T, \cir... | By [[Identity of Monoid is Unique]]:
: there is only one [[Definition:Identity Element|identity element]] of $\struct {S, \circ}$
and:
: there is only one [[Definition:Identity Element|identity element]] of $\struct {T, \circ}$.
Let $e_S$ be the [[Definition:Identity Element|identity]] of $\struct {S, \circ}$, and $e... | Identity of Cancellable Monoid is Identity of Submonoid | https://proofwiki.org/wiki/Identity_of_Cancellable_Monoid_is_Identity_of_Submonoid | https://proofwiki.org/wiki/Identity_of_Cancellable_Monoid_is_Identity_of_Submonoid | [
"Cancellable Monoids",
"Identity Elements"
] | [
"Definition:Monoid",
"Definition:Element",
"Definition:Cancellable Element",
"Definition:Submonoid",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Element",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Identity of Monoid is Unique",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Identity is only Idempotent Cancellable Element",
"Definition:Cancellable Element",
"Def... |
proofwiki-528 | Cancellable Elements of Monoid form Submonoid | The cancellable elements of a monoid $\struct {S, \circ}$ form a submonoid of $\struct {S, \circ}$. | Let $C$ be the set of cancellable elements of $\struct {S, \circ}$.
Obviously $C \subseteq S$.
From Cancellable Elements of Semigroup form Subsemigroup, $\struct {C, \circ}$ is a subsemigroup of $S$.
Let $e_S$ be the identity of $\struct {S, \circ}$.
From Identity of Monoid is Cancellable, $e_S$ is cancellable, therefo... | The [[Definition:Cancellable Element|cancellable elements]] of a [[Definition:Monoid|monoid]] $\struct {S, \circ}$ form a [[Definition:Submonoid|submonoid]] of $\struct {S, \circ}$. | Let $C$ be the [[Definition:Set|set]] of [[Definition:Cancellable Element|cancellable elements]] of $\struct {S, \circ}$.
Obviously $C \subseteq S$.
From [[Cancellable Elements of Semigroup form Subsemigroup]], $\struct {C, \circ}$ is a [[Definition:Subsemigroup|subsemigroup]] of $S$.
Let $e_S$ be the [[Definition:I... | Cancellable Elements of Monoid form Submonoid | https://proofwiki.org/wiki/Cancellable_Elements_of_Monoid_form_Submonoid | https://proofwiki.org/wiki/Cancellable_Elements_of_Monoid_form_Submonoid | [
"Monoids"
] | [
"Definition:Cancellable Element",
"Definition:Monoid",
"Definition:Submonoid"
] | [
"Definition:Set",
"Definition:Cancellable Element",
"Cancellable Elements of Semigroup form Subsemigroup",
"Definition:Subsemigroup",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Identity of Monoid is Cancellable",
"Definition:Monoid",
"Category:Monoids"
] |
proofwiki-529 | Product of Semigroup Element with Left Inverse is Idempotent | Let $\struct {S, \circ}$ be a semigroup with a left identity $e_L$.
Let $x \in S$ such that $\exists x_L: x_L \circ x = e_L$, that is $x$ has a left inverse with respect to the left identity.
Then:
:$\paren {x \circ x_L} \circ \paren {x \circ x_L} = x \circ x_L$
That is, $x \circ x_L$ is idempotent. | {{begin-eqn}}
{{eqn | l = \paren {x \circ x_L} \circ \paren {x \circ x_L}
| r = x \circ \paren {x_L \circ x} \circ x_L
| c = {{Semigroup-axiom|1}}
}}
{{eqn | r = x \circ e_L \circ x_L
| c = {{Defof|Left Inverse Element}}
}}
{{eqn | r = x \circ x_L
| c = {{Defof|Left Identity}}
}}
{{end-eqn}}
{{q... | Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]] with a [[Definition:Left Identity|left identity]] $e_L$.
Let $x \in S$ such that $\exists x_L: x_L \circ x = e_L$, that is $x$ has a [[Definition:Left Inverse Element|left inverse]] with respect to the [[Definition:Left Identity|left identity]].
Then:
... | {{begin-eqn}}
{{eqn | l = \paren {x \circ x_L} \circ \paren {x \circ x_L}
| r = x \circ \paren {x_L \circ x} \circ x_L
| c = {{Semigroup-axiom|1}}
}}
{{eqn | r = x \circ e_L \circ x_L
| c = {{Defof|Left Inverse Element}}
}}
{{eqn | r = x \circ x_L
| c = {{Defof|Left Identity}}
}}
{{end-eqn}}
{{q... | Product of Semigroup Element with Left Inverse is Idempotent | https://proofwiki.org/wiki/Product_of_Semigroup_Element_with_Left_Inverse_is_Idempotent | https://proofwiki.org/wiki/Product_of_Semigroup_Element_with_Left_Inverse_is_Idempotent | [
"Semigroups",
"Idempotence"
] | [
"Definition:Semigroup",
"Definition:Identity (Abstract Algebra)/Left Identity",
"Definition:Inverse (Abstract Algebra)/Left Inverse",
"Definition:Identity (Abstract Algebra)/Left Identity",
"Definition:Idempotence/Element"
] | [] |
proofwiki-530 | Product of Semigroup Element with Right Inverse is Idempotent | Let $\struct {S, \circ}$ be a semigroup with a right identity $e_R$.
Let $x \in S$ such that $\exists x_R: x \circ x_R = e_R$, that is, $x$ has a right inverse with respect to the right identity.
Then:
:$\paren {x_R \circ x} \circ \paren {x_R \circ x} = x_R \circ x$
That is, $x_R \circ x$ is idempotent. | {{begin-eqn}}
{{eqn | l = \paren {x_R \circ x} \circ \paren {x_R \circ x}
| r = x_R \circ \paren {x \circ x_R} \circ x
| c = {{Semigroup-axiom|1}}
}}
{{eqn | r = x_R \circ e_R \circ x
| c = {{Defof|Right Inverse Element}}
}}
{{eqn | r = x_R \circ x
| c = {{Defof|Right Identity}}
}}
{{end-eqn}}
{... | Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]] with a [[Definition:Right Identity|right identity]] $e_R$.
Let $x \in S$ such that $\exists x_R: x \circ x_R = e_R$, that is, $x$ has a [[Definition:Right Inverse Element|right inverse]] with respect to the [[Definition:Right Identity|right identity]].
... | {{begin-eqn}}
{{eqn | l = \paren {x_R \circ x} \circ \paren {x_R \circ x}
| r = x_R \circ \paren {x \circ x_R} \circ x
| c = {{Semigroup-axiom|1}}
}}
{{eqn | r = x_R \circ e_R \circ x
| c = {{Defof|Right Inverse Element}}
}}
{{eqn | r = x_R \circ x
| c = {{Defof|Right Identity}}
}}
{{end-eqn}}
{... | Product of Semigroup Element with Right Inverse is Idempotent | https://proofwiki.org/wiki/Product_of_Semigroup_Element_with_Right_Inverse_is_Idempotent | https://proofwiki.org/wiki/Product_of_Semigroup_Element_with_Right_Inverse_is_Idempotent | [
"Product of Semigroup Element with Right Inverse is Idempotent",
"Semigroups",
"Idempotence"
] | [
"Definition:Semigroup",
"Definition:Identity (Abstract Algebra)/Right Identity",
"Definition:Inverse (Abstract Algebra)/Right Inverse",
"Definition:Identity (Abstract Algebra)/Right Identity",
"Definition:Idempotence/Element"
] | [] |
proofwiki-531 | Left Inverse for All is Right Inverse | Let $\struct {S, \circ}$ be a semigroup with a left identity $e_L$ such that:
:$\forall x \in S: \exists x_L: x_L \circ x = e_L$
That is, every element of $S$ has a left inverse with respect to the left identity.
Then $x \circ x_L = e_L$, that is, $x_L$ is also a right inverse with respect to the left identity. | Let $y = x \circ x_L$. Then:
{{begin-eqn}}
{{eqn | l = e_L \circ y
| r = \paren {y_L \circ y} \circ y
| c = {{Defof|Left Inverse Element}}
}}
{{eqn | r = y_L \circ \paren {y \circ y}
| c = {{Semigroup-axiom|1}}
}}
{{eqn | r = y_L \circ y
| c = Product of Semigroup Element with Left Inverse is Id... | Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]] with a [[Definition:Left Identity|left identity]] $e_L$ such that:
:$\forall x \in S: \exists x_L: x_L \circ x = e_L$
That is, every [[Definition:Element|element]] of $S$ has a [[Definition:Left Inverse Element|left inverse]] with respect to the [[Defin... | Let $y = x \circ x_L$. Then:
{{begin-eqn}}
{{eqn | l = e_L \circ y
| r = \paren {y_L \circ y} \circ y
| c = {{Defof|Left Inverse Element}}
}}
{{eqn | r = y_L \circ \paren {y \circ y}
| c = {{Semigroup-axiom|1}}
}}
{{eqn | r = y_L \circ y
| c = [[Product of Semigroup Element with Left Inverse is... | Left Inverse for All is Right Inverse | https://proofwiki.org/wiki/Left_Inverse_for_All_is_Right_Inverse | https://proofwiki.org/wiki/Left_Inverse_for_All_is_Right_Inverse | [
"Semigroups",
"Inverse Elements"
] | [
"Definition:Semigroup",
"Definition:Identity (Abstract Algebra)/Left Identity",
"Definition:Element",
"Definition:Inverse (Abstract Algebra)/Left Inverse",
"Definition:Identity (Abstract Algebra)/Left Identity",
"Definition:Inverse (Abstract Algebra)/Right Inverse",
"Definition:Identity (Abstract Algebr... | [
"Product of Semigroup Element with Left Inverse is Idempotent",
"Definition:Inverse (Abstract Algebra)/Right Inverse",
"Definition:Inverse (Abstract Algebra)/Left Inverse",
"Definition:Identity (Abstract Algebra)/Left Identity"
] |
proofwiki-532 | Left Inverse and Right Inverse is Inverse | Let $\struct {S, \circ}$ be a monoid with identity element $e_S$.
Let $x \in S$ such that $x$ has both a left inverse and a right inverse. That is:
:$\exists x_L \in S: x_L \circ x = e_S$
:$\exists x_R \in S: x \circ x_R = e_S$
Then $x_L = x_R$, that is, $x$ has an inverse.
Furthermore, that element is the ''only'' inv... | We note that as $\struct {S, \circ}$ is a monoid, $\circ$ is associative by definition.
{{begin-eqn}}
{{eqn | l = x_L
| r = x_L \circ e_S
| c = {{Defof|Identity Element}}
}}
{{eqn | r = x_L \circ \paren {x \circ x_R}
| c = {{Defof|Right Inverse Element}}
}}
{{eqn | r = \paren {x_L \circ x} \circ x_R
... | Let $\struct {S, \circ}$ be a [[Definition:Monoid|monoid]] with [[Definition:Identity Element|identity element]] $e_S$.
Let $x \in S$ such that $x$ has both a [[Definition:Left Inverse Element|left inverse]] and a [[Definition:Right Inverse Element|right inverse]]. That is:
:$\exists x_L \in S: x_L \circ x = e_S$
:$\... | We note that as $\struct {S, \circ}$ is a [[Definition:Monoid|monoid]], $\circ$ is [[Definition:Associative Operation|associative]] by definition.
{{begin-eqn}}
{{eqn | l = x_L
| r = x_L \circ e_S
| c = {{Defof|Identity Element}}
}}
{{eqn | r = x_L \circ \paren {x \circ x_R}
| c = {{Defof|Right Inver... | Left Inverse and Right Inverse is Inverse | https://proofwiki.org/wiki/Left_Inverse_and_Right_Inverse_is_Inverse | https://proofwiki.org/wiki/Left_Inverse_and_Right_Inverse_is_Inverse | [
"Monoids",
"Inverse Elements"
] | [
"Definition:Monoid",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Inverse (Abstract Algebra)/Left Inverse",
"Definition:Inverse (Abstract Algebra)/Right Inverse",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Inv... | [
"Definition:Monoid",
"Definition:Associative Operation",
"Inverse in Monoid is Unique"
] |
proofwiki-533 | Left and Right Inverses of Product | Let $\struct {S, \circ}$ be a monoid whose identity is $e_S$.
Let $x, y \in S$.
Let:
:$(1): \quad x \circ y$ have a left inverse for $\circ$
:$(2): \quad y \circ x$ have a right inverse for $\circ$.
Then both $x$ and $y$ are invertible for $\circ$. | Let $z_L$ be the left inverse of $x \circ y$ and $z_R$ be the right inverse of $y \circ x$. Then:
{{begin-eqn}}
{{eqn | l = z_L \circ \paren {x \circ y}
| r = e_S
| c = {{hypothesis}}
}}
{{eqn | ll= \leadsto
| l = \paren {z_L \circ x} \circ y
| r = e_S
| c = Associativity of $\circ$
}}
{{e... | Let $\struct {S, \circ}$ be a [[Definition:Monoid|monoid]] whose [[Definition:Identity Element|identity]] is $e_S$.
Let $x, y \in S$.
Let:
:$(1): \quad x \circ y$ have a [[Definition:Left Inverse Element|left inverse]] for $\circ$
:$(2): \quad y \circ x$ have a [[Definition:Right Inverse Element|right inverse]] for $... | Let $z_L$ be the [[Definition:Left Inverse Element|left inverse]] of $x \circ y$ and $z_R$ be the [[Definition:Right Inverse Element|right inverse]] of $y \circ x$. Then:
{{begin-eqn}}
{{eqn | l = z_L \circ \paren {x \circ y}
| r = e_S
| c = {{hypothesis}}
}}
{{eqn | ll= \leadsto
| l = \paren {z_L \c... | Left and Right Inverses of Product | https://proofwiki.org/wiki/Left_and_Right_Inverses_of_Product | https://proofwiki.org/wiki/Left_and_Right_Inverses_of_Product | [
"Monoids"
] | [
"Definition:Monoid",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Inverse (Abstract Algebra)/Left Inverse",
"Definition:Inverse (Abstract Algebra)/Right Inverse",
"Definition:Invertible Element"
] | [
"Definition:Inverse (Abstract Algebra)/Left Inverse",
"Definition:Inverse (Abstract Algebra)/Right Inverse",
"Definition:Associative Operation",
"Definition:Associative Operation",
"Definition:Inverse (Abstract Algebra)/Left Inverse",
"Definition:Inverse (Abstract Algebra)/Right Inverse",
"Left Inverse ... |
proofwiki-534 | Equivalence of Definitions of Self-Inverse | Let $\struct {S, \circ}$ be a monoid whose identity is $e_S$.
Let $x \in S$.
{{TFAE|def = Self-Inverse Element|context = Abstract Algebra}} | Let $x \in S$.
{{begin-eqn}}
{{eqn | l = x \circ x
| r = e_S
| c = {{hypothesis}}
}}
{{eqn | ll= \leadstoandfrom
| l = \paren {x \circ x} \circ x^{-1}
| r = e_S \circ x^{-1}
| c = {{MonoidAxiom|0}}
}}
{{eqn | ll= \leadstoandfrom
| l = x \circ \paren {x \circ x^{-1} }
| r = e_S ... | Let $\struct {S, \circ}$ be a [[Definition:Monoid|monoid]] whose [[Definition:Identity Element|identity]] is $e_S$.
Let $x \in S$.
{{TFAE|def = Self-Inverse Element|context = Abstract Algebra}} | Let $x \in S$.
{{begin-eqn}}
{{eqn | l = x \circ x
| r = e_S
| c = {{hypothesis}}
}}
{{eqn | ll= \leadstoandfrom
| l = \paren {x \circ x} \circ x^{-1}
| r = e_S \circ x^{-1}
| c = {{MonoidAxiom|0}}
}}
{{eqn | ll= \leadstoandfrom
| l = x \circ \paren {x \circ x^{-1} }
| r = e_S... | Equivalence of Definitions of Self-Inverse | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Self-Inverse | https://proofwiki.org/wiki/Equivalence_of_Definitions_of_Self-Inverse | [
"Monoids",
"Inverse Elements"
] | [
"Definition:Monoid",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Category:Monoids",
"Category:Inverse Elements"
] |
proofwiki-535 | Inverse of Identity Element is Itself | Let $\struct {S, \circ}$ be an algebraic structure with an identity element $e$.
Let the inverse of $e$ be $e^{-1}$.
Then:
:$e^{-1} = e$
That is, $e$ is self-inverse. | From Identity Element is Idempotent:
:$e \circ e = e$
Hence the result by definition of identity element.
{{qed}} | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]] with an [[Definition:Identity Element|identity element]] $e$.
Let the [[Definition:Inverse Element|inverse]] of $e$ be $e^{-1}$.
Then:
:$e^{-1} = e$
That is, $e$ is [[Definition:Self-Inverse Element|self-inverse]... | From [[Identity Element is Idempotent]]:
:$e \circ e = e$
Hence the result by definition of [[Definition:Identity Element|identity element]].
{{qed}} | Inverse of Identity Element is Itself | https://proofwiki.org/wiki/Inverse_of_Identity_Element_is_Itself | https://proofwiki.org/wiki/Inverse_of_Identity_Element_is_Itself | [
"Inverse Elements",
"Identity Elements"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Self-Inverse Element"
] | [
"Identity Element is Idempotent",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] |
proofwiki-536 | Invertible Element of Associative Structure is Cancellable | Let $\struct {S, \circ}$ be an algebraic structure where $\circ$ is associative.
Let $\struct {S, \circ}$ have an identity element $e_S$.
An element of $\struct {S, \circ}$ which is invertible is also cancellable. | Let $a \in S$ be invertible.
Suppose $a \circ x = a \circ y$.
Then:
{{begin-eqn}}
{{eqn | l = x
| r = e_S \circ x
| c = {{Defof|Identity Element}}
}}
{{eqn | r = \paren {a^{-1} \circ a} \circ x
| c = {{Defof|Inverse Element}}
}}
{{eqn | r = a^{-1} \circ \paren {a \circ x}
| c = Associativity of ... | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]] where $\circ$ is [[Definition:Associative Operation|associative]].
Let $\struct {S, \circ}$ have an [[Definition:Identity Element|identity element]] $e_S$.
An [[Definition:Element|element]] of $\struct {S, \circ}$... | Let $a \in S$ be [[Definition:Invertible Element|invertible]].
Suppose $a \circ x = a \circ y$.
Then:
{{begin-eqn}}
{{eqn | l = x
| r = e_S \circ x
| c = {{Defof|Identity Element}}
}}
{{eqn | r = \paren {a^{-1} \circ a} \circ x
| c = {{Defof|Inverse Element}}
}}
{{eqn | r = a^{-1} \circ \paren {a ... | Invertible Element of Associative Structure is Cancellable | https://proofwiki.org/wiki/Invertible_Element_of_Associative_Structure_is_Cancellable | https://proofwiki.org/wiki/Invertible_Element_of_Associative_Structure_is_Cancellable | [
"Associativity",
"Inverse Elements",
"Cancellability",
"Invertible Element of Associative Structure is Cancellable"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Associative Operation",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Element",
"Definition:Invertible Element",
"Definition:Cancellable Element"
] | [
"Definition:Invertible Element",
"Definition:Associative Operation",
"Definition:Associative Operation"
] |
proofwiki-537 | Regular Representation of Invertible Element is Permutation | Let $\struct {S, \circ}$ be a monoid.
Let $a \in S$ be invertible.
Then the left regular representation $\lambda_a$ and the right regular representation $\rho_a$ are permutations of $S$. | Suppose $a \in \struct {S, \circ}$ is invertible.
A permutation is a bijection from a set to itself.
As $\lambda_a: S \to S$ and $\rho_a: S \to S$ are defined from $S$ to $S$, all we need to do is show that they are bijections.
To do that we can show that they are both injective and surjective. | Let $\struct {S, \circ}$ be a [[Definition:Monoid|monoid]].
Let $a \in S$ be [[Definition:Invertible Element|invertible]].
Then the [[Definition:Left Regular Representation|left regular representation]] $\lambda_a$ and the [[Definition:Right Regular Representation|right regular representation]] $\rho_a$ are [[Defini... | Suppose $a \in \struct {S, \circ}$ is [[Definition:Invertible Element|invertible]].
A [[Definition:Permutation|permutation]] is a [[Definition:Bijection|bijection]] from a [[Definition:Set|set]] to itself.
As $\lambda_a: S \to S$ and $\rho_a: S \to S$ are defined from $S$ to $S$, all we need to do is show that they a... | Regular Representation of Invertible Element is Permutation | https://proofwiki.org/wiki/Regular_Representation_of_Invertible_Element_is_Permutation | https://proofwiki.org/wiki/Regular_Representation_of_Invertible_Element_is_Permutation | [
"Monoids",
"Regular Representations"
] | [
"Definition:Monoid",
"Definition:Invertible Element",
"Definition:Regular Representations/Left Regular Representation",
"Definition:Regular Representations/Right Regular Representation",
"Definition:Permutation"
] | [
"Definition:Invertible Element",
"Definition:Permutation",
"Definition:Bijection",
"Definition:Set",
"Definition:Bijection",
"Definition:Injection",
"Definition:Surjection",
"Definition:Invertible Element",
"Definition:Injection",
"Definition:Invertible Element",
"Definition:Invertible Element",... |
proofwiki-538 | Group is Inverse Semigroup with Identity | A group is an inverse semigroup with an identity. | Let $\struct {S, \circ}$ be a group.
Let $a \in S$.
Then:
{{begin-eqn}}
{{eqn | l = e
| r = a \circ a^{-1}
| c = {{Group-axiom|3}}
}}
{{eqn | ll = \leadsto
| l = e \circ a
| r = a \circ a^{-1} \circ a
| c =
}}
{{eqn | ll= \leadsto
| l = a
| r = a \circ a^{-1} \circ a
| c ... | A [[Definition:Group|group]] is an [[Definition:Inverse Semigroup|inverse semigroup]] with an [[Definition:Identity (Abstract Algebra)|identity]]. | Let $\struct {S, \circ}$ be a [[Definition:Group|group]].
Let $a \in S$.
Then:
{{begin-eqn}}
{{eqn | l = e
| r = a \circ a^{-1}
| c = {{Group-axiom|3}}
}}
{{eqn | ll = \leadsto
| l = e \circ a
| r = a \circ a^{-1} \circ a
| c =
}}
{{eqn | ll= \leadsto
| l = a
| r = a \circ a... | Group is Inverse Semigroup with Identity | https://proofwiki.org/wiki/Group_is_Inverse_Semigroup_with_Identity | https://proofwiki.org/wiki/Group_is_Inverse_Semigroup_with_Identity | [
"Group Theory"
] | [
"Definition:Group",
"Definition:Inverse Semigroup",
"Definition:Identity (Abstract Algebra)"
] | [
"Definition:Group",
"Definition:Inverse Semigroup",
"Category:Group Theory"
] |
proofwiki-539 | Invertible Elements of Monoid form Subgroup of Cancellable Elements | Let $\struct {S, \circ}$ be an monoid whose identity is $e_S$.
Let $C$ be the set of all cancellable elements of $S$.
Let $T$ be the set of all invertible elements of $S$.
Then $\struct {T, \circ}$ is a subgroup of $\struct {C, \circ}$. | From Cancellable Elements of Monoid form Submonoid, $\struct {C, \circ}$ is a submonoid of $\struct {S, \circ}$.
Let its identity be $e_C$ (which may or may not be the same as $e_S$).
Let $T$ be the set of all invertible elements of $S$.
From Invertible Element of Monoid is Cancellable, all the invertible elements of $... | Let $\struct {S, \circ}$ be an [[Definition:Monoid|monoid]] whose [[Definition:Identity Element|identity]] is $e_S$.
Let $C$ be the [[Definition:Set|set]] of all [[Definition:Cancellable Element|cancellable elements]] of $S$.
Let $T$ be the [[Definition:Set|set]] of all [[Definition:Invertible Element|invertible elem... | From [[Cancellable Elements of Monoid form Submonoid]], $\struct {C, \circ}$ is a [[Definition:Submonoid|submonoid]] of $\struct {S, \circ}$.
Let its [[Definition:Identity Element|identity]] be $e_C$ (which may or may not be the same as $e_S$).
Let $T$ be the [[Definition:Set|set]] of all [[Definition:Invertible Elem... | Invertible Elements of Monoid form Subgroup of Cancellable Elements | https://proofwiki.org/wiki/Invertible_Elements_of_Monoid_form_Subgroup_of_Cancellable_Elements | https://proofwiki.org/wiki/Invertible_Elements_of_Monoid_form_Subgroup_of_Cancellable_Elements | [
"Group Theory",
"Monoids",
"Cancellability"
] | [
"Definition:Monoid",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Set",
"Definition:Cancellable Element",
"Definition:Set",
"Definition:Invertible Element",
"Definition:Subgroup"
] | [
"Cancellable Elements of Monoid form Submonoid",
"Definition:Submonoid",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Set",
"Definition:Invertible Element",
"Invertible Element of Associative Structure is Cancellable/Corollary",
"Definition:Invertible Element",
"Definition:... |
proofwiki-540 | Structure Induced by Associative Operation is Associative | Let $\struct {T, \circ}$ be an algebraic structure.
Let $S$ be a set.
Let $\struct {T^S, \oplus}$ be the structure on $T^S$ induced by $\circ$.
Let $\circ$ be associative.
Then the pointwise operation $\oplus$ induced on $T^S$ by $\circ$ is also associative. | Let $f, g, h \in T^S$.
Let $\struct {T, \circ}$ be an associative algebraic structure.
Then:
{{begin-eqn}}
{{eqn | l = \map {\paren {\paren {f \oplus g} \oplus h} } x
| r = \map {\paren {f \oplus g} } x \circ \map h x
| c = {{Defof|Pointwise Operation}}
}}
{{eqn | r = \paren {\map f x \circ \map g x} \circ ... | Let $\struct {T, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]].
Let $S$ be a [[Definition:Set|set]].
Let $\struct {T^S, \oplus}$ be the [[Definition:Induced Structure|structure on $T^S$ induced]] by $\circ$.
Let $\circ$ be [[Definition:Associative Operation|associative]].
... | Let $f, g, h \in T^S$.
Let $\struct {T, \circ}$ be an [[Definition:Associative Algebraic Structure|associative algebraic structure]].
Then:
{{begin-eqn}}
{{eqn | l = \map {\paren {\paren {f \oplus g} \oplus h} } x
| r = \map {\paren {f \oplus g} } x \circ \map h x
| c = {{Defof|Pointwise Operation}}
}}
{... | Structure Induced by Associative Operation is Associative | https://proofwiki.org/wiki/Structure_Induced_by_Associative_Operation_is_Associative | https://proofwiki.org/wiki/Structure_Induced_by_Associative_Operation_is_Associative | [
"Associativity",
"Pointwise Operations"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Set",
"Definition:Pointwise Operation/Induced Structure",
"Definition:Associative Operation",
"Definition:Pointwise Operation",
"Definition:Associative Operation"
] | [
"Definition:Semigroup",
"Definition:Associative Operation"
] |
proofwiki-541 | Structure Induced by Commutative Operation is Commutative | Let $\struct {T, \circ}$ be an algebraic structure, and let $S$ be a set.
Let $\struct {T^S, \oplus}$ be the structure on $T^S$ induced by $\circ$.
Let $\circ$ be a commutative operation.
Then the pointwise operation $\oplus$ induced on $T^S$ by $\circ$ is also commutative. | Let $\struct {T, \circ}$ be a commutative algebraic structure.
Let $f, g \in T^S$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\paren {f \oplus g} } x
| r = \map f x \circ \map g x
| c = {{Defof|Pointwise Operation}}
}}
{{eqn | r = \map g x \circ \map f x
| c = $\circ$ is commutative
}}
{{eqn | r = \map {\p... | Let $\struct {T, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]], and let $S$ be a [[Definition:Set|set]].
Let $\struct {T^S, \oplus}$ be the [[Definition:Induced Structure|structure on $T^S$ induced]] by $\circ$.
Let $\circ$ be a [[Definition:Commutative Operation|commutative ... | Let $\struct {T, \circ}$ be a [[Definition:Commutative Algebraic Structure|commutative algebraic structure]].
Let $f, g \in T^S$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\paren {f \oplus g} } x
| r = \map f x \circ \map g x
| c = {{Defof|Pointwise Operation}}
}}
{{eqn | r = \map g x \circ \map f x
|... | Structure Induced by Commutative Operation is Commutative | https://proofwiki.org/wiki/Structure_Induced_by_Commutative_Operation_is_Commutative | https://proofwiki.org/wiki/Structure_Induced_by_Commutative_Operation_is_Commutative | [
"Pointwise Operations",
"Commutativity"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Set",
"Definition:Pointwise Operation/Induced Structure",
"Definition:Commutative/Operation",
"Definition:Pointwise Operation",
"Definition:Commutative/Operation"
] | [
"Definition:Commutative/Algebraic Structure",
"Definition:Commutative/Operation"
] |
proofwiki-542 | Induced Structure Identity | Let $\struct {T, \circ}$ be an algebraic structure, and let $S$ be a set.
Let $\struct {T^S, \oplus}$ be the structure on $T^S$ induced by $\circ$.
Let $e$ be an identity for $\circ$.
Then the constant mapping $f_e: S \to T$ defined as:
:$\forall x \in S: \map {f_e} x = e$
is the identity for the pointwise operation $\... | Let $\struct {T, \circ}$ be an algebraic structure with an identity $e$.
Let $f \in T^S$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\paren {f \oplus f_e} } x
| r = \map f x \circ \map {f_e} x
| c = {{Defof|Pointwise Operation}}
}}
{{eqn | r = \map f x \circ e
| c = {{Defof|Constant Mapping}} $f_e$
}}
{{eq... | Let $\struct {T, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]], and let $S$ be a [[Definition:Set|set]].
Let $\struct {T^S, \oplus}$ be the [[Definition:Induced Structure|structure on $T^S$ induced]] by $\circ$.
Let $e$ be an [[Definition:Identity Element|identity]] for $\cir... | Let $\struct {T, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]] with an [[Definition:Identity Element|identity]] $e$.
Let $f \in T^S$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\paren {f \oplus f_e} } x
| r = \map f x \circ \map {f_e} x
| c = {{Defof|Pointwise Operat... | Induced Structure Identity | https://proofwiki.org/wiki/Induced_Structure_Identity | https://proofwiki.org/wiki/Induced_Structure_Identity | [
"Identity Elements",
"Pointwise Operations"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Set",
"Definition:Pointwise Operation/Induced Structure",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Constant Mapping",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Pointwise Operation"
... | [
"Definition:Algebraic Structure/One Operation",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] |
proofwiki-543 | Structure Induced by Abelian Group Operation is Abelian Group | Let $\struct {G, \circ}$ be an abelian group whose identity is $e$.
Let $S$ be a set.
Let $\struct {G^S, \oplus}$ be the structure on $G^S$ induced by $\circ$.
Then $\struct {G^S, \oplus}$ is an abelian group. | From Structure Induced by Group Operation is Group, $\struct {G^S, \oplus}$ is a group.
From Structure Induced by Commutative Operation is Commutative, so is the pointwise operation it induces on $G^S$.
Hence the result.
{{qed}}
Category:Abelian Groups
Category:Pointwise Operations
gffz66j74cdmaqb4seldfypi6jcr7vr | Let $\struct {G, \circ}$ be an [[Definition:Abelian Group|abelian group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $S$ be a [[Definition:Set|set]].
Let $\struct {G^S, \oplus}$ be the [[Definition:Induced Structure|structure on $G^S$ induced]] by $\circ$.
Then $\struct {G^S, \oplus}$ is an [[Defin... | From [[Structure Induced by Group Operation is Group]], $\struct {G^S, \oplus}$ is a [[Definition:Group|group]].
From [[Structure Induced by Commutative Operation is Commutative]], so is the [[Definition:Pointwise Operation|pointwise operation it induces]] on $G^S$.
Hence the result.
{{qed}}
[[Category:Abelian Grou... | Structure Induced by Abelian Group Operation is Abelian Group | https://proofwiki.org/wiki/Structure_Induced_by_Abelian_Group_Operation_is_Abelian_Group | https://proofwiki.org/wiki/Structure_Induced_by_Abelian_Group_Operation_is_Abelian_Group | [
"Abelian Groups",
"Pointwise Operations"
] | [
"Definition:Abelian Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Set",
"Definition:Pointwise Operation/Induced Structure",
"Definition:Abelian Group"
] | [
"Structure Induced by Group Operation is Group",
"Definition:Group",
"Structure Induced by Commutative Operation is Commutative",
"Definition:Pointwise Operation",
"Category:Abelian Groups",
"Category:Pointwise Operations"
] |
proofwiki-544 | Power Set with Union is Commutative Monoid | Let $S$ be a set and let $\powerset S$ be its power set.
Then $\struct {\powerset S, \cup}$ is a commutative monoid whose identity is $\O$.
The only invertible element of this structure is $\O$.
Thus (except in the degenerate case $S = \O$) $\struct {\powerset S, \cup}$ cannot be a group. | From Power Set is Closed under Union:
:$\forall A, B \in \powerset S: A \cup B \in \powerset S$
From Set System Closed under Union is Commutative Semigroup, $\struct {\powerset S, \cup}$ is a commutative semigroup.
From Identity of Power Set with Union, $\O$ acts as the identity of $\struct {\powerset S, \cup}$.
It rem... | Let $S$ be a [[Definition:Set|set]] and let $\powerset S$ be its [[Definition:Power Set|power set]].
Then $\struct {\powerset S, \cup}$ is a [[Definition:Commutative Monoid|commutative monoid]] whose [[Definition:Identity Element|identity]] is $\O$.
The only [[Definition:Invertible Element|invertible element]] of th... | From [[Power Set is Closed under Union]]:
:$\forall A, B \in \powerset S: A \cup B \in \powerset S$
From [[Set System Closed under Union is Commutative Semigroup]], $\struct {\powerset S, \cup}$ is a [[Definition:Commutative Semigroup|commutative semigroup]].
From [[Identity of Power Set with Union]], $\O$ acts as ... | Power Set with Union is Commutative Monoid | https://proofwiki.org/wiki/Power_Set_with_Union_is_Commutative_Monoid | https://proofwiki.org/wiki/Power_Set_with_Union_is_Commutative_Monoid | [
"Examples of Commutative Monoids",
"Set Union",
"Power Set"
] | [
"Definition:Set",
"Definition:Power Set",
"Definition:Commutative Monoid",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Invertible Element",
"Definition:Algebraic Structure",
"Definition:Degenerate Case",
"Definition:Group"
] | [
"Power Set is Closed under Union",
"Set System Closed under Union is Commutative Semigroup",
"Definition:Commutative Semigroup",
"Identity of Power Set with Union",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Inverse (Abstract... |
proofwiki-545 | Power Set with Intersection is Commutative Monoid | Let $S$ be a set and let $\powerset S$ be its power set.
Then $\struct {\powerset S, \cap}$ is a commutative monoid whose identity is $S$.
The only invertible element of this structure is $S$.
Thus (except in the degenerate case $S = \O$) $\struct {\powerset S, \cap}$ cannot be a group. | From Power Set is Closed under Intersection:
:$\forall A, B \in \powerset S: A \cap B \in \powerset S$
From Set System Closed under Intersection is Commutative Semigroup, $\struct {\powerset S, \cap}$ is a commutative semigroup.
From Identity of Power Set with Intersection, we have that $S$ acts as the identity.
It rem... | Let $S$ be a [[Definition:Set|set]] and let $\powerset S$ be its [[Definition:Power Set|power set]].
Then $\struct {\powerset S, \cap}$ is a [[Definition:Commutative Monoid|commutative monoid]] whose [[Definition:Identity Element|identity]] is $S$.
The only [[Definition:Invertible Element|invertible element]] of thi... | From [[Power Set is Closed under Intersection]]:
:$\forall A, B \in \powerset S: A \cap B \in \powerset S$
From [[Set System Closed under Intersection is Commutative Semigroup]], $\struct {\powerset S, \cap}$ is a [[Definition:Commutative Semigroup|commutative semigroup]].
From [[Identity of Power Set with Intersec... | Power Set with Intersection is Commutative Monoid | https://proofwiki.org/wiki/Power_Set_with_Intersection_is_Commutative_Monoid | https://proofwiki.org/wiki/Power_Set_with_Intersection_is_Commutative_Monoid | [
"Examples of Commutative Monoids",
"Power Set",
"Set Intersection"
] | [
"Definition:Set",
"Definition:Power Set",
"Definition:Commutative Monoid",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Invertible Element",
"Definition:Algebraic Structure",
"Definition:Degenerate Case",
"Definition:Group"
] | [
"Power Set is Closed under Intersection",
"Set System Closed under Intersection is Commutative Semigroup",
"Definition:Commutative Semigroup",
"Identity of Power Set with Intersection",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definit... |
proofwiki-546 | Diagonal Relation is Universally Compatible | The diagonal relation $\Delta_S$ on a set $S$ is universally compatible on $S$. | {{questionable|It needs to be pointed out that the definition of universally compatible relation may not be correct as such. I will address it when I work up to that chapter in my book.}}
Let $\struct {S, \circ}$ be any algebraic structure.
{{begin-eqn}}
{{eqn | o =
| r = x_1 \Delta_S x_2 \land y_1 \Delta_S y_2
... | The [[Definition:Diagonal Relation|diagonal relation]] $\Delta_S$ on a [[Definition:Set|set]] $S$ is [[Definition:Universally Compatible Relation|universally compatible]] on $S$. | {{questionable|It needs to be pointed out that the definition of [[Definition:Universally Compatible Relation|universally compatible relation]] may not be correct as such. I will address it when I work up to that chapter in my book.}}
Let $\struct {S, \circ}$ be any [[Definition:Algebraic Structure|algebraic structure... | Diagonal Relation is Universally Compatible | https://proofwiki.org/wiki/Diagonal_Relation_is_Universally_Compatible | https://proofwiki.org/wiki/Diagonal_Relation_is_Universally_Compatible | [
"Compatible Relations"
] | [
"Definition:Diagonal Relation",
"Definition:Set",
"Definition:Universally Compatible Relation"
] | [
"Definition:Universally Compatible Relation",
"Definition:Algebraic Structure",
"Category:Compatible Relations"
] |
proofwiki-547 | Trivial Relation is Universally Congruent | The trivial relation $\RR = S \times S$ on a set $S$ is universally congruent on $S$. | Let $\struct {S, \circ}$ be any algebraic structure which is closed for $\circ$.
By definition of trivial relation:
:$x \in S \land y \in S \implies x \mathrel \RR y$
So:
{{begin-eqn}}
{{eqn | l = x_1, x_2, y_1, y_2
| o = \in
| r = S
| c =
}}
{{eqn | ll= \leadsto
| l = x_1 \circ y_1, x_2 \circ ... | The [[Definition:Trivial Relation|trivial relation]] $\RR = S \times S$ on a [[Definition:Set|set]] $S$ is [[Definition:Universally Congruent|universally congruent]] on $S$. | Let $\struct {S, \circ}$ be any [[Definition:Algebraic Structure|algebraic structure]] which is [[Definition:Closed Algebraic Structure|closed]] for $\circ$.
By definition of [[Definition:Trivial Relation|trivial relation]]:
:$x \in S \land y \in S \implies x \mathrel \RR y$
So:
{{begin-eqn}}
{{eqn | l = x_1, x_2, ... | Trivial Relation is Universally Congruent | https://proofwiki.org/wiki/Trivial_Relation_is_Universally_Congruent | https://proofwiki.org/wiki/Trivial_Relation_is_Universally_Congruent | [
"Examples of Congruence Relations",
"Trivial Relation"
] | [
"Definition:Trivial Relation",
"Definition:Set",
"Definition:Universally Congruent"
] | [
"Definition:Algebraic Structure",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Trivial Relation"
] |
proofwiki-548 | Equivalence Relation is Congruence for Constant Operation | Every equivalence relation is a congruence relation for the constant operation. | Let $c \in S$.
By the definition of the constant operation:
:$\forall x, y \in S: x \sqbrk c y = c$
Let $\RR$ be an equivalence relation on $S$.
Every equivalence relation is reflexive, so:
:$c \mathrel \RR c$
So:
{{begin-eqn}}
{{eqn | l = x_1 \mathrel \RR x_2
| o = \land
| r = y_1 \mathrel \RR y_2
| ... | Every [[Definition:Equivalence Relation|equivalence relation]] is a [[Definition:Congruence Relation|congruence relation]] for the [[Definition:Constant Operation|constant operation]]. | Let $c \in S$.
By the definition of the [[Definition:Constant Operation|constant operation]]:
:$\forall x, y \in S: x \sqbrk c y = c$
Let $\RR$ be an [[Definition:Equivalence Relation|equivalence relation]] on $S$.
Every [[Definition:Equivalence Relation|equivalence relation]] is [[Definition:Reflexive Relation|ref... | Equivalence Relation is Congruence for Constant Operation | https://proofwiki.org/wiki/Equivalence_Relation_is_Congruence_for_Constant_Operation | https://proofwiki.org/wiki/Equivalence_Relation_is_Congruence_for_Constant_Operation | [
"Constant Operation",
"Examples of Equivalence Relations",
"Examples of Congruence Relations"
] | [
"Definition:Equivalence Relation",
"Definition:Congruence Relation",
"Definition:Constant Operation"
] | [
"Definition:Constant Operation",
"Definition:Equivalence Relation",
"Definition:Equivalence Relation",
"Definition:Reflexive Relation",
"True Statement is implied by Every Statement"
] |
proofwiki-549 | Quotient Structure is Well-Defined | Let $\struct {S, \circ}$ be an algebraic structure.
Let $\RR$ be a congruence relation on $\struct {S, \circ}$.
Let $S / \RR$ be the quotient set of $S$ by $\RR$.
Let $\circ_\RR$ be the operation induced on $S / \RR$ by $\circ$.
Then $\circ_\RR$ is a well-defined operation in the quotient structure $\struct {S / \RR, \... | {{begin-eqn}}
{{eqn | l = \eqclass {x_1} \RR = \eqclass {x_2} \RR
| o = \land
| r = \eqclass {y_1} \RR = \eqclass {y_2} \RR
| c =
}}
{{eqn | ll= \leadsto
| l = x_1 \mathop \RR x_2
| o = \land
| r = y_1 \mathop \RR y_2
| c = {{Defof|Equivalence Class}}
}}
{{eqn | ll= \leadsto
... | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]].
Let $\RR$ be a [[Definition:Congruence Relation|congruence relation]] on $\struct {S, \circ}$.
Let $S / \RR$ be the [[Definition:Quotient Set|quotient set of $S$ by $\RR$]].
Let $\circ_\RR$ be the [[Definition:Operation Induced on... | {{begin-eqn}}
{{eqn | l = \eqclass {x_1} \RR = \eqclass {x_2} \RR
| o = \land
| r = \eqclass {y_1} \RR = \eqclass {y_2} \RR
| c =
}}
{{eqn | ll= \leadsto
| l = x_1 \mathop \RR x_2
| o = \land
| r = y_1 \mathop \RR y_2
| c = {{Defof|Equivalence Class}}
}}
{{eqn | ll= \leadsto
... | Quotient Structure is Well-Defined | https://proofwiki.org/wiki/Quotient_Structure_is_Well-Defined | https://proofwiki.org/wiki/Quotient_Structure_is_Well-Defined | [
"Quotient Structures"
] | [
"Definition:Algebraic Structure",
"Definition:Congruence Relation",
"Definition:Quotient Set",
"Definition:Operation Induced on Quotient Set",
"Definition:Well-Defined/Operation",
"Definition:Quotient Structure"
] | [] |
proofwiki-550 | Quotient Structure on Subset Product | Let $\struct {S, \circ}$ be an algebraic structure.
Let $\RR$ be a congruence for $\circ$ on $S$.
Then:
:$\forall X, Y \in S / \RR: X \circ_\PP Y \subseteq X \circ_\RR Y$
where:
:$S / \RR$ is the quotient of $S$ by $\RR$
:$\circ_\PP$ is the operation induced on $\powerset S$ by $\circ$
:$\circ_\RR$ is the operation ind... | By definition of subset product:
:$X \circ_\PP Y = \set {x \circ y: x \in X, y \in Y}$
{{explain|Thus? That is not the definition of $X \circ_\RR Y$}}
Thus:
:$X \circ_\RR Y = \set {x \circ y: x \in X, y \in Y} \cup \set {x \circ y: x \in \eqclass X \RR, y \in \eqclass Y \RR}$
The result follows from Subset of Union.
{{... | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]].
Let $\RR$ be a [[Definition:Congruence Relation|congruence]] for $\circ$ on $S$.
Then:
:$\forall X, Y \in S / \RR: X \circ_\PP Y \subseteq X \circ_\RR Y$
where:
:$S / \RR$ is the [[Definition:Quotient Set|quotient of $S$ by $\RR... | By definition of [[Definition:Subset Product|subset product]]:
:$X \circ_\PP Y = \set {x \circ y: x \in X, y \in Y}$
{{explain|Thus? That is not the definition of $X \circ_\RR Y$}}
Thus:
:$X \circ_\RR Y = \set {x \circ y: x \in X, y \in Y} \cup \set {x \circ y: x \in \eqclass X \RR, y \in \eqclass Y \RR}$
The resul... | Quotient Structure on Subset Product | https://proofwiki.org/wiki/Quotient_Structure_on_Subset_Product | https://proofwiki.org/wiki/Quotient_Structure_on_Subset_Product | [
"Quotient Structures",
"Subset Products"
] | [
"Definition:Algebraic Structure",
"Definition:Congruence Relation",
"Definition:Quotient Set",
"Definition:Subset Product",
"Definition:Operation Induced on Quotient Set"
] | [
"Definition:Subset Product",
"Set is Subset of Union"
] |
proofwiki-551 | External Direct Product Closure | Let $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ be algebraic structures.
Let $\struct {S \times T, \circ}$ be the external direct product of $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.
Then:
:$\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ be closed
{{iff}}:
:$\struct {S \times T, \circ}$ is also closed. | === Sufficient Condition ===
{{:External Direct Product Closure/Sufficient Condition}}{{qed|lemma}} | Let $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ be [[Definition:Algebraic Structure|algebraic structures]].
Let $\struct {S \times T, \circ}$ be the [[Definition:External Direct Product|external direct product]] of $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.
Then:
:$\struct {S, \circ_1}$ and $\struct {T,... | === [[External Direct Product Closure/Sufficient Condition|Sufficient Condition]] ===
{{:External Direct Product Closure/Sufficient Condition}}{{qed|lemma}} | External Direct Product Closure | https://proofwiki.org/wiki/External_Direct_Product_Closure | https://proofwiki.org/wiki/External_Direct_Product_Closure | [
"Closed Algebraic Structures",
"External Direct Products",
"External Direct Product Closure"
] | [
"Definition:Algebraic Structure",
"Definition:External Direct Product",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Closure (Abstract Algebra)/Algebraic Structure"
] | [
"External Direct Product Closure/Sufficient Condition"
] |
proofwiki-552 | External Direct Product Associativity | Let $\struct {S \times T, \circ}$ be the external direct product of the two algebraic structures $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.
Then:
:$\circ_1$ and $\circ_2$ are associative
{{iff}}:
:$\circ$ is associative. | === Sufficient Condition ===
{{:External Direct Product Associativity/Sufficient Condition}}{{qed|lemma}} | Let $\struct {S \times T, \circ}$ be the [[Definition:External Direct Product|external direct product]] of the two [[Definition:Algebraic Structure with One Operation|algebraic structures]] $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.
Then:
:$\circ_1$ and $\circ_2$ are [[Definition:Associative Operation|associa... | === [[External Direct Product Associativity/Sufficient Condition|Sufficient Condition]] ===
{{:External Direct Product Associativity/Sufficient Condition}}{{qed|lemma}} | External Direct Product Associativity | https://proofwiki.org/wiki/External_Direct_Product_Associativity | https://proofwiki.org/wiki/External_Direct_Product_Associativity | [
"External Direct Products",
"Associativity",
"External Direct Product Associativity"
] | [
"Definition:External Direct Product",
"Definition:Algebraic Structure/One Operation",
"Definition:Associative Operation",
"Definition:Associative Operation"
] | [
"External Direct Product Associativity/Sufficient Condition"
] |
proofwiki-553 | External Direct Product of Semigroups | The external direct product of two semigroups is itself a semigroup. | * From External Direct Product Closure, the external direct product of two closed algebraic structures is closed.
* From External Direct Product Associativity, the external direct product of two associative algebraic structures is associative.
The result follows.
{{qed}}
Category:External Direct Products
Category:Semig... | The [[Definition:External Direct Product|external direct product]] of two [[Definition:Semigroup|semigroups]] is itself a [[Definition:Semigroup|semigroup]]. | * From [[External Direct Product Closure]], the [[Definition:External Direct Product|external direct product]] of two closed [[Definition:Algebraic Structure|algebraic structures]] is [[Definition:Closure (Abstract Algebra)|closed]].
* From [[External Direct Product Associativity]], the [[Definition:External Direct Pr... | External Direct Product of Semigroups | https://proofwiki.org/wiki/External_Direct_Product_of_Semigroups | https://proofwiki.org/wiki/External_Direct_Product_of_Semigroups | [
"External Direct Products",
"Semigroups"
] | [
"Definition:External Direct Product",
"Definition:Semigroup",
"Definition:Semigroup"
] | [
"External Direct Product Closure",
"Definition:External Direct Product",
"Definition:Algebraic Structure",
"Definition:Closure (Abstract Algebra)",
"External Direct Product Associativity",
"Definition:External Direct Product",
"Definition:Semigroup",
"Definition:Associative Operation",
"Category:Ext... |
proofwiki-554 | External Direct Product Commutativity | Let $\struct {S \times T, \circ}$ be the external direct product of the two algebraic structures $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.
Then:
:$\circ_1$ and $\circ_2$ are commutative
{{iff}}:
:$\circ$ is commutative. | === Sufficient Condition ===
{{:External Direct Product Commutativity/Sufficient Condition}}{{qed|lemma}} | Let $\struct {S \times T, \circ}$ be the [[Definition:External Direct Product|external direct product]] of the two [[Definition:Algebraic Structure with One Operation|algebraic structures]] $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.
Then:
:$\circ_1$ and $\circ_2$ are [[Definition:Commutative Operation|commutat... | === [[External Direct Product Commutativity/Sufficient Condition|Sufficient Condition]] ===
{{:External Direct Product Commutativity/Sufficient Condition}}{{qed|lemma}} | External Direct Product Commutativity | https://proofwiki.org/wiki/External_Direct_Product_Commutativity | https://proofwiki.org/wiki/External_Direct_Product_Commutativity | [
"External Direct Products",
"Commutativity",
"External Direct Product Commutativity"
] | [
"Definition:External Direct Product",
"Definition:Algebraic Structure/One Operation",
"Definition:Commutative/Operation",
"Definition:Commutative/Operation"
] | [
"External Direct Product Commutativity/Sufficient Condition"
] |
proofwiki-555 | External Direct Product Identity | Let $\struct {S \times T, \circ}$ be the external direct product of two algebraic structures $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.
Then:
:$\struct {S, \circ_1}$ has identity element $e_S$ and $\struct {T, \circ_2}$ has identity element $e_T$
{{iff}}:
:$\tuple {e_S, e_T}$ is the identity element for $\struc... | === Sufficient Condition ===
{{:External Direct Product Identity/Sufficient Condition}}{{qed|lemma}} | Let $\struct {S \times T, \circ}$ be the [[Definition:External Direct Product|external direct product]] of two [[Definition:Algebraic Structure with One Operation|algebraic structures]] $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.
Then:
:$\struct {S, \circ_1}$ has [[Definition:Identity Element|identity element]... | === [[External Direct Product Identity/Sufficient Condition|Sufficient Condition]] ===
{{:External Direct Product Identity/Sufficient Condition}}{{qed|lemma}} | External Direct Product Identity | https://proofwiki.org/wiki/External_Direct_Product_Identity | https://proofwiki.org/wiki/External_Direct_Product_Identity | [
"External Direct Products",
"Identity Elements",
"External Direct Product Identity"
] | [
"Definition:External Direct Product",
"Definition:Algebraic Structure/One Operation",
"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/Sufficient Condition"
] |
proofwiki-556 | External Direct Product Inverses | Let $\struct {S \times T, \circ}$ be the external direct product of the two algebraic structures $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.
Then:
:$s^{-1}$ and $t^{-1}$ are inverse elements of $s \in \struct {S, \circ_1}$ and $t \in \struct {T, \circ_2}$ respectively
{{iff}}:
:$\tuple {s^{-1}, t^{-1} }$ is an i... | === Sufficient Condition ===
{{:External Direct Product Inverses/Sufficient Condition}}{{qed|lemma}} | Let $\struct {S \times T, \circ}$ be the [[Definition:External Direct Product|external direct product]] of the two [[Definition:Algebraic Structure with One Operation|algebraic structures]] $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$.
Then:
:$s^{-1}$ and $t^{-1}$ are [[Definition:Inverse Element|inverse elements... | === [[External Direct Product Inverses/Sufficient Condition|Sufficient Condition]] ===
{{:External Direct Product Inverses/Sufficient Condition}}{{qed|lemma}} | External Direct Product Inverses | https://proofwiki.org/wiki/External_Direct_Product_Inverses | https://proofwiki.org/wiki/External_Direct_Product_Inverses | [
"External Direct Products",
"Inverse Elements",
"External Direct Product Inverses"
] | [
"Definition:External Direct Product",
"Definition:Algebraic Structure/One Operation",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Inverse (Abstract Algebra)/Inverse"
] | [
"External Direct Product Inverses/Sufficient Condition"
] |
proofwiki-557 | Morphism Property Preserves Closure | Let $\phi: \struct {S, \circ_1, \circ_2, \ldots, \circ_n} \to \struct {T, *_1, *_2, \ldots, *_n}$ be a mapping from one algebraic structure $\struct {S, \circ_1, \circ_2, \ldots, \circ_n}$ to another $\struct {T, *_1, *_2, \ldots, *_n}$.
Let $\circ_k$ have the morphism property under $\phi$ for some operation $\circ_k$... | Suppose that $\circ_k$ has the morphism property under $\phi$.
{{ProofWanted|It remains to be shown that the Theorem is true where S is the empty set <br/>The empty case seems considered, doesn't it?<br/>Please feel free to take this debate to the talk page. I also have no idea why we need explicitly to consider the em... | Let $\phi: \struct {S, \circ_1, \circ_2, \ldots, \circ_n} \to \struct {T, *_1, *_2, \ldots, *_n}$ be a [[Definition:Mapping|mapping]] from one [[Definition:Algebraic Structure|algebraic structure]] $\struct {S, \circ_1, \circ_2, \ldots, \circ_n}$ to another $\struct {T, *_1, *_2, \ldots, *_n}$.
Let $\circ_k$ have the ... | Suppose that $\circ_k$ has the [[Definition:Morphism Property|morphism property]] under $\phi$.
{{ProofWanted|It remains to be shown that the Theorem is true where S is the empty set <br/>The empty case seems considered, doesn't it?<br/>Please feel free to take this debate to the talk page. I also have no idea why we ... | Morphism Property Preserves Closure | https://proofwiki.org/wiki/Morphism_Property_Preserves_Closure | https://proofwiki.org/wiki/Morphism_Property_Preserves_Closure | [
"Morphism Property"
] | [
"Definition:Mapping",
"Definition:Algebraic Structure",
"Definition:Morphism Property",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Closure (Abstract Alg... | [
"Definition:Morphism Property",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Non-Empty Set",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Non-Empty Set",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Closure (Abstract Alg... |
proofwiki-558 | Morphism Property Preserves Cancellability | Let:
:$\phi: \struct {S, \circ_1, \circ_2, \ldots, \circ_n} \to \struct {T, *_1, *_2, \ldots, *_n}$
be a mapping from one algebraic structure:
:$\struct {S, \circ_1, \circ_2, \ldots, \circ_n}$
to another:
:$\struct {T, *_1, *_2, \ldots, *_n}$
Let $\circ_k$ have the morphism property under $\phi$ for some operation $\ci... | First let $S$ be the empty set.
It follows from the definition of an Definition:Morphism Property that $\circ_1, \circ_2, \ldots, \circ_n$ are all empty maps.
It also follows from the definition of an Definition:Morphism Property that $*_1, *_2, \ldots, *_n$ are all empty maps.
By Image of Empty Set is Empty Set, $T$ i... | Let:
:$\phi: \struct {S, \circ_1, \circ_2, \ldots, \circ_n} \to \struct {T, *_1, *_2, \ldots, *_n}$
be a [[Definition:Mapping|mapping]] from one [[Definition:Algebraic Structure|algebraic structure]]:
:$\struct {S, \circ_1, \circ_2, \ldots, \circ_n}$
to another:
:$\struct {T, *_1, *_2, \ldots, *_n}$
Let $\circ_k$ have... | First let $S$ be the [[Definition:Empty Set|empty set]].
It follows from the definition of an [[Definition:Morphism Property]] that $\circ_1, \circ_2, \ldots, \circ_n$ are all [[Definition:Empty Mapping|empty maps]].
It also follows from the definition of an [[Definition:Morphism Property]] that $*_1, *_2, \ldots, *_... | Morphism Property Preserves Cancellability | https://proofwiki.org/wiki/Morphism_Property_Preserves_Cancellability | https://proofwiki.org/wiki/Morphism_Property_Preserves_Cancellability | [
"Morphism Property"
] | [
"Definition:Mapping",
"Definition:Algebraic Structure",
"Definition:Morphism Property",
"Definition:Element",
"Definition:Cancellable Element/Left Cancellable",
"Definition:Cancellable Element/Right Cancellable",
"Definition:Cancellable Element/Left Cancellable",
"Definition:Cancellable Element/Right ... | [
"Definition:Empty Set",
"Definition:Morphism Property",
"Definition:Empty Mapping",
"Definition:Morphism Property",
"Definition:Empty Mapping",
"Image of Empty Set is Empty Set",
"Definition:Empty Set",
"Definition:Morphism Property",
"Definition:Cancellable Element",
"Definition:Empty Set",
"De... |
proofwiki-559 | Quotient Mapping on Structure is Epimorphism | Let $\RR$ be a congruence relation on an algebraic structure $\struct {S, \circ}$.
Then the quotient mapping from $\struct {S, \circ}$ to the quotient structure $\struct {S / \RR, \circ_\RR}$ is an epimorphism:
:$q_\RR: \struct {S, \circ} \to \struct {S / \RR, \circ_\RR}: \forall x, y \in S: \map {q_\RR} {x \circ y} = ... | The quotient mapping $q_\RR: S \to S / \RR$ is the canonical surjection from $S$ to $S / \RR$.
Next we show that this is a homomorphism:
{{begin-eqn}}
{{eqn | l = \map {q_\RR} x
| r = \eqclass x \RR
| c = {{Defof|Quotient Mapping}}
}}
{{eqn | l = \map {q_\RR} y
| r = \eqclass y \RR
| c = {{Defof... | Let $\RR$ be a [[Definition:Congruence Relation|congruence relation]] on an [[Definition:Algebraic Structure|algebraic structure]] $\struct {S, \circ}$.
Then the [[Definition:Quotient Mapping|quotient mapping]] from $\struct {S, \circ}$ to the [[Definition:Quotient Structure|quotient structure]] $\struct {S / \RR, \c... | The [[Definition:Quotient Mapping|quotient mapping]] $q_\RR: S \to S / \RR$ is the [[Quotient Mapping is Surjection|canonical surjection]] from $S$ to $S / \RR$.
Next we show that this is a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]]:
{{begin-eqn}}
{{eqn | l = \map {q_\RR} x
| r = \eqclass x \R... | Quotient Mapping on Structure is Epimorphism | https://proofwiki.org/wiki/Quotient_Mapping_on_Structure_is_Epimorphism | https://proofwiki.org/wiki/Quotient_Mapping_on_Structure_is_Epimorphism | [
"Quotient Epimorphisms",
"Quotient Mappings",
"Quotient Structures",
"Epimorphisms (Abstract Algebra)"
] | [
"Definition:Congruence Relation",
"Definition:Algebraic Structure",
"Definition:Quotient Mapping",
"Definition:Quotient Structure",
"Definition:Epimorphism (Abstract Algebra)"
] | [
"Definition:Quotient Mapping",
"Quotient Mapping is Surjection",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Morphism Property",
"Definition:Quotient Mapping",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Surjection",
"Definition:Epimorphism (Abstract Algebra)"
] |
proofwiki-560 | Restriction of Homomorphism to Image is Epimorphism | Let $S$ and $T$ be algebraic structures.
Let $\phi: S \to T$ be a homomorphism.
Then a surjective restriction of $\phi$ can be produced by limiting the codomain of $\phi$ to its image $\Img \phi$. | Let $\phi: S \to T$ be a homomorphism.
Let $\Img \phi = T'$
By Morphism Property Preserves Closure, $T'$ is closed.
From Restriction of Mapping to Image is Surjection, $\phi \to \Img \phi$ is a surjection.
Thus $\phi: S \to T$ is an epimorphism.
Therefore, by suitably restricting the codomain of a homomorphism, it is p... | Let $S$ and $T$ be [[Definition:Algebraic Structure|algebraic structures]].
Let $\phi: S \to T$ be a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]].
Then a [[Definition:Surjective Restriction|surjective restriction]] of $\phi$ can be produced by limiting the [[Definition:Codomain of Mapping|codomain]] o... | Let $\phi: S \to T$ be a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]].
Let $\Img \phi = T'$
By [[Morphism Property Preserves Closure]], $T'$ is [[Definition:Closed Algebraic Structure|closed]].
From [[Restriction of Mapping to Image is Surjection]], $\phi \to \Img \phi$ is a [[Definition:Surjection|su... | Restriction of Homomorphism to Image is Epimorphism | https://proofwiki.org/wiki/Restriction_of_Homomorphism_to_Image_is_Epimorphism | https://proofwiki.org/wiki/Restriction_of_Homomorphism_to_Image_is_Epimorphism | [
"Epimorphisms (Abstract Algebra)"
] | [
"Definition:Algebraic Structure",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Surjective Restriction",
"Definition:Codomain (Set Theory)/Mapping",
"Definition:Image (Set Theory)/Mapping/Mapping"
] | [
"Definition:Homomorphism (Abstract Algebra)",
"Morphism Property Preserves Closure",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Restriction of Mapping to Image is Surjection",
"Definition:Surjection",
"Definition:Epimorphism (Abstract Algebra)",
"Definition:Homomorphism (Abstract Alge... |
proofwiki-561 | Epimorphism Preserves Associativity | Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism.
Let $\circ$ be an associative operation.
Then $*$ is also an associative operation. | {{improve|How necessary really is it to include all this clutter? Is this really the direction we want to go or would the rest of the community be all right with me reversing it out?}}
Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ is associative.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ b... | Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure with One Operation|algebraic structures]].
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]].
Let $\circ$ be an [[Definition:Associative Operation|associative operation]].
... | {{improve|How necessary really is it to include all this clutter? Is this really the direction we want to go or would the rest of the community be all right with me reversing it out?}}
Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]] in which $\circ$ is [[Definit... | Epimorphism Preserves Associativity | https://proofwiki.org/wiki/Epimorphism_Preserves_Associativity | https://proofwiki.org/wiki/Epimorphism_Preserves_Associativity | [
"Epimorphisms (Abstract Algebra)",
"Associativity"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Epimorphism (Abstract Algebra)",
"Definition:Associative Operation",
"Definition:Associative Operation"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Associative Operation",
"Definition:Epimorphism (Abstract Algebra)",
"Definition:Empty Set",
"Definition:Epimorphism (Abstract Algebra)",
"Definition:Surjection",
"Definition:Homomorphism (Abstract Algebra)",
"Empty Mapping to Empty Set is Bi... |
proofwiki-562 | Epimorphism Preserves Semigroups | Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism.
Let $\struct {S, \circ}$ be a semigroup.
Then $\struct {T, *}$ is also a semigroup. | As $\struct {S, \circ}$ is a semigroup, then by definition it is closed.
As $\phi$ is an epimorphism, it is by definition surjective.
That is:
:$T = \phi \sqbrk S$
where $\phi \sqbrk S$ denotes the image of $S$ under $\phi$.
From Morphism Property Preserves Closure it follows that $\struct {T, *}$ is closed.
As $\struc... | Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure with One Operation|algebraic structures]].
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]].
Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]].
Then $\str... | As $\struct {S, \circ}$ is a [[Definition:Semigroup|semigroup]], then by definition it is [[Definition:Closed Algebraic Structure|closed]].
As $\phi$ is an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]], it is by definition [[Definition:Surjection|surjective]].
That is:
:$T = \phi \sqbrk S$
where $\phi \sq... | Epimorphism Preserves Semigroups | https://proofwiki.org/wiki/Epimorphism_Preserves_Semigroups | https://proofwiki.org/wiki/Epimorphism_Preserves_Semigroups | [
"Epimorphisms (Abstract Algebra)",
"Semigroups"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Epimorphism (Abstract Algebra)",
"Definition:Semigroup",
"Definition:Semigroup"
] | [
"Definition:Semigroup",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Epimorphism (Abstract Algebra)",
"Definition:Surjection",
"Definition:Image (Set Theory)/Mapping/Subset",
"Morphism Property Preserves Closure",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
... |
proofwiki-563 | Homomorphism Preserves Subsemigroups | Let $\struct {S, \circ}$ and $\struct {T, *}$ be semigroups.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a homomorphism.
Let $S'$ be a subsemigroup of $S$.
Then $\phi \paren {S'}$ is a subsemigroup of $T$. | By Restriction of Homomorphism to Image is Epimorphism, $\phi$ is an epimorphism onto its image .
Then by Epimorphism Preserves Semigroups, it follows that the image of $S'$ is a semigroup.
The result follows.
{{Qed}}
Category:Homomorphisms (Abstract Algebra)
Category:Subsemigroups
rcr8g5my8td1oh6ytpnzeo82lt644i6 | Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Semigroup|semigroups]].
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]].
Let $S'$ be a [[Definition:Subsemigroup|subsemigroup]] of $S$.
Then $\phi \paren {S'}$ is a [[Definition:Subsemigr... | By [[Restriction of Homomorphism to Image is Epimorphism]], $\phi$ is an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]] onto its [[Definition:Image of Mapping|image]] .
Then by [[Epimorphism Preserves Semigroups]], it follows that the [[Definition:Image of Mapping|image]] of $S'$ is a [[Definition:Semigrou... | Homomorphism Preserves Subsemigroups | https://proofwiki.org/wiki/Homomorphism_Preserves_Subsemigroups | https://proofwiki.org/wiki/Homomorphism_Preserves_Subsemigroups | [
"Homomorphisms (Abstract Algebra)",
"Subsemigroups"
] | [
"Definition:Semigroup",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Subsemigroup",
"Definition:Subsemigroup"
] | [
"Restriction of Homomorphism to Image is Epimorphism",
"Definition:Epimorphism (Abstract Algebra)",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Epimorphism Preserves Semigroups",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Semigroup",
"Category:Homomorphisms (Abstract Algebra)",
... |
proofwiki-564 | Epimorphism Preserves Commutativity | Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism.
Let $\circ$ be a commutative operation.
Then $*$ is also a commutative operation. | {{Proofread| Check validity of proof for the case of $S$ being the empty set}}
Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ is commutative.
Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ is associative.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism.
Supp... | Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure with One Operation|algebraic structures]].
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]].
Let $\circ$ be a [[Definition:Commutative Operation|commutative operation]].
... | {{Proofread| Check validity of proof for the case of $S$ being the empty set}}
Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]] in which $\circ$ is [[Definition:Commutative Operation|commutative]].
Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure w... | Epimorphism Preserves Commutativity | https://proofwiki.org/wiki/Epimorphism_Preserves_Commutativity | https://proofwiki.org/wiki/Epimorphism_Preserves_Commutativity | [
"Epimorphisms (Abstract Algebra)",
"Commutativity"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Epimorphism (Abstract Algebra)",
"Definition:Commutative/Operation",
"Definition:Commutative/Operation"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Commutative/Operation",
"Definition:Algebraic Structure/One Operation",
"Definition:Associative Operation",
"Definition:Epimorphism (Abstract Algebra)",
"Definition:Empty Set",
"Definition:Epimorphism (Abstract Algebra)",
"Definition:Surjecti... |
proofwiki-565 | Epimorphism Preserves Identity | Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism.
Let $\struct {S, \circ}$ have an identity element $e_S$.
Then $\struct {T, *}$ has the identity element $\map \phi {e_S}$. | Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ has an identity $e_S$.
Then:
:$\forall x \in S: x \circ e_S = x = e_S \circ x$
Thus:
:$\forall x \in S: x \circ e_S, e_S \circ x \in \Dom \phi$
Hence:
{{begin-eqn}}
{{eqn | l = \map \phi x
| r = \map \phi {x \circ e_S}
| c = {{Defof|Identit... | Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure with One Operation|algebraic structures]].
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]].
Let $\struct {S, \circ}$ have an [[Definition:Identity Element|identity elemen... | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]] in which $\circ$ has an [[Definition:Identity Element|identity]] $e_S$.
Then:
:$\forall x \in S: x \circ e_S = x = e_S \circ x$
Thus:
:$\forall x \in S: x \circ e_S, e_S \circ x \in \Dom \phi$
Hence:
{{begin-eqn}... | Epimorphism Preserves Identity | https://proofwiki.org/wiki/Epimorphism_Preserves_Identity | https://proofwiki.org/wiki/Epimorphism_Preserves_Identity | [
"Epimorphisms (Abstract Algebra)",
"Identity Elements"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Epimorphism (Abstract Algebra)",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Element"
] |
proofwiki-566 | Epimorphism Preserves Inverses | Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an epimorphism.
Let $\struct {S, \circ}$ have an identity $e_S$.
Let $x^{-1}$ be an inverse element of $x$ for $\circ$.
Then $\map \phi {x^{-1} }$ is an inverse element of $\map \phi x$ for $*$.
T... | Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ has an identity element $e_S$.
From Epimorphism Preserves Identity, it follows that $\struct {T, *}$ also has an identity element, which is $\map \phi {e_S}$.
Let $y$ be an inverse of $x$ in $\struct {S, \circ}$.
By definition of inverse element:
:$x ... | Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure with One Operation|algebraic structures]].
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]].
Let $\struct {S, \circ}$ have an [[Definition:Identity Element|identity]] $e_S... | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]] in which $\circ$ has an [[Definition:Identity Element|identity element]] $e_S$.
From [[Epimorphism Preserves Identity]], it follows that $\struct {T, *}$ also has an [[Definition:Identity Element|identity element]]... | Epimorphism Preserves Inverses | https://proofwiki.org/wiki/Epimorphism_Preserves_Inverses | https://proofwiki.org/wiki/Epimorphism_Preserves_Inverses | [
"Epimorphisms (Abstract Algebra)",
"Inverse Elements"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Epimorphism (Abstract Algebra)",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Inverse (Abstract Algebra)/Inverse"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Epimorphism Preserves Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Inverse (Abstract Algebra)/Inverse",
"D... |
proofwiki-567 | Homomorphism with Cancellable Codomain Preserves Identity | Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a homomorphism.
Let $\struct {S, \circ}$ have an identity $e_S$.
Let $\struct {T, *}$ have an identity $e_T$.
Let every element of $\struct {T, *}$ be cancellable.
Then $\map \phi {e_S}$ is the id... | Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ has an identity $e_S$.
Let $\struct {T, *}$ be an algebraic structure in which $*$ has an identity $e_T$.
Let $\struct {T, *}$ be such that every element is cancellable.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a homomorphism.
Every element... | Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure with One Operation|algebraic structures]].
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]].
Let $\struct {S, \circ}$ have an [[Definition:Identity Element|identity]] $e_... | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]] in which $\circ$ has an [[Definition:Identity Element|identity]] $e_S$.
Let $\struct {T, *}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]] in which $*$ has an [[Definition:Identity... | Homomorphism with Cancellable Codomain Preserves Identity | https://proofwiki.org/wiki/Homomorphism_with_Cancellable_Codomain_Preserves_Identity | https://proofwiki.org/wiki/Homomorphism_with_Cancellable_Codomain_Preserves_Identity | [
"Homomorphisms (Abstract Algebra)",
"Identity Elements"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Cancellable Element",
"Definition:Identity (Abstract Algebra)/Two-Sided Identi... | [
"Definition:Algebraic Structure/One Operation",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Algebraic Structure/One Operation",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Cancellable Element",
"Definition:Homomorphism (Abstract Algebra)",
"Defin... |
proofwiki-568 | Homomorphism with Identity Preserves Inverses | Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a homomorphism.
Let $\struct {S, \circ}$ have an identity $e_S$.
Let $\struct {T, *}$ also have an identity $e_T = \map \phi {e_S}$.
If $x^{-1}$ is an inverse of $x$ for $\circ$, then $\map \phi {... | Let $\struct {S, \circ}$ be an algebraic structure in which $\circ$ has an identity $e_S$.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a homomorphism.
Let $\struct {T, *}$ be an algebraic structure in which $*$ has an identity $e_T = \map \phi {e_S}$.
Let $x^{-1}$ be an inverse of $x$ for $\circ$.
Then by exis... | Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure with One Operation|algebraic structures]].
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]].
Let $\struct {S, \circ}$ have an [[Definition:Identity Element|identity]] $e_... | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]] in which $\circ$ has an [[Definition:Identity Element|identity]] $e_S$.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]].
Let $\struct {T, *}$ be a... | Homomorphism with Identity Preserves Inverses | https://proofwiki.org/wiki/Homomorphism_with_Identity_Preserves_Inverses | https://proofwiki.org/wiki/Homomorphism_with_Identity_Preserves_Inverses | [
"Homomorphisms (Abstract Algebra)",
"Inverse Elements"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Inverse (Abstract Algebra)/Inverse"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Algebraic Structure/One Operation",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Inverse (Abstract Algebra)/Inv... |
proofwiki-569 | Homomorphism to Group Preserves Identity | Let $\struct {S, \circ}$ be an algebraic structure.
Let $\struct {T, *}$ be a group.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a homomorphism.
Let $\struct {S, \circ}$ have an identity $e_S$.
Then:
:$\map \phi {e_S} = e_T$ | By hypothesis, $\struct {T, *}$ is a group.
By the Cancellation Laws, all elements of $T$ are cancellable.
Thus Homomorphism with Cancellable Codomain Preserves Identity can be applied.
{{Qed}} | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]].
Let $\struct {T, *}$ be a [[Definition:Group|group]].
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]].
Let $\struct {S, \circ}$ have an [[Defini... | By hypothesis, $\struct {T, *}$ is a [[Definition:Group|group]].
By the [[Cancellation Laws]], all [[Definition:Element|elements]] of $T$ are [[Definition:Cancellable Element|cancellable]].
Thus [[Homomorphism with Cancellable Codomain Preserves Identity]] can be applied.
{{Qed}} | Homomorphism to Group Preserves Identity | https://proofwiki.org/wiki/Homomorphism_to_Group_Preserves_Identity | https://proofwiki.org/wiki/Homomorphism_to_Group_Preserves_Identity | [
"Homomorphisms (Abstract Algebra)",
"Group Theory"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Group",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Definition:Group",
"Cancellation Laws",
"Definition:Element",
"Definition:Cancellable Element",
"Homomorphism with Cancellable Codomain Preserves Identity"
] |
proofwiki-570 | Homomorphism of External Direct Products | Let:
:$\struct {S_1 \times S_2, \circ}$ be the external direct product of two algebraic structures $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$
:$\struct {T_1 \times T_2, *}$ be the external direct product of two algebraic structures $\struct {T_1, *_1}$ and $\struct {T_2, *_2}$
:$\phi_1$ be a homomorphism fro... | Let $\tuple {x_1, x_2}, \tuple {y_1, y_2} \in S_1 \circ S_2$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\paren {\phi_1 \times \phi_2} } {\tuple {x_1, x_2} \circ \tuple {y_1, y_2} }
| r = \map {\paren {\phi_1 \times \phi_2} } {x_1 \circ_1 y_1, x_2 \circ_2 y_2}
| c =
}}
{{eqn | r = \tuple {\map {\phi_1} {x_1 \ci... | Let:
:$\struct {S_1 \times S_2, \circ}$ be the [[Definition:External Direct Product|external direct product]] of two [[Definition:Algebraic Structure|algebraic structures]] $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$
:$\struct {T_1 \times T_2, *}$ be the [[Definition:External Direct Product|external direct p... | Let $\tuple {x_1, x_2}, \tuple {y_1, y_2} \in S_1 \circ S_2$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\paren {\phi_1 \times \phi_2} } {\tuple {x_1, x_2} \circ \tuple {y_1, y_2} }
| r = \map {\paren {\phi_1 \times \phi_2} } {x_1 \circ_1 y_1, x_2 \circ_2 y_2}
| c =
}}
{{eqn | r = \tuple {\map {\phi_1} {x_1 \... | Homomorphism of External Direct Products | https://proofwiki.org/wiki/Homomorphism_of_External_Direct_Products | https://proofwiki.org/wiki/Homomorphism_of_External_Direct_Products | [
"Homomorphism of External Direct Products",
"Homomorphisms (Abstract Algebra)",
"External Direct Products"
] | [
"Definition:External Direct Product",
"Definition:Algebraic Structure",
"Definition:External Direct Product",
"Definition:Algebraic Structure",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Mapping",
"Definition:Homomorphism (Abstract Algebra)"... | [
"Definition:Morphism Property",
"Category:Homomorphism of External Direct Products",
"Category:Homomorphisms (Abstract Algebra)",
"Category:External Direct Products"
] |
proofwiki-571 | Monomorphism Image is Isomorphic to Domain | The image of a monomorphism is isomorphic to its domain.
That is, if $\phi: S_1 \to S_2$ is a monomorphism, then:
:$\phi: S_1 \to \Img \phi$
is an isomorphism. | Let $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$ be closed algebraic structures.
Let $\phi$ be a monomorphism from $\struct {S_1, \circ_1}$ to $\struct {S_2, \circ_2}$.
Let $T = \Img \phi$ be the image of $\phi$.
By Morphism Property Preserves Closure, $\struct {T, \circ_2}$ is closed.
As $\phi$ is a monomorph... | The [[Definition:Image of Mapping|image]] of a [[Definition:Monomorphism (Abstract Algebra)|monomorphism]] is [[Definition:Isomorphism (Abstract Algebra)|isomorphic]] to its [[Definition:Domain of Mapping|domain]].
That is, if $\phi: S_1 \to S_2$ is a [[Definition:Monomorphism (Abstract Algebra)|monomorphism]], then:... | Let $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$ be [[Definition:Closed Algebraic Structure|closed algebraic structures]].
Let $\phi$ be a [[Definition:Monomorphism (Abstract Algebra)|monomorphism]] from $\struct {S_1, \circ_1}$ to $\struct {S_2, \circ_2}$.
Let $T = \Img \phi$ be the [[Definition:Image of Ma... | Monomorphism Image is Isomorphic to Domain | https://proofwiki.org/wiki/Monomorphism_Image_is_Isomorphic_to_Domain | https://proofwiki.org/wiki/Monomorphism_Image_is_Isomorphic_to_Domain | [
"Monomorphisms (Abstract Algebra)"
] | [
"Definition:Image (Set Theory)/Mapping/Mapping",
"Definition:Monomorphism (Abstract Algebra)",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Domain (Set Theory)/Mapping",
"Definition:Monomorphism (Abstract Algebra)",
"Definition:Isomorphism (Abstract Algebra)"
] | [
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Monomorphism (Abstract Algebra)",
"Definition:Image (Set Theory)/Mapping/Mapping",
"Morphism Property Preserves Closure",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Monomorphism (Abstract Algebra)",
"D... |
proofwiki-572 | Inverse of Algebraic Structure Isomorphism is Isomorphism | Let $\struct {S, \circ}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a mapping.
Then $\phi$ is an isomorphism {{iff}} $\phi^{-1}: \struct {T, *} \to \struct {S, \circ}$ is also an isomorphism. | Let $\phi$ be an isomorphism.
Then by definition $\phi$ is a bijection.
Thus $\exists \phi^{-1}$ such that $\phi^{-1}$ is also a bijection from Bijection iff Inverse is Bijection.
That is:
:$\exists \phi^{-1}: \struct {T, *} \to \struct {S, \circ}$
It follows that:
{{begin-eqn}}
{{eqn | q = \forall s \in S, t \in T
... | Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure with One Operation|algebraic structures]].
Let $\phi: \struct {S, \circ} \to \struct {T, *}$ be a [[Definition:Mapping|mapping]].
Then $\phi$ is an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]] {{iff}} $\phi^{-1}: \struct {... | Let $\phi$ be an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]].
Then by definition $\phi$ is a [[Definition:Bijection|bijection]].
Thus $\exists \phi^{-1}$ such that $\phi^{-1}$ is also a [[Definition:Bijection|bijection]] from [[Bijection iff Inverse is Bijection]].
That is:
:$\exists \phi^{-1}: \struct... | Inverse of Algebraic Structure Isomorphism is Isomorphism | https://proofwiki.org/wiki/Inverse_of_Algebraic_Structure_Isomorphism_is_Isomorphism | https://proofwiki.org/wiki/Inverse_of_Algebraic_Structure_Isomorphism_is_Isomorphism | [
"Inverse of Algebraic Structure Isomorphism is Isomorphism",
"Isomorphisms (Abstract Algebra)"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Mapping",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Isomorphism (Abstract Algebra)"
] | [
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Bijection",
"Definition:Bijection",
"Inverse of Bijection is Bijection",
"Inverse Element of Bijection",
"Inverse Element of Bijection",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Bijection",
"Definition:Isomorphism (Abstract Al... |
proofwiki-573 | Isomorphism is Equivalence Relation | Isomorphism is an equivalence on a set of magmas.
This result applies to all magmas: rings, groups, R-algebraic structures etc. | From the formal definitions:
=== Simple Graph ===
A (simple) graph $G$ is a non-empty set $V$ together with an antireflexive, symmetric relation $\RR$ on $V$.
=== Digraph ===
A digraph $D$ is a non-empty set $V$ together with an antireflexive relation $\RR$ on $V$.
=== Loop-graph ===
A loop-graph $P$ is a non-empty set... | [[Definition:Isomorphism (Abstract Algebra)|Isomorphism]] is an [[Definition:Equivalence Relation|equivalence]] on a [[Definition:Set|set]] of [[Definition:Magma|magmas]].
This result applies to all [[Definition:Magma|magmas]]: [[Definition:Ring (Abstract Algebra)|rings]], [[Definition:Group|groups]], [[Definition:R-... | From the formal definitions:
=== Simple Graph ===
A [[Definition:Simple Graph/Formal Definition|(simple) graph]] $G$ is a [[Definition:Non-Empty Set|non-empty]] [[Definition:Set|set]] $V$ together with an [[Definition:Antireflexive Relation|antireflexive]], [[Definition:Symmetric Relation|symmetric]] [[Definition:Re... | Graph Isomorphism is Equivalence Relation/Proof 1 | https://proofwiki.org/wiki/Isomorphism_is_Equivalence_Relation | https://proofwiki.org/wiki/Graph_Isomorphism_is_Equivalence_Relation/Proof_1 | [
"Isomorphisms (Abstract Algebra)",
"Examples of Equivalence Relations"
] | [
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Equivalence Relation",
"Definition:Set",
"Definition:Magma",
"Definition:Magma",
"Definition:Ring (Abstract Algebra)",
"Definition:Group",
"Definition:R-Algebraic Structure"
] | [
"Definition:Simple Graph/Formal Definition",
"Definition:Non-Empty Set",
"Definition:Set",
"Definition:Antireflexive Relation",
"Definition:Symmetric Relation",
"Definition:Relation",
"Definition:Digraph/Formal Definition",
"Definition:Non-Empty Set",
"Definition:Set",
"Definition:Antireflexive Re... |
proofwiki-574 | Isomorphism is Equivalence Relation | Isomorphism is an equivalence on a set of magmas.
This result applies to all magmas: rings, groups, R-algebraic structures etc. | In the following, let:
{{begin-eqn}}
{{eqn | l = G_1
| r = \struct {\map V {G_1}, \map E {G_1} }
}}
{{eqn | l = G_2
| r = \struct {\map V {G_2}, \map E {G_2} }
}}
{{eqn | l = G_3
| r = \struct {\map V {G_3}, \map E {G_3} }
}}
{{end-eqn}}
be arbitrary graphs.
Checking in turn each of the criteria for e... | [[Definition:Isomorphism (Abstract Algebra)|Isomorphism]] is an [[Definition:Equivalence Relation|equivalence]] on a [[Definition:Set|set]] of [[Definition:Magma|magmas]].
This result applies to all [[Definition:Magma|magmas]]: [[Definition:Ring (Abstract Algebra)|rings]], [[Definition:Group|groups]], [[Definition:R-... | In the following, let:
{{begin-eqn}}
{{eqn | l = G_1
| r = \struct {\map V {G_1}, \map E {G_1} }
}}
{{eqn | l = G_2
| r = \struct {\map V {G_2}, \map E {G_2} }
}}
{{eqn | l = G_3
| r = \struct {\map V {G_3}, \map E {G_3} }
}}
{{end-eqn}}
be arbitrary [[Definition:Graph (Graph Theory)|graphs]].
Chec... | Graph Isomorphism is Equivalence Relation/Proof 2 | https://proofwiki.org/wiki/Isomorphism_is_Equivalence_Relation | https://proofwiki.org/wiki/Graph_Isomorphism_is_Equivalence_Relation/Proof_2 | [
"Isomorphisms (Abstract Algebra)",
"Examples of Equivalence Relations"
] | [
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Equivalence Relation",
"Definition:Set",
"Definition:Magma",
"Definition:Magma",
"Definition:Ring (Abstract Algebra)",
"Definition:Group",
"Definition:R-Algebraic Structure"
] | [
"Definition:Graph (Graph Theory)",
"Definition:Equivalence Relation",
"Definition:Identity Mapping",
"Definition:Bijection",
"Definition:Graph (Graph Theory)/Vertex",
"Definition:Adjacent (Graph Theory)/Vertices",
"Definition:Isomorphism (Graph Theory)",
"Definition:Reflexive Relation",
"Definition:... |
proofwiki-575 | Isomorphism is Equivalence Relation | Isomorphism is an equivalence on a set of magmas.
This result applies to all magmas: rings, groups, R-algebraic structures etc. | To prove a relation is an equivalence, we need to prove it is reflexive, symmetric and transitive.
So, checking in turn each of the criteria for equivalence: | [[Definition:Isomorphism (Abstract Algebra)|Isomorphism]] is an [[Definition:Equivalence Relation|equivalence]] on a [[Definition:Set|set]] of [[Definition:Magma|magmas]].
This result applies to all [[Definition:Magma|magmas]]: [[Definition:Ring (Abstract Algebra)|rings]], [[Definition:Group|groups]], [[Definition:R-... | To prove a [[Definition:Relation|relation]] is an [[Definition:Equivalence Relation|equivalence]], we need to prove it is [[Definition:Reflexive Relation|reflexive]], [[Definition:Symmetric Relation|symmetric]] and [[Definition:Transitive Relation|transitive]].
So, checking in turn each of the criteria for [[Definitio... | Isomorphism is Equivalence Relation | https://proofwiki.org/wiki/Isomorphism_is_Equivalence_Relation | https://proofwiki.org/wiki/Isomorphism_is_Equivalence_Relation | [
"Isomorphisms (Abstract Algebra)",
"Examples of Equivalence Relations"
] | [
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Equivalence Relation",
"Definition:Set",
"Definition:Magma",
"Definition:Magma",
"Definition:Ring (Abstract Algebra)",
"Definition:Group",
"Definition:R-Algebraic Structure"
] | [
"Definition:Relation",
"Definition:Equivalence Relation",
"Definition:Reflexive Relation",
"Definition:Symmetric Relation",
"Definition:Transitive Relation",
"Definition:Equivalence Relation",
"Definition:Reflexive Relation",
"Definition:Symmetric Relation",
"Definition:Transitive Relation",
"Defi... |
proofwiki-576 | Isomorphism is Equivalence Relation | Isomorphism is an equivalence on a set of magmas.
This result applies to all magmas: rings, groups, R-algebraic structures etc. | Let $\struct {S_1, \preccurlyeq_1} \cong \struct {S_2, \preccurlyeq_2}$ denote that $\struct {S_1, \preccurlyeq_1}$ is isomorphic to $\struct {S_2, \preccurlyeq_2}$.
Checking in turn each of the criteria for equivalence:
=== Reflexivity ===
{{:Order Isomorphism is Reflexive}}
Thus $\cong$ is seen to be reflexive.
{{qed... | [[Definition:Isomorphism (Abstract Algebra)|Isomorphism]] is an [[Definition:Equivalence Relation|equivalence]] on a [[Definition:Set|set]] of [[Definition:Magma|magmas]].
This result applies to all [[Definition:Magma|magmas]]: [[Definition:Ring (Abstract Algebra)|rings]], [[Definition:Group|groups]], [[Definition:R-... | Let $\struct {S_1, \preccurlyeq_1} \cong \struct {S_2, \preccurlyeq_2}$ denote that $\struct {S_1, \preccurlyeq_1}$ is [[Definition:Isomorphic Ordered Sets|isomorphic]] to $\struct {S_2, \preccurlyeq_2}$.
Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]:
=== [[Order Isomorph... | Order Isomorphism is Equivalence Relation/Proof 1 | https://proofwiki.org/wiki/Isomorphism_is_Equivalence_Relation | https://proofwiki.org/wiki/Order_Isomorphism_is_Equivalence_Relation/Proof_1 | [
"Isomorphisms (Abstract Algebra)",
"Examples of Equivalence Relations"
] | [
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Equivalence Relation",
"Definition:Set",
"Definition:Magma",
"Definition:Magma",
"Definition:Ring (Abstract Algebra)",
"Definition:Group",
"Definition:R-Algebraic Structure"
] | [
"Definition:Order Isomorphism/Isomorphic Sets",
"Definition:Equivalence Relation",
"Order Isomorphism is Reflexive",
"Definition:Reflexive Relation",
"Order Isomorphism is Symmetric",
"Definition:Symmetric Relation",
"Order Isomorphism is Transitive",
"Definition:Transitive Relation",
"Definition:Re... |
proofwiki-577 | Isomorphism is Equivalence Relation | Isomorphism is an equivalence on a set of magmas.
This result applies to all magmas: rings, groups, R-algebraic structures etc. | An ordered set is a relational structure where order isomorphism is a special case of relation isomorphism.
The result follows directly from Relation Isomorphism is Equivalence Relation.
{{qed}} | [[Definition:Isomorphism (Abstract Algebra)|Isomorphism]] is an [[Definition:Equivalence Relation|equivalence]] on a [[Definition:Set|set]] of [[Definition:Magma|magmas]].
This result applies to all [[Definition:Magma|magmas]]: [[Definition:Ring (Abstract Algebra)|rings]], [[Definition:Group|groups]], [[Definition:R-... | An [[Definition:Ordered Set|ordered set]] is a [[Definition:Relational Structure|relational structure]] where [[Definition:Order Isomorphism|order isomorphism]] is a special case of [[Definition:Relation Isomorphism|relation isomorphism]].
The result follows directly from [[Relation Isomorphism is Equivalence Relation... | Order Isomorphism is Equivalence Relation/Proof 2 | https://proofwiki.org/wiki/Isomorphism_is_Equivalence_Relation | https://proofwiki.org/wiki/Order_Isomorphism_is_Equivalence_Relation/Proof_2 | [
"Isomorphisms (Abstract Algebra)",
"Examples of Equivalence Relations"
] | [
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Equivalence Relation",
"Definition:Set",
"Definition:Magma",
"Definition:Magma",
"Definition:Ring (Abstract Algebra)",
"Definition:Group",
"Definition:R-Algebraic Structure"
] | [
"Definition:Ordered Set",
"Definition:Relational Structure",
"Definition:Order Isomorphism",
"Definition:Relation Isomorphism",
"Relation Isomorphism is Equivalence Relation"
] |
proofwiki-578 | Projection is Epimorphism | Let $\struct {\SS, \circ}$ be the external direct product of the algebraic structures $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$.
Then:
:$\pr_1$ is an epimorphism from $\struct {\SS, \circ}$ to $\struct {S_1, \circ_1}$
:$\pr_2$ is an epimorphism from $\struct {\SS, \circ}$ to $\struct {S_2, \circ_2}$
where $... | From Projection is Surjection, $\pr_1$ and $\pr_2$ are surjections.
We now need to show they are homomorphisms.
Let $s, t \in \struct {\SS, \circ}$ where $s = \tuple {s_1, s_2}$ and $t = \tuple {t_1, t_2$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\pr_1} {s \circ t}
| r = \map {\pr_1} {\tuple {s_1, s_2} \circ \tuple ... | Let $\struct {\SS, \circ}$ be the [[Definition:External Direct Product|external direct product]] of the [[Definition:Algebraic Structure with One Operation|algebraic structures]] $\struct {S_1, \circ_1}$ and $\struct {S_2, \circ_2}$.
Then:
:$\pr_1$ is an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]] from ... | From [[Projection is Surjection]], $\pr_1$ and $\pr_2$ are [[Definition:Surjection|surjections]].
We now need to show they are [[Definition:Homomorphism (Abstract Algebra)|homomorphisms]].
Let $s, t \in \struct {\SS, \circ}$ where $s = \tuple {s_1, s_2}$ and $t = \tuple {t_1, t_2$.
Then:
{{begin-eqn}}
{{eqn | l = \... | Projection is Epimorphism | https://proofwiki.org/wiki/Projection_is_Epimorphism | https://proofwiki.org/wiki/Projection_is_Epimorphism | [
"Projection is Epimorphism",
"Epimorphisms (Abstract Algebra)",
"Projections"
] | [
"Definition:External Direct Product",
"Definition:Algebraic Structure/One Operation",
"Definition:Epimorphism (Abstract Algebra)",
"Definition:Epimorphism (Abstract Algebra)",
"Definition:Projection (Mapping Theory)/First Projection",
"Definition:Projection (Mapping Theory)/Second Projection"
] | [
"Projection is Surjection",
"Definition:Surjection",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Morphism Property"
] |
proofwiki-579 | Quotient Theorem for Epimorphisms | Let $\struct {S, \oplus}$ and $\struct {T, *}$ be algebraic structures.
Let $\phi: \struct {S, \oplus} \to \struct {T, *}$ be an epimorphism.
Let $\RR_\phi$ be the equivalence induced by $\phi$.
Let $S / \RR_\phi$ be the quotient of $S$ by $\RR_\phi$.
Let $q_{\RR_\phi}: S \to S / \RR_\phi$ be the quotient mapping induc... | === Proof of Congruence Relation ===
{{:Equivalence Induced by Epimorphism is Congruence Relation}}{{qed|lemma}} | Let $\struct {S, \oplus}$ and $\struct {T, *}$ be [[Definition:Algebraic Structure with One Operation|algebraic structures]].
Let $\phi: \struct {S, \oplus} \to \struct {T, *}$ be an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]].
Let $\RR_\phi$ be the [[Definition:Equivalence Relation Induced by Mapping|e... | === [[Equivalence Induced by Epimorphism is Congruence Relation|Proof of Congruence Relation]] ===
{{:Equivalence Induced by Epimorphism is Congruence Relation}}{{qed|lemma}} | Quotient Theorem for Epimorphisms | https://proofwiki.org/wiki/Quotient_Theorem_for_Epimorphisms | https://proofwiki.org/wiki/Quotient_Theorem_for_Epimorphisms | [
"Isomorphisms (Abstract Algebra)",
"Quotient Epimorphisms",
"Named Theorems",
"Quotient Theorems"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Epimorphism (Abstract Algebra)",
"Definition:Equivalence Relation Induced by Mapping",
"Definition:Quotient Set",
"Definition:Quotient Mapping",
"Definition:Quotient Structure",
"Definition:Equivalence Relation Induced by Mapping",
"Definitio... | [
"Equivalence Induced by Epimorphism is Congruence Relation"
] |
proofwiki-580 | Homomorphism on Induced Structure to Commutative Semigroup | Let $\struct {S, \circ}$ be an algebraic structure.
Let $\struct {T, \oplus}$ be a commutative semigroup.
Let $T^S$ be the set of all mappings from $S$ to $T$.
Let $f$ and $g$ be homomorphisms from $S$ into $T$.
Let $f \oplus' g$ be the pointwise operation on $T^S$ induced by $\oplus$.
Then $f \oplus' g$ is a homomorph... | Let $\struct {T, \oplus}$ be a commutative semigroup.
Let $x, y \in S$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\paren {f \oplus' g} } {x \circ y}
| r = \map f {x \circ y} \oplus \map g {x \circ y}
| c = {{Defof|Pointwise Operation}}
}}
{{eqn | r = \paren {\map f x \oplus \map f y} \oplus \paren {\map g x \op... | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]].
Let $\struct {T, \oplus}$ be a [[Definition:Commutative Semigroup|commutative semigroup]].
Let $T^S$ be the [[Definition:Set of All Mappings|set of all mappings]] from $S$ to $T$.
Let $f$ and $g$ be [[Definition... | Let $\struct {T, \oplus}$ be a [[Definition:Commutative Semigroup|commutative semigroup]].
Let $x, y \in S$.
Then:
{{begin-eqn}}
{{eqn | l = \map {\paren {f \oplus' g} } {x \circ y}
| r = \map f {x \circ y} \oplus \map g {x \circ y}
| c = {{Defof|Pointwise Operation}}
}}
{{eqn | r = \paren {\map f x \opl... | Homomorphism on Induced Structure to Commutative Semigroup | https://proofwiki.org/wiki/Homomorphism_on_Induced_Structure_to_Commutative_Semigroup | https://proofwiki.org/wiki/Homomorphism_on_Induced_Structure_to_Commutative_Semigroup | [
"Homomorphisms (Abstract Algebra)",
"Pointwise Operations",
"Commutative Semigroups"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Commutative Semigroup",
"Definition:Set of All Mappings",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Pointwise Operation",
"Definition:Homomorphism (Abstract Algebra)"
] | [
"Definition:Commutative Semigroup",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Commutative/Operation",
"Definition:Associative Operation",
"Definition:Associative Operation",
"Definition:Commutative/Operation"
] |
proofwiki-581 | Inverse Mapping in Induced Structure of Homomorphism to Abelian Group | Let $\struct {S, \circ}$ be an algebraic structure.
Let $\struct {T, \oplus}$ be an abelian group.
Let $f$ be a homomorphism from $S$ into $T$.
Let $f^*$ be the pointwise inverse of $f$.
Then $f^*$ is a homomorphism from $\struct {S, \circ}$ into $\struct {T, \oplus}$. | Let $\struct {T, \oplus}$ be an abelian group.
Let $x, y \in S$.
Then:
{{begin-eqn}}
{{eqn | l = \map {f^*} {x \circ y}
| r = \paren {\map f {x \circ y} }^{-1}
| c = {{Defof|Pointwise Inverse}}
}}
{{eqn | r = \paren {\map f x \oplus \map f y}^{-1}
| c = $f$ is a homomorphism
}}
{{eqn | r = \paren {\ma... | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]].
Let $\struct {T, \oplus}$ be an [[Definition:Abelian Group|abelian group]].
Let $f$ be a [[Definition:Homomorphism (Abstract Algebra)|homomorphism]] from $S$ into $T$.
Let $f^*$ be the [[Definition:Pointwise Inv... | Let $\struct {T, \oplus}$ be an [[Definition:Abelian Group|abelian group]].
Let $x, y \in S$.
Then:
{{begin-eqn}}
{{eqn | l = \map {f^*} {x \circ y}
| r = \paren {\map f {x \circ y} }^{-1}
| c = {{Defof|Pointwise Inverse}}
}}
{{eqn | r = \paren {\map f x \oplus \map f y}^{-1}
| c = $f$ is a [[Defin... | Inverse Mapping in Induced Structure of Homomorphism to Abelian Group | https://proofwiki.org/wiki/Inverse_Mapping_in_Induced_Structure_of_Homomorphism_to_Abelian_Group | https://proofwiki.org/wiki/Inverse_Mapping_in_Induced_Structure_of_Homomorphism_to_Abelian_Group | [
"Homomorphisms (Abstract Algebra)",
"Abelian Groups",
"Pointwise Operations"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Abelian Group",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Pointwise Inverse",
"Definition:Homomorphism (Abstract Algebra)"
] | [
"Definition:Abelian Group",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Commutative/Operation",
"Inverse of Group Product"
] |
proofwiki-582 | Set of Homomorphisms to Abelian Group is Subgroup of All Mappings | Let $\struct {S, \circ}$ be an algebraic structure.
Let $\struct {T, \oplus}$ be an abelian group.
Let $\struct {T^S, \oplus}$ be the algebraic structure on $T^S$ induced by $\oplus$.
Then the set of all homomorphisms from $\struct {S, \circ}$ into $\struct {T, \oplus}$ is a subgroup of $\struct {T^S, \oplus}$. | Let $H$ be the set of all homomorphisms from $\struct {S, \circ}$ into $\struct {T, \oplus}$. | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure|algebraic structure]].
Let $\struct {T, \oplus}$ be an [[Definition:Abelian Group|abelian group]].
Let $\struct {T^S, \oplus}$ be the [[Definition:Induced Structure|algebraic structure on $T^S$ induced by $\oplus$]].
Then the set of all [[Definition:Ho... | Let $H$ be the set of all [[Definition:Homomorphism (Abstract Algebra)|homomorphisms]] from $\struct {S, \circ}$ into $\struct {T, \oplus}$. | Set of Homomorphisms to Abelian Group is Subgroup of All Mappings | https://proofwiki.org/wiki/Set_of_Homomorphisms_to_Abelian_Group_is_Subgroup_of_All_Mappings | https://proofwiki.org/wiki/Set_of_Homomorphisms_to_Abelian_Group_is_Subgroup_of_All_Mappings | [
"Homomorphisms (Abstract Algebra)",
"Subgroups",
"Abelian Groups"
] | [
"Definition:Algebraic Structure",
"Definition:Abelian Group",
"Definition:Pointwise Operation/Induced Structure",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Subgroup"
] | [
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Homomorphism (Abstract Algebra)"
] |
proofwiki-583 | Transplanting Theorem | Let $\struct {S, \circ}$ be an algebraic structure.
Let $f: S \to T$ be a bijection.
Then there exists one and only one operation $\oplus$ such that $f: \struct {S, \circ} \to \struct {T, \oplus}$ is an isomorphism.
The operation $\oplus$ is defined by:
:$\forall x, y \in T: x \oplus y = \map f {\map {f^{-1} } x \circ ... | === Existence ===
To show that $\oplus$ as defined above exists:
Let $u, v \in S$, and let $x = \map f u, y = \map f v$.
Then as $f$ is a bijection, $u = \map {f^{-1} } x, v = \map {f^{-1} } y$.
Thus:
{{begin-eqn}}
{{eqn | l = \map f {u \circ v}
| r = \map f {\map {f^{-1} } x \circ \map {f^{-1} } y}
| c = D... | Let $\struct {S, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]].
Let $f: S \to T$ be a [[Definition:Bijection|bijection]].
Then there exists [[Definition:Unique|one and only one]] [[Definition:Binary Operation|operation]] $\oplus$ such that $f: \struct {S, \circ} \to \struct ... | === Existence ===
To show that $\oplus$ as defined above exists:
Let $u, v \in S$, and let $x = \map f u, y = \map f v$.
Then as $f$ is a [[Definition:Bijection|bijection]], $u = \map {f^{-1} } x, v = \map {f^{-1} } y$.
Thus:
{{begin-eqn}}
{{eqn | l = \map f {u \circ v}
| r = \map f {\map {f^{-1} } x \circ \... | Transplanting Theorem | https://proofwiki.org/wiki/Transplanting_Theorem | https://proofwiki.org/wiki/Transplanting_Theorem | [
"Transplanting Theorem",
"Isomorphisms (Abstract Algebra)",
"Transplants",
"Named Theorems"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Bijection",
"Definition:Unique",
"Definition:Operation/Binary Operation",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Operation/Binary Operation"
] | [
"Definition:Bijection",
"Inverse of Inverse of Bijection",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Isomorphism (Abstract Algebra)"
] |
proofwiki-584 | Exists Bijection to a Disjoint Set | Let $S$ and $T$ be sets.
Then there exists a bijection from $T$ onto a set $T'$ disjoint from $S$. | Consider the set:
:$X = \text{Ran}(S) = \set {y: \paren {\exists x : \tuple {x, y} \in S} }$
From Relation is Set implies Domain and Image are Sets, this is indeed a set.
That is, $X$ consists of all the elements which are the second coordinate of some ordered pair which happens to be in $S$.
From Exists Element Not i... | Let $S$ and $T$ be [[Definition:Set|sets]].
Then there exists a [[Definition:Bijection|bijection]] from $T$ onto a set $T'$ [[Definition:Disjoint Sets|disjoint]] from $S$. | Consider the [[Definition:Set|set]]:
:$X = \text{Ran}(S) = \set {y: \paren {\exists x : \tuple {x, y} \in S} }$
From [[Relation is Set implies Domain and Image are Sets]], this is indeed a [[Definition:Set|set]].
That is, $X$ consists of all the [[Definition:Element|elements]] which are the [[Definition:Coordinate|... | Exists Bijection to a Disjoint Set | https://proofwiki.org/wiki/Exists_Bijection_to_a_Disjoint_Set | https://proofwiki.org/wiki/Exists_Bijection_to_a_Disjoint_Set | [
"Bijections",
"Disjoint Sets"
] | [
"Definition:Set",
"Definition:Bijection",
"Definition:Disjoint Sets"
] | [
"Definition:Set",
"Relation is Set implies Domain and Image are Sets",
"Definition:Set",
"Definition:Element",
"Definition:Coordinate",
"Definition:Ordered Pair",
"Exists Element Not in Set",
"Definition:Cartesian Product",
"Definition:Bijection"
] |
proofwiki-585 | Embedding Theorem | Let:
:$(1): \quad \struct {T_2, \oplus_2}$ be a submagma of $\struct {S_2, *_2}$
:$(2): \quad f: \struct {T_1, \oplus_1} \to \struct {T_2, \oplus_2}$ be an isomorphism
then there exists:
:$(1): \quad$ a magma $\struct {S_1, *_1}$ which algebraically contains $\struct {T_1, \oplus_1}$
:$(2): \quad g: \struct {S_1, *_1} ... | There are two cases to consider: when $T_1$ and $S_2$ are disjoint, and when they are not.
Suppose $T_1$ and $S_2$ are disjoint.
Let $S_1$ be the set $T_1 \cup \paren {S_2 \setminus T_2}$.
Then we can define the mapping $h: S_2 \to S_1$ as:
:$\forall x \in S_2: \map h x = \begin{cases}
x & : x \in S_2 \setminus T_2 \\
... | Let:
:$(1): \quad \struct {T_2, \oplus_2}$ be a [[Definition:Submagma|submagma]] of $\struct {S_2, *_2}$
:$(2): \quad f: \struct {T_1, \oplus_1} \to \struct {T_2, \oplus_2}$ be an [[Definition:Isomorphism (Abstract Algebra)|isomorphism]]
then there exists:
:$(1): \quad$ a [[Definition:Magma|magma]] $\struct {S_1, *_1}... | There are two cases to consider: when $T_1$ and $S_2$ are [[Definition:Disjoint Sets|disjoint]], and when they are not.
Suppose $T_1$ and $S_2$ are [[Definition:Disjoint Sets|disjoint]].
Let $S_1$ be the [[Definition:Set|set]] $T_1 \cup \paren {S_2 \setminus T_2}$.
Then we can define the [[Definition:Mapping|mappin... | Embedding Theorem | https://proofwiki.org/wiki/Embedding_Theorem | https://proofwiki.org/wiki/Embedding_Theorem | [
"Embedding Theorem",
"Isomorphisms (Abstract Algebra)",
"Named Theorems"
] | [
"Definition:Submagma",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Magma",
"Definition:Submagma",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Extension of Mapping"
] | [
"Definition:Disjoint Sets",
"Definition:Disjoint Sets",
"Definition:Set",
"Definition:Mapping",
"Definition:Disjoint Sets",
"Definition:Bijection",
"Definition:Submagma",
"Definition:Operation/Binary Operation",
"Definition:Transplant (Abstract Algebra)",
"Transplanting Theorem",
"Transplanting ... |
proofwiki-586 | Quotient Structure of Inverse Completion | Let $\struct {T, \circ'}$ be an inverse completion of a commutative semigroup $\struct {S, \circ}$, where $C$ is the set of cancellable elements of $S$.
Let $f: S \times C: T$ be the mapping defined as:
:$\forall x \in S, y \in C: \map f {x, y} = x \circ' y^{-1}$
Then the mapping $g: \paren {S \times C} / \RR_f \to T$ ... | $T$ is commutative, from Inverse Completion is Commutative Semigroup.
The mapping $\map f {x, y} = x \circ' y^{-1}$ is an epimorphism from the cartesian product of $\struct {S, \circ}$ and $\struct {C, \circ \restriction_C}$ onto $\struct {T, \circ'}$.
By Quotient Theorem for Epimorphisms, the proof follows.
{{Explain|... | Let $\struct {T, \circ'}$ be an [[Definition:Inverse Completion|inverse completion]] of a [[Definition:Commutative Semigroup|commutative semigroup]] $\struct {S, \circ}$, where $C$ is the set of [[Definition:Cancellable Element|cancellable elements]] of $S$.
Let $f: S \times C: T$ be the [[Definition:Mapping|mapping]]... | $T$ is [[Definition:Commutative Semigroup|commutative]], from [[Inverse Completion is Commutative Semigroup]].
The mapping $\map f {x, y} = x \circ' y^{-1}$ is an [[Definition:Epimorphism (Abstract Algebra)|epimorphism]] from the [[Definition:Cartesian Product|cartesian product]] of $\struct {S, \circ}$ and $\struct {... | Quotient Structure of Inverse Completion | https://proofwiki.org/wiki/Quotient_Structure_of_Inverse_Completion | https://proofwiki.org/wiki/Quotient_Structure_of_Inverse_Completion | [
"Inverse Completions",
"Quotient Structures"
] | [
"Definition:Inverse Completion",
"Definition:Commutative Semigroup",
"Definition:Cancellable Element",
"Definition:Mapping",
"Definition:Quotient Structure",
"Definition:Isomorphism (Abstract Algebra)"
] | [
"Definition:Commutative Semigroup",
"Inverse Completion is Commutative Semigroup",
"Definition:Epimorphism (Abstract Algebra)",
"Definition:Cartesian Product",
"Quotient Theorem for Epimorphisms",
"Definition:Epimorphism (Abstract Algebra)",
"Definition:Cartesian Product",
"Category:Inverse Completion... |
proofwiki-587 | Quotient Mapping of Inverse Completion | Let $\struct {T, \circ'}$ be an inverse completion of a commutative semigroup $\struct {S, \circ}$, where $C$ is the set of cancellable elements of $S$.
Let $f: S \times C: T$ be the mapping defined as:
:$\forall x \in S, y \in C: \map f {x, y} = x \circ' y^{-1}$
Let $\RR_f$ be the equivalence relation induced by $f$.
... | By the definition of $\RR_f$:
:$\tuple {x_1, y_1} \mathrel {\RR_f} \tuple {x_2, y_2} \iff x_1 \circ' y_1^{-1} = x_2 \circ' y_2^{-1}$
Now:
{{begin-eqn}}
{{eqn | l = x_1 \circ' y_1^{-1}
| r = x_2 \circ' y_2^{-1}
| c =
}}
{{eqn | ll= \leadsto
| l = x_1 \circ' y_1^{-1} \circ' y_1 \circ' y_2
| r = x... | Let $\struct {T, \circ'}$ be an [[Definition:Inverse Completion|inverse completion]] of a [[Definition:Commutative Semigroup|commutative semigroup]] $\struct {S, \circ}$, where $C$ is the set of [[Definition:Cancellable Element|cancellable elements]] of $S$.
Let $f: S \times C: T$ be the [[Definition:Mapping|mapping]]... | By the definition of $\RR_f$:
:$\tuple {x_1, y_1} \mathrel {\RR_f} \tuple {x_2, y_2} \iff x_1 \circ' y_1^{-1} = x_2 \circ' y_2^{-1}$
Now:
{{begin-eqn}}
{{eqn | l = x_1 \circ' y_1^{-1}
| r = x_2 \circ' y_2^{-1}
| c =
}}
{{eqn | ll= \leadsto
| l = x_1 \circ' y_1^{-1} \circ' y_1 \circ' y_2
| r... | Quotient Mapping of Inverse Completion | https://proofwiki.org/wiki/Quotient_Mapping_of_Inverse_Completion | https://proofwiki.org/wiki/Quotient_Mapping_of_Inverse_Completion | [
"Inverse Completions"
] | [
"Definition:Inverse Completion",
"Definition:Commutative Semigroup",
"Definition:Cancellable Element",
"Definition:Mapping",
"Definition:Equivalence Relation Induced by Mapping"
] | [
"Definition:Cancellable Element",
"Definition:Commutative/Operation",
"Definition:Extension of Operation",
"Category:Inverse Completions"
] |
proofwiki-588 | Subset Product defining Inverse Completion of Commutative Semigroup is Commutative Semigroup | Let $\struct {S, \circ}$ be a commutative semigroup.
Let $\struct {C, \circ} \subseteq \struct {S, \circ}$ be the subsemigroup of cancellable elements of $\struct {S, \circ}$.
Let $\struct {T, \circ'}$ be an inverse completion of $\struct {S, \circ}$.
Then:
:$S \circ' C^{-1}$ is a commutative semigroup
where $S \circ' ... | Note that by definition of inverse completion, $\struct {T, \circ'}$ is a semigroup.
Thus $\circ'$ is associative.
First it is demonstrated that $S \circ' C^{-1}$ is a semigroup.
Let $x, z \in S$.
Let $y, w \in C$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {x \circ' y^{-1} } \circ' \paren {z \circ' w^{-1} }
| r = x ... | Let $\struct {S, \circ}$ be a [[Definition:Commutative Semigroup|commutative semigroup]].
Let $\struct {C, \circ} \subseteq \struct {S, \circ}$ be the [[Definition:Subsemigroup|subsemigroup]] of [[Definition:Cancellable Element|cancellable elements]] of $\struct {S, \circ}$.
Let $\struct {T, \circ'}$ be an [[Definiti... | Note that by definition of [[Definition:Inverse Completion|inverse completion]], $\struct {T, \circ'}$ is a [[Definition:Semigroup|semigroup]].
Thus $\circ'$ is [[Definition:Associative Operation|associative]].
First it is demonstrated that $S \circ' C^{-1}$ is a [[Definition:Semigroup|semigroup]].
Let $x, z \in S$... | Subset Product defining Inverse Completion of Commutative Semigroup is Commutative Semigroup | https://proofwiki.org/wiki/Subset_Product_defining_Inverse_Completion_of_Commutative_Semigroup_is_Commutative_Semigroup | https://proofwiki.org/wiki/Subset_Product_defining_Inverse_Completion_of_Commutative_Semigroup_is_Commutative_Semigroup | [
"Inverse Completions"
] | [
"Definition:Commutative Semigroup",
"Definition:Subsemigroup",
"Definition:Cancellable Element",
"Definition:Inverse Completion",
"Definition:Commutative Semigroup",
"Definition:Subset Product"
] | [
"Definition:Inverse Completion",
"Definition:Semigroup",
"Definition:Associative Operation",
"Definition:Semigroup",
"Definition:Associative Operation",
"Commutation with Inverse in Monoid",
"Definition:Associative Operation",
"Inverse of Product/Monoid",
"Definition:Extension of Operation",
"Defi... |
proofwiki-589 | Inverse Completion of Commutative Semigroup is Inverse Completion of Itself | Let $\struct {S, \circ}$ be a commutative semigroup.
Let $\struct {C, \circ} \subseteq \struct {S, \circ}$ be the subsemigroup of cancellable elements of $\struct {S, \circ}$.
Let $\struct {T, \circ'}$ be an inverse completion of $\struct {S, \circ}$.
Then $\struct {T, \circ'}$ is its own inverse completion. | Let $x \circ' y^{-1}$ be cancellable for $\circ'$, where $x \in S$ and $y \in C$.
We have that $y$ is invertible for $\circ'$.
So by Invertible Element of Associative Structure is Cancellable:
:$y$ is cancellable for $\circ'$.
Now by definition of inverse element:
:$x = \paren {x \circ' y^{-1} } \circ' y$
Thus $x$ is a... | Let $\struct {S, \circ}$ be a [[Definition:Commutative Semigroup|commutative semigroup]].
Let $\struct {C, \circ} \subseteq \struct {S, \circ}$ be the [[Definition:Subsemigroup|subsemigroup]] of [[Definition:Cancellable Element|cancellable]] elements of $\struct {S, \circ}$.
Let $\struct {T, \circ'}$ be an [[Definiti... | Let $x \circ' y^{-1}$ be [[Definition:Cancellable Element|cancellable]] for $\circ'$, where $x \in S$ and $y \in C$.
We have that $y$ is [[Definition:Invertible Element|invertible]] for $\circ'$.
So by [[Invertible Element of Associative Structure is Cancellable]]:
:$y$ is [[Definition:Cancellable Element|cancellable... | Inverse Completion of Commutative Semigroup is Inverse Completion of Itself | https://proofwiki.org/wiki/Inverse_Completion_of_Commutative_Semigroup_is_Inverse_Completion_of_Itself | https://proofwiki.org/wiki/Inverse_Completion_of_Commutative_Semigroup_is_Inverse_Completion_of_Itself | [
"Inverse Completions"
] | [
"Definition:Commutative Semigroup",
"Definition:Subsemigroup",
"Definition:Cancellable Element",
"Definition:Inverse Completion",
"Definition:Inverse Completion"
] | [
"Definition:Cancellable Element",
"Definition:Invertible Element",
"Invertible Element of Associative Structure is Cancellable",
"Definition:Cancellable Element",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Cancellable Element",
"Cancellable Elements of Semigroup form Subsemigroup",
"... |
proofwiki-590 | Identity of Inverse Completion of Commutative Monoid | Let $\struct {S, \circ}$ be a commutative monoid whose identity is $e$.
Let $\struct {C, \circ} \subseteq \struct {S, \circ}$ be the subsemigroup of cancellable elements of $\struct {S, \circ}$.
Let $\struct {T, \circ'}$ be an inverse completion of $\struct {S, \circ}$.
Then $e \in T$ is the identity for $\circ'$. | Let $e$ be the identity for $\circ$.
Let $e = x \circ' y^{-1}$, where $x \in S, y \in C$.
Then:
{{begin-eqn}}
{{eqn | l = y
| r = e \circ y
| c = {{Defof|Identity Element}}
}}
{{eqn | r = \paren {x \circ' y^{-1} } \circ' y
| c = {{Defof|Inverse Element}}
}}
{{eqn | r = x \circ' \paren {y^{-1} \circ' y... | Let $\struct {S, \circ}$ be a [[Definition:Commutative Monoid|commutative monoid]] whose [[Definition:Identity Element|identity]] is $e$.
Let $\struct {C, \circ} \subseteq \struct {S, \circ}$ be the [[Definition:Subsemigroup|subsemigroup]] of [[Definition:Cancellable Element|cancellable elements]] of $\struct {S, \cir... | Let $e$ be the [[Definition:Identity Element|identity]] for $\circ$.
Let $e = x \circ' y^{-1}$, where $x \in S, y \in C$.
Then:
{{begin-eqn}}
{{eqn | l = y
| r = e \circ y
| c = {{Defof|Identity Element}}
}}
{{eqn | r = \paren {x \circ' y^{-1} } \circ' y
| c = {{Defof|Inverse Element}}
}}
{{eqn | ... | Identity of Inverse Completion of Commutative Monoid | https://proofwiki.org/wiki/Identity_of_Inverse_Completion_of_Commutative_Monoid | https://proofwiki.org/wiki/Identity_of_Inverse_Completion_of_Commutative_Monoid | [
"Inverse Completions"
] | [
"Definition:Commutative Monoid",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Subsemigroup",
"Definition:Cancellable Element",
"Definition:Inverse Completion",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] | [
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Associative Operation"
] |
proofwiki-591 | Inverse Completion of Commutative Semigroup is Abelian Group | Let $\struct {S, \circ}$ be a commutative semigroup
Let all the elements of $\struct {S, \circ}$ be cancellable.
Then an inverse completion of $\struct {S, \circ}$ is an abelian group. | Let $\struct {S, \circ}$ be a commutative semigroup, all of whose elements are cancellable.
Let $\struct {T, \circ'}$ be an inverse completion of $\struct {S, \circ}$.
From Inverse Completion is Commutative Monoid:
:$\struct {T, \circ'} = \struct {S \circ' S^{-1}, \circ'}$
has been shown to be a commutative monoid.
Tak... | Let $\struct {S, \circ}$ be a [[Definition:Commutative Semigroup|commutative semigroup]]
Let all the [[Definition:Element|elements]] of $\struct {S, \circ}$ be [[Definition:Cancellable Element|cancellable]].
Then an [[Definition:Inverse Completion|inverse completion]] of $\struct {S, \circ}$ is an [[Definition:Abeli... | Let $\struct {S, \circ}$ be a [[Definition:Commutative Semigroup|commutative semigroup]], all of whose [[Definition:Element|elements]] are [[Definition:Cancellable Element|cancellable]].
Let $\struct {T, \circ'}$ be an [[Definition:Inverse Completion|inverse completion]] of $\struct {S, \circ}$.
From [[Inverse Compl... | Inverse Completion of Commutative Semigroup is Abelian Group | https://proofwiki.org/wiki/Inverse_Completion_of_Commutative_Semigroup_is_Abelian_Group | https://proofwiki.org/wiki/Inverse_Completion_of_Commutative_Semigroup_is_Abelian_Group | [
"Inverse Completions",
"Abelian Groups"
] | [
"Definition:Commutative Semigroup",
"Definition:Element",
"Definition:Cancellable Element",
"Definition:Inverse Completion",
"Definition:Abelian Group"
] | [
"Definition:Commutative Semigroup",
"Definition:Element",
"Definition:Cancellable Element",
"Definition:Inverse Completion",
"Inverse Completion is Commutative Monoid",
"Definition:Commutative Monoid",
"Axiom:Group Axioms",
"Definition:Commutative Semigroup",
"Definition:Commutative Semigroup",
"D... |
proofwiki-592 | Taylor's Theorem | Every infinitely differentiable function can be approximated by a series of polynomials. | Let $G$ be a real-valued function continuous on $\closedint a x$ and differentiable with non-vanishing derivative on $\openint a x$.
Let:
:$\map F t = \map f t + \dfrac {\map {f'} t} {1!} \paren {x - t} + \dotsb + \dfrac {\map {f^{\paren n} } t} {n!} \paren {x - t}^n$
By the Cauchy Mean Value Theorem:
:$(1): \quad \dfr... | Every [[Definition:Smooth Real Function|infinitely differentiable function]] can be approximated by a [[Definition:Series|series]] of [[Definition:Real Polynomial Function|polynomials]]. | Let $G$ be a [[Definition:Real-Valued Function|real-valued function]] [[Definition:Continuous Real Function|continuous]] on $\closedint a x$ and [[Definition:Differentiable Real Function|differentiable]] with non-vanishing [[Definition:Derivative|derivative]] on $\openint a x$.
Let:
:$\map F t = \map f t + \dfrac {\ma... | Taylor's Theorem/One Variable/Proof by Cauchy Mean Value Theorem | https://proofwiki.org/wiki/Taylor's_Theorem | https://proofwiki.org/wiki/Taylor's_Theorem/One_Variable/Proof_by_Cauchy_Mean_Value_Theorem | [
"Calculus",
"Taylor's Theorem"
] | [
"Definition:Smooth Real Function",
"Definition:Series",
"Definition:Polynomial Function/Real"
] | [
"Definition:Real-Valued Function",
"Definition:Continuous Real Function",
"Definition:Differentiable Mapping/Real Function",
"Definition:Derivative",
"Cauchy Mean Value Theorem",
"Definition:Fraction/Numerator",
"Definition:Taylor Series/Remainder/Lagrange Form",
"Definition:Taylor Series/Remainder/Ca... |
proofwiki-593 | Taylor's Theorem | Every infinitely differentiable function can be approximated by a series of polynomials. | Let the function $g$ be defined as:
:$\map g t = \map {R_n} t - \dfrac {\paren {t - a}^{n + 1} } {\paren {x - a}^{n + 1} } \map {R_n} x$
Then:
:$\map {g^{\paren k} } a = 0$
for $k = 0, \dotsc, n$, and $\map g x = 0$.
Apply Rolle's Theorem successively to $g, g', \dotsc, g^{\paren n}$.
Then there exist:
:$\xi_1, \ldots,... | Every [[Definition:Smooth Real Function|infinitely differentiable function]] can be approximated by a [[Definition:Series|series]] of [[Definition:Real Polynomial Function|polynomials]]. | Let the [[Definition:Real Function|function]] $g$ be defined as:
:$\map g t = \map {R_n} t - \dfrac {\paren {t - a}^{n + 1} } {\paren {x - a}^{n + 1} } \map {R_n} x$
Then:
:$\map {g^{\paren k} } a = 0$
for $k = 0, \dotsc, n$, and $\map g x = 0$.
Apply [[Rolle's Theorem]] successively to $g, g', \dotsc, g^{\paren n}$... | Taylor's Theorem/One Variable/Proof by Rolle's Theorem | https://proofwiki.org/wiki/Taylor's_Theorem | https://proofwiki.org/wiki/Taylor's_Theorem/One_Variable/Proof_by_Rolle's_Theorem | [
"Calculus",
"Taylor's Theorem"
] | [
"Definition:Smooth Real Function",
"Definition:Series",
"Definition:Polynomial Function/Real"
] | [
"Definition:Real Function",
"Rolle's Theorem"
] |
proofwiki-594 | Inverse Completion Theorem | Every commutative semigroup containing cancellable elements admits an inverse completion. | Let $\struct {S, \circ}$ be a commutative semigroup which has cancellable elements.
From Construction of Inverse Completion, we can construct an inverse completion $\struct {T', \oplus'}$ of $\struct {S', \oplus'}$, which is an isomorphic copy of $S$ under the mapping $\psi: S \to S'$.
{{explain|$\psi: S \to S'$ has no... | Every [[Definition:Commutative Semigroup|commutative semigroup]] containing [[Definition:Cancellable Element|cancellable elements]] admits an [[Definition:Inverse Completion|inverse completion]]. | Let $\struct {S, \circ}$ be a [[Definition:Commutative Semigroup|commutative semigroup]] which has [[Definition:Cancellable Element|cancellable elements]].
From [[Construction of Inverse Completion]], we can construct an [[Definition:Inverse Completion|inverse completion]] $\struct {T', \oplus'}$ of $\struct {S', \op... | Inverse Completion Theorem | https://proofwiki.org/wiki/Inverse_Completion_Theorem | https://proofwiki.org/wiki/Inverse_Completion_Theorem | [
"Inverse Completions",
"Semigroups",
"Named Theorems"
] | [
"Definition:Commutative Semigroup",
"Definition:Cancellable Element",
"Definition:Inverse Completion"
] | [
"Definition:Commutative Semigroup",
"Definition:Cancellable Element",
"Construction of Inverse Completion",
"Definition:Inverse Completion",
"Definition:Isomorphism (Abstract Algebra)",
"Embedding Theorem",
"Definition:Semigroup",
"Definition:Isomorphism (Abstract Algebra)",
"Definition:Inverse Comp... |
proofwiki-595 | Extension Theorem for Homomorphisms | Let $\struct {S, \circ}$ be a commutative semigroup with cancellable elements
Let $\struct {C, \circ} \subseteq \struct {S, \circ}$ be the subsemigroup of all cancellable elements of $S$
Let $\struct {S', \circ'}$ be an inverse completion of $\struct {S, \circ}$
Let $\phi$ be a (semigroup) homomorphism from $\struct {S... | It is proved that $\struct {C, \circ}$ is a subsemigroup of $\struct {S, \circ}$ by Cancellable Elements of Semigroup form Subsemigroup. | Let $\struct {S, \circ}$ be a [[Definition:Commutative Semigroup|commutative semigroup]] with [[Definition:Cancellable Element|cancellable elements]]
Let $\struct {C, \circ} \subseteq \struct {S, \circ}$ be the [[Definition:Subsemigroup|subsemigroup]] of all [[Definition:Cancellable Element|cancellable elements]] of $... | It is proved that $\struct {C, \circ}$ is a [[Definition:Subsemigroup|subsemigroup]] of $\struct {S, \circ}$ by [[Cancellable Elements of Semigroup form Subsemigroup]]. | Extension Theorem for Homomorphisms | https://proofwiki.org/wiki/Extension_Theorem_for_Homomorphisms | https://proofwiki.org/wiki/Extension_Theorem_for_Homomorphisms | [
"Homomorphisms (Abstract Algebra)",
"Monomorphisms (Abstract Algebra)",
"Semigroups",
"Inverse Completions",
"Named Theorems"
] | [
"Definition:Commutative Semigroup",
"Definition:Cancellable Element",
"Definition:Subsemigroup",
"Definition:Cancellable Element",
"Definition:Inverse Completion",
"Definition:Semigroup Homomorphism",
"Definition:Semigroup",
"Definition:Invertible Element",
"Definition:Homomorphism (Abstract Algebra... | [
"Definition:Subsemigroup",
"Cancellable Elements of Semigroup form Subsemigroup",
"Definition:Subsemigroup"
] |
proofwiki-596 | Extension Theorem for Isomorphisms | Let the following conditions be fulfilled:
:Let $\struct {S, \circ}$ be a commutative semigroup with cancellable elements
:Let $\phi$ be an isomorphism from $\struct {S, \circ}$ into a semigroup $\struct {T, *}$
:Let $\struct {S', \circ'}$ be an inverse completion of $\struct {S, \circ}$
:Let $\struct {T', \circ'}$ be ... | Let $C$ be the subsemigroup of cancellable elements of $S$.
It is proved that this is a semigroup by Cancellable Elements of Semigroup form Subsemigroup.
The set of cancellable elements of $T$ is $\phi \sqbrk C$.
By the Extension Theorem for Homomorphisms, there is:
:A unique homomorphism $\phi'$ from $S'$ into $T'$ ex... | Let the following conditions be fulfilled:
:Let $\struct {S, \circ}$ be a [[Definition:Commutative Semigroup|commutative semigroup]] with [[Definition:Cancellable Element|cancellable elements]]
:Let $\phi$ be an [[Definition:Semigroup Isomorphism|isomorphism]] from $\struct {S, \circ}$ into a [[Definition:Semigroup|s... | Let $C$ be the [[Definition:Subsemigroup|subsemigroup]] of [[Definition:Cancellable Element|cancellable elements]] of $S$.
It is proved that this is a [[Definition:Semigroup|semigroup]] by [[Cancellable Elements of Semigroup form Subsemigroup]].
The set of [[Definition:Cancellable Element|cancellable elements]] of $... | Extension Theorem for Isomorphisms | https://proofwiki.org/wiki/Extension_Theorem_for_Isomorphisms | https://proofwiki.org/wiki/Extension_Theorem_for_Isomorphisms | [
"Semigroup Isomorphisms",
"Inverse Completions",
"Semigroups",
"Named Theorems"
] | [
"Definition:Commutative Semigroup",
"Definition:Cancellable Element",
"Definition:Isomorphism (Abstract Algebra)/Semigroup Isomorphism",
"Definition:Semigroup",
"Definition:Inverse Completion",
"Definition:Inverse Completion",
"Definition:Isomorphism (Abstract Algebra)/Semigroup Isomorphism"
] | [
"Definition:Subsemigroup",
"Definition:Cancellable Element",
"Definition:Semigroup",
"Cancellable Elements of Semigroup form Subsemigroup",
"Definition:Cancellable Element",
"Extension Theorem for Homomorphisms",
"Definition:Homomorphism (Abstract Algebra)",
"Definition:Homomorphism (Abstract Algebra)... |
proofwiki-597 | Inverse Completion is Unique | An inverse completion of a commutative semigroup is unique up to isomorphism. | Let $T$ and $T'$ both be inverse completions of a commutative semigroup $S$ having cancellable elements.
Then from the Extension Theorem for Isomorphisms, there is a unique isomorphism $\phi: T \to T'$ satisfying $\forall x \in S: \map \phi x = x$.
Hence the result.
{{Qed}} | An [[Definition:Inverse Completion|inverse completion]] of a [[Definition:Commutative Semigroup|commutative semigroup]] is unique up to [[Definition:Isomorphism (Abstract Algebra)|isomorphism]]. | Let $T$ and $T'$ both be [[Definition:Inverse Completion|inverse completions]] of a [[Definition:Commutative Semigroup|commutative semigroup]] $S$ having [[Definition:Cancellable Element|cancellable elements]].
Then from the [[Extension Theorem for Isomorphisms]], there is a unique [[Definition:Isomorphism (Abstract A... | Inverse Completion is Unique | https://proofwiki.org/wiki/Inverse_Completion_is_Unique | https://proofwiki.org/wiki/Inverse_Completion_is_Unique | [
"Inverse Completions",
"Semigroups"
] | [
"Definition:Inverse Completion",
"Definition:Commutative Semigroup",
"Definition:Isomorphism (Abstract Algebra)"
] | [
"Definition:Inverse Completion",
"Definition:Commutative Semigroup",
"Definition:Cancellable Element",
"Extension Theorem for Isomorphisms",
"Definition:Isomorphism (Abstract Algebra)"
] |
proofwiki-598 | Monoid is not Empty | A monoid cannot be empty. | Let $\left({S, \circ}\right)$ be a the monoid.
By definition:
:Identity: $\exists e_S \in S: \forall a \in S: a \circ e_S = a = e_S \circ a$
So a monoid must ''at least'' have an identity.
Therefore $e_S \in S$ and so $S$ is not the empty set.
{{qed}} | A [[Definition:Monoid|monoid]] cannot be [[Definition:Empty Set|empty]]. | Let $\left({S, \circ}\right)$ be a the [[Definition:Monoid|monoid]].
By [[Definition:Monoid|definition]]:
:[[Definition:Identity Element|Identity]]: $\exists e_S \in S: \forall a \in S: a \circ e_S = a = e_S \circ a$
So a monoid must ''at least'' have an [[Definition:Identity Element|identity]].
Therefore $e_S \in... | Monoid is not Empty | https://proofwiki.org/wiki/Monoid_is_not_Empty | https://proofwiki.org/wiki/Monoid_is_not_Empty | [
"Monoids"
] | [
"Definition:Monoid",
"Definition:Empty Set"
] | [
"Definition:Monoid",
"Definition:Monoid",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Empty Set"
] |
proofwiki-599 | Group is not Empty | A group cannot be empty. | A group is defined as a monoid for which every element has an inverse.
Thus, as a group is already a monoid, it must ''at least'' have an identity, therefore can not be empty.
{{qed}} | A [[Definition:Group|group]] cannot be [[Definition:Empty Set|empty]]. | A [[Definition:Group|group]] is defined as a [[Definition:Monoid|monoid]] for which every [[Definition:Element|element]] has an [[Definition:Inverse (Abstract Algebra)|inverse]].
Thus, as a group is already a [[Definition:Monoid|monoid]], it must ''at least'' have an [[Definition:Identity (Abstract Algebra)|identity]... | Group is not Empty | https://proofwiki.org/wiki/Group_is_not_Empty | https://proofwiki.org/wiki/Group_is_not_Empty | [
"Groups",
"Empty Set"
] | [
"Definition:Group",
"Definition:Empty Set"
] | [
"Definition:Group",
"Definition:Monoid",
"Definition:Element",
"Definition:Inverse (Abstract Algebra)",
"Definition:Monoid",
"Definition:Identity (Abstract Algebra)"
] |
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