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-1000 | Real Number is between Floor Functions | :$\forall x \in \R: \floor x \le x < \floor {x + 1}$ | $\floor x$ is defined as:
:$\floor x = \sup \set {m \in \Z: m \le x}$
So $\floor x \le x$ by definition.
From Floor plus One:
:$\floor {x + 1} > \floor x$
Hence by the definition of the supremum:
:$\floor {x + 1} > x$
The result follows.
{{qed}}
Category:Floor Function
akn94fzyqk42jhnqcxjm9xqmvzfon8w | :$\forall x \in \R: \floor x \le x < \floor {x + 1}$ | $\floor x$ is defined as:
:$\floor x = \sup \set {m \in \Z: m \le x}$
So $\floor x \le x$ by definition.
From [[Floor plus One]]:
:$\floor {x + 1} > \floor x$
Hence by the definition of the [[Definition:Supremum of Set|supremum]]:
:$\floor {x + 1} > x$
The result follows.
{{qed}}
[[Category:Floor Function]]
akn... | Real Number is between Floor Functions | https://proofwiki.org/wiki/Real_Number_is_between_Floor_Functions | https://proofwiki.org/wiki/Real_Number_is_between_Floor_Functions | [
"Floor Function"
] | [] | [
"Floor plus One",
"Definition:Supremum of Set",
"Category:Floor Function"
] |
proofwiki-1001 | Real Number is between Ceiling Functions | :$\forall x \in \R: \ceiling {x - 1} \le x < \ceiling x$ | $\ceiling x$ is defined as:
:$\ceiling x = \map \inf {\set {m \in \Z: m \ge x} }$
So $\ceiling x \ge x$ by definition.
Now $\ceiling {x - 1} < \ceiling x$, so by the definition of the infimum:
:$\ceiling {x - 1} > x$
The result follows.
{{qed}}
Category:Ceiling Function
p1ftr5soshr01dofp1spqabila6zl24 | :$\forall x \in \R: \ceiling {x - 1} \le x < \ceiling x$ | $\ceiling x$ is defined as:
:$\ceiling x = \map \inf {\set {m \in \Z: m \ge x} }$
So $\ceiling x \ge x$ by definition.
Now $\ceiling {x - 1} < \ceiling x$, so by the definition of the [[Definition:Infimum of Set|infimum]]:
:$\ceiling {x - 1} > x$
The result follows.
{{qed}}
[[Category:Ceiling Function]]
p1ftr5sos... | Real Number is between Ceiling Functions | https://proofwiki.org/wiki/Real_Number_is_between_Ceiling_Functions | https://proofwiki.org/wiki/Real_Number_is_between_Ceiling_Functions | [
"Ceiling Function"
] | [] | [
"Definition:Infimum of Set",
"Category:Ceiling Function"
] |
proofwiki-1002 | Real Number minus Floor | :$x - \floor x \in \hointr 0 1$ | {{begin-eqn}}
{{eqn | l = \floor x
| o = \le
| m = x
| mo= <
| r = \floor x + 1
| c = {{Defof|Floor Function}}
}}
{{eqn | ll= \leadsto
| l = \floor x - \floor x
| o = \le
| m = x - \floor x
| mo= <
| r = \floor x + 1 - \floor x
| c = subtracting $\floor ... | :$x - \floor x \in \hointr 0 1$ | {{begin-eqn}}
{{eqn | l = \floor x
| o = \le
| m = x
| mo= <
| r = \floor x + 1
| c = {{Defof|Floor Function}}
}}
{{eqn | ll= \leadsto
| l = \floor x - \floor x
| o = \le
| m = x - \floor x
| mo= <
| r = \floor x + 1 - \floor x
| c = subtracting $\floor ... | Real Number minus Floor | https://proofwiki.org/wiki/Real_Number_minus_Floor | https://proofwiki.org/wiki/Real_Number_minus_Floor | [
"Floor Function"
] | [] | [] |
proofwiki-1003 | Ceiling minus Real Number | :$\forall x \in \R: \ceiling x - x \in \hointr 0 1$ | {{begin-eqn}}
{{eqn | l = \ceiling x - 1
| o = <
| r = x \le \ceiling x
| c = Real Number is between Ceiling Functions
}}
{{eqn | ll= \leadsto
| l = \ceiling x - 1 - \ceiling x
| o = <
| r = x - \ceiling x \le \ceiling x - \ceiling x
| c =
}}
{{eqn | ll= \leadsto
| l = -... | :$\forall x \in \R: \ceiling x - x \in \hointr 0 1$ | {{begin-eqn}}
{{eqn | l = \ceiling x - 1
| o = <
| r = x \le \ceiling x
| c = [[Real Number is between Ceiling Functions]]
}}
{{eqn | ll= \leadsto
| l = \ceiling x - 1 - \ceiling x
| o = <
| r = x - \ceiling x \le \ceiling x - \ceiling x
| c =
}}
{{eqn | ll= \leadsto
| l... | Ceiling minus Real Number | https://proofwiki.org/wiki/Ceiling_minus_Real_Number | https://proofwiki.org/wiki/Ceiling_minus_Real_Number | [
"Ceiling Function"
] | [] | [
"Real Number is between Ceiling Functions",
"Category:Ceiling Function"
] |
proofwiki-1004 | Real Number is Floor plus Difference | :There exists an integer $n \in \Z$ such that for some $t \in \hointr 0 1$:
::$x = n + t$
{{iff}}:
:$n = \floor x$ | === Sufficient Condition ===
Let there exist $n \in \Z$ such that $x = n + t$, where $t \in \hointr 0 1$.
We have that $1 - t > 0$.
Thus:
:$0 \le x - n < 1$
Thus:
: $n \le x < n + 1$
That is, $n$ is the floor of $x$.
{{qed|lemma}} | :There exists an [[Definition:Integer|integer]] $n \in \Z$ such that for some $t \in \hointr 0 1$:
::$x = n + t$
{{iff}}:
:$n = \floor x$ | === Sufficient Condition ===
Let there exist $n \in \Z$ such that $x = n + t$, where $t \in \hointr 0 1$.
We have that $1 - t > 0$.
Thus:
:$0 \le x - n < 1$
Thus:
: $n \le x < n + 1$
That is, $n$ is the [[Definition:Floor Function|floor]] of $x$.
{{qed|lemma}} | Real Number is Floor plus Difference | https://proofwiki.org/wiki/Real_Number_is_Floor_plus_Difference | https://proofwiki.org/wiki/Real_Number_is_Floor_plus_Difference | [
"Floor Function"
] | [
"Definition:Integer"
] | [
"Definition:Floor Function"
] |
proofwiki-1005 | Floor plus One | Let $x \in \R$.
Then:
:$\floor {x + 1} = \floor x + 1$
where $\floor x$ is the floor function of $x$. | {{begin-eqn}}
{{eqn | l = \floor {x + 1}
| r = n
| c =
}}
{{eqn | ll= \leadsto
| l = n
| o = \le
| m = x + 1
| mo= <
| r = n + 1
| c = {{Defof|Floor Function}}
}}
{{eqn | ll= \leadsto
| l = n - 1
| o = \le
| m = x
| mo= <
| r = n
| c = ... | Let $x \in \R$.
Then:
:$\floor {x + 1} = \floor x + 1$
where $\floor x$ is the [[Definition:Floor Function|floor function]] of $x$. | {{begin-eqn}}
{{eqn | l = \floor {x + 1}
| r = n
| c =
}}
{{eqn | ll= \leadsto
| l = n
| o = \le
| m = x + 1
| mo= <
| r = n + 1
| c = {{Defof|Floor Function}}
}}
{{eqn | ll= \leadsto
| l = n - 1
| o = \le
| m = x
| mo= <
| r = n
| c = ... | Floor plus One | https://proofwiki.org/wiki/Floor_plus_One | https://proofwiki.org/wiki/Floor_plus_One | [
"Floor Function"
] | [
"Definition:Floor Function"
] | [] |
proofwiki-1006 | Real Number is Integer iff equals Floor | :$x = \floor x \iff x \in \Z$ | Let $x = \floor x$.
As $\floor x \in \Z$, then so must $x$ be.
Now let $x \in \Z$.
We have:
:$\floor x = \sup \set {m \in \Z: m \le x}$
As $x \in \sup \set {m \in \Z: m \le x}$, and there can be no greater $n \in \Z$ such that $n \in \sup \set {m \in \Z: m \le x}$, it follows that:
:$x = \floor x$
{{qed}} | :$x = \floor x \iff x \in \Z$ | Let $x = \floor x$.
As $\floor x \in \Z$, then so must $x$ be.
Now let $x \in \Z$.
We have:
:$\floor x = \sup \set {m \in \Z: m \le x}$
As $x \in \sup \set {m \in \Z: m \le x}$, and there can be no greater $n \in \Z$ such that $n \in \sup \set {m \in \Z: m \le x}$, it follows that:
:$x = \floor x$
{{qed}} | Real Number is Integer iff equals Floor | https://proofwiki.org/wiki/Real_Number_is_Integer_iff_equals_Floor | https://proofwiki.org/wiki/Real_Number_is_Integer_iff_equals_Floor | [
"Floor Function"
] | [] | [] |
proofwiki-1007 | Sum of Floor and Floor of Negative | Let $x \in \R$. Then:
:$\floor x + \floor {-x} = \begin{cases} 0 & : x \in \Z \\ -1 & : x \notin \Z \end{cases}$
where $\floor x$ denotes the floor of $x$. | Let $x \in \Z$.
Then from Real Number is Integer iff equals Floor:
:$x = \floor x$
Now $x \in \Z \implies -x \in \Z$, so:
:$\floor {-x} = -x$
Thus:
:$\floor x + \floor {-x} = x + \paren {-x} = x - x = 0$
Now let $x \notin \Z$.
From Real Number is Floor plus Difference:
:$x = n + t$
where $n = \floor x$ and $t \in \hoin... | Let $x \in \R$. Then:
:$\floor x + \floor {-x} = \begin{cases} 0 & : x \in \Z \\ -1 & : x \notin \Z \end{cases}$
where $\floor x$ denotes the [[Definition:Floor Function|floor]] of $x$. | Let $x \in \Z$.
Then from [[Real Number is Integer iff equals Floor]]:
:$x = \floor x$
Now $x \in \Z \implies -x \in \Z$, so:
:$\floor {-x} = -x$
Thus:
:$\floor x + \floor {-x} = x + \paren {-x} = x - x = 0$
Now let $x \notin \Z$.
From [[Real Number is Floor plus Difference]]:
:$x = n + t$
where $n = \floor x$ an... | Sum of Floor and Floor of Negative | https://proofwiki.org/wiki/Sum_of_Floor_and_Floor_of_Negative | https://proofwiki.org/wiki/Sum_of_Floor_and_Floor_of_Negative | [
"Floor Function"
] | [
"Definition:Floor Function"
] | [
"Real Number is Integer iff equals Floor",
"Real Number is Floor plus Difference"
] |
proofwiki-1008 | Floor defines Equivalence Relation | Let $x \in \R$ be a real number.
Let $\floor x$ denote the floor function of $x$.
Let $\RR$ be the relation defined on $\R$ such that:
:$\forall x, y, \in \R: \tuple {x, y} \in \RR \iff \floor x = \floor y$
Then $\RR$ is an equivalence, and $\forall n \in \Z$, the $\RR$-class of $n$ is the half-open interval $\hointr n... | Checking in turn each of the criteria for equivalence: | Let $x \in \R$ be a [[Definition:Real Number|real number]].
Let $\floor x$ denote the [[Definition:Floor Function|floor function]] of $x$.
Let $\RR$ be the [[Definition:Relation|relation]] defined on $\R$ such that:
:$\forall x, y, \in \R: \tuple {x, y} \in \RR \iff \floor x = \floor y$
Then $\RR$ is an [[Definitio... | Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]: | Floor defines Equivalence Relation | https://proofwiki.org/wiki/Floor_defines_Equivalence_Relation | https://proofwiki.org/wiki/Floor_defines_Equivalence_Relation | [
"Floor Function",
"Examples of Equivalence Relations"
] | [
"Definition:Real Number",
"Definition:Floor Function",
"Definition:Relation",
"Definition:Equivalence Relation",
"Definition:Equivalence Class",
"Definition:Real Interval/Half-Open"
] | [
"Definition:Equivalence Relation",
"Definition:Equivalence Relation"
] |
proofwiki-1009 | Ceiling defines Equivalence Relation | Let $\RR$ be the relation defined on $\R$ such that:
:$\forall x, y, \in \R: \tuple {x, y} \in \RR \iff \ceiling x = \ceiling y$
where $\ceiling x$ is the ceiling of $x$.
Then $\RR$ is an equivalence, and $\forall n \in \Z$, the $\RR$-class of $n$ is the half-open interval $\hointl {n - 1} n$. | Checking in turn each of the criteria for equivalence: | Let $\RR$ be the [[Definition:Relation|relation]] defined on $\R$ such that:
:$\forall x, y, \in \R: \tuple {x, y} \in \RR \iff \ceiling x = \ceiling y$
where $\ceiling x$ is the [[Definition:Ceiling Function|ceiling]] of $x$.
Then $\RR$ is an [[Definition:Equivalence Relation|equivalence]], and $\forall n \in \Z$, t... | Checking in turn each of the criteria for [[Definition:Equivalence Relation|equivalence]]: | Ceiling defines Equivalence Relation | https://proofwiki.org/wiki/Ceiling_defines_Equivalence_Relation | https://proofwiki.org/wiki/Ceiling_defines_Equivalence_Relation | [
"Ceiling Function",
"Examples of Equivalence Relations"
] | [
"Definition:Relation",
"Definition:Ceiling Function",
"Definition:Equivalence Relation",
"Definition:Equivalence Class",
"Definition:Real Interval/Half-Open"
] | [
"Definition:Equivalence Relation",
"Definition:Equivalence Relation"
] |
proofwiki-1010 | Real Number is Ceiling minus Difference | Let $n$ be a integer.
{{TFAE}}
:$(1): \quad$ There exists $t \in \hointr 0 1$ such that $x = n - t$
:$(2): \quad n = \ceiling x$ | === 1 implies 2 ===
Let $x = n - t$, where $t \in \hointr 0 1$.
Because $0 \le t < 1$, we have:
:$0 \leq n - x < 1$
Thus:
:$n - 1 < x \le n$
That is, $n$ is the ceiling of $x$.
{{qed|lemma}} | Let $n$ be a [[Definition:Integer|integer]].
{{TFAE}}
:$(1): \quad$ There exists $t \in \hointr 0 1$ such that $x = n - t$
:$(2): \quad n = \ceiling x$ | === 1 implies 2 ===
Let $x = n - t$, where $t \in \hointr 0 1$.
Because $0 \le t < 1$, we have:
:$0 \leq n - x < 1$
Thus:
:$n - 1 < x \le n$
That is, $n$ is the [[Definition:Ceiling Function|ceiling]] of $x$.
{{qed|lemma}} | Real Number is Ceiling minus Difference | https://proofwiki.org/wiki/Real_Number_is_Ceiling_minus_Difference | https://proofwiki.org/wiki/Real_Number_is_Ceiling_minus_Difference | [
"Ceiling Function"
] | [
"Definition:Integer"
] | [
"Definition:Ceiling Function"
] |
proofwiki-1011 | Cauchy's Inequality | :$\ds \sum {r_i}^2 \sum {s_i}^2 \ge \paren {\sum {r_i s_i} }^2$
where all of $r_i, s_i \in \R$. | For any $\lambda \in \R$, we define $f: \R \to \R$ as the function:
:$\ds \map f \lambda = \sum {\paren {r_i + \lambda s_i}^2}$
Now:
:$\map f \lambda \ge 0$
because it is the sum of squares of real numbers.
Hence:
{{begin-eqn}}
{{eqn | q = \forall \lambda \in \R
| l = \map f \lambda
| o = \equiv
| m =... | :$\ds \sum {r_i}^2 \sum {s_i}^2 \ge \paren {\sum {r_i s_i} }^2$
where all of $r_i, s_i \in \R$. | For any $\lambda \in \R$, we define $f: \R \to \R$ as the [[Definition:Real Function|function]]:
:$\ds \map f \lambda = \sum {\paren {r_i + \lambda s_i}^2}$
Now:
:$\map f \lambda \ge 0$
because it is the sum of [[Definition:Square Function|squares]] of [[Definition:Real Number|real numbers]].
Hence:
{{begin-eqn}}
... | Cauchy's Inequality/Proof 1 | https://proofwiki.org/wiki/Cauchy's_Inequality | https://proofwiki.org/wiki/Cauchy's_Inequality/Proof_1 | [
"Cauchy's Inequality",
"Inequalities",
"Cauchy-Bunyakovsky-Schwarz Inequality"
] | [] | [
"Definition:Real Function",
"Definition:Square/Function",
"Definition:Real Number",
"Definition:Quadratic Equation",
"Solution to Quadratic Equation",
"Discriminant of Quadratic Equation",
"Definition:Discriminant of Polynomial",
"Definition:Strictly Positive/Real Number",
"Definition:Distinct/Plura... |
proofwiki-1012 | Cauchy's Inequality | :$\ds \sum {r_i}^2 \sum {s_i}^2 \ge \paren {\sum {r_i s_i} }^2$
where all of $r_i, s_i \in \R$. | From the Complex Number form of the Cauchy-Schwarz Inequality, we have:
{{:Cauchy-Schwarz Inequality/Complex Numbers}}
As elements of $\R$ are also elements of $\C$, it follows that:
:$\ds \sum \size {r_i}^2 \sum \size {s_i}^2 \ge \size {\sum r_i s_i}^2$
where all of $r_i, s_i \in \R$.
But from the definition of modulu... | :$\ds \sum {r_i}^2 \sum {s_i}^2 \ge \paren {\sum {r_i s_i} }^2$
where all of $r_i, s_i \in \R$. | From the [[Cauchy-Schwarz Inequality/Complex Numbers|Complex Number form of the Cauchy-Schwarz Inequality]], we have:
{{:Cauchy-Schwarz Inequality/Complex Numbers}}
As elements of $\R$ are also elements of $\C$, it follows that:
:$\ds \sum \size {r_i}^2 \sum \size {s_i}^2 \ge \size {\sum r_i s_i}^2$
where all of $r_i,... | Cauchy's Inequality/Proof 2 | https://proofwiki.org/wiki/Cauchy's_Inequality | https://proofwiki.org/wiki/Cauchy's_Inequality/Proof_2 | [
"Cauchy's Inequality",
"Inequalities",
"Cauchy-Bunyakovsky-Schwarz Inequality"
] | [] | [
"Cauchy-Schwarz Inequality/Complex Numbers",
"Definition:Complex Modulus"
] |
proofwiki-1013 | Cancellable Finite Semigroup is Group | Let $\struct {S, \circ}$ be a non-empty finite semigroup in which all elements are cancellable.
Then $\struct {S, \circ}$ is a group. | As $\struct {S, \circ}$ is a semigroup, it is {{afortiori}} closed and associative.
It remains to be shown that:
:$\struct {S, \circ}$ has an identity
:every element of $S$ has an inverse in $S$.
Let $a \in S$ be arbitrary.
Let the mapping $\lambda_a: S \to S$ be the left regular representation of $\struct {S, \circ}$ ... | Let $\struct {S, \circ}$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Finite Semigroup|finite semigroup]] in which all [[Definition:Element|elements]] are [[Definition:Cancellable Element|cancellable]].
Then $\struct {S, \circ}$ is a [[Definition:Group|group]]. | As $\struct {S, \circ}$ is a [[Definition:Semigroup|semigroup]], it is {{afortiori}} [[Definition:Closed Algebraic Structure|closed]] and [[Definition:Associative Algebraic Structure|associative]].
It remains to be shown that:
:$\struct {S, \circ}$ has an [[Definition:Identity Element|identity]]
:every [[Definition:El... | Cancellable Finite Semigroup is Group | https://proofwiki.org/wiki/Cancellable_Finite_Semigroup_is_Group | https://proofwiki.org/wiki/Cancellable_Finite_Semigroup_is_Group | [
"Semigroups",
"Finite Groups",
"Cancellability"
] | [
"Definition:Non-Empty Set",
"Definition:Finite Semigroup",
"Definition:Element",
"Definition:Cancellable Element",
"Definition:Group"
] | [
"Definition:Semigroup",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Semigroup",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Element",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Definition:Mapping",
"Definition:Regular Representations/Lef... |
proofwiki-1014 | Finite Semigroup Equal Elements for Different Powers | Let $\left({S, \circ}\right)$ be a finite semigroup.
Then:
: $\forall x \in S: \exists m, n \in \N: m \ne n: x^m = x^n$ | List the positive powers $x, x^2, x^3, \ldots$ of any element $x$ of a finite semigroup $\left({S, \circ}\right)$.
Since all are elements of $S$, and the semigroup has a finite number of elements, it follows from the Pigeonhole Principle this list must contain repetitions.
So there must be at least one instance where $... | Let $\left({S, \circ}\right)$ be a [[Definition:Finite|finite]] [[Definition:Semigroup|semigroup]].
Then:
: $\forall x \in S: \exists m, n \in \N: m \ne n: x^m = x^n$ | List the [[Definition:Power of Element of Semigroup|positive powers]] $x, x^2, x^3, \ldots$ of any element $x$ of a finite semigroup $\left({S, \circ}\right)$.
Since all are elements of $S$, and the [[Definition:Semigroup|semigroup]] has a [[Definition:Finite|finite]] number of elements, it follows from the [[Pigeonho... | Finite Semigroup Equal Elements for Different Powers | https://proofwiki.org/wiki/Finite_Semigroup_Equal_Elements_for_Different_Powers | https://proofwiki.org/wiki/Finite_Semigroup_Equal_Elements_for_Different_Powers | [
"Semigroups"
] | [
"Definition:Finite",
"Definition:Semigroup"
] | [
"Definition:Power of Element/Semigroup",
"Definition:Semigroup",
"Definition:Finite",
"Dirichlet's Box Principle/Corollary",
"Category:Semigroups"
] |
proofwiki-1015 | Element has Idempotent Power in Finite Semigroup | Let $\struct {S, \circ}$ be a finite semigroup.
For every element in $\struct {S, \circ}$, there is a power of that element which is idempotent.
That is:
:$\forall x \in S: \exists i \in \N: x^i = x^i \circ x^i$ | From Finite Semigroup Equal Elements for Different Powers, we have:
:$\forall x \in S: \exists m, n \in \N: m \ne n: x^m = x^n$
Let $m > n$.
Let $n = k, m = k + l$.
Then:
: $\forall x \in S: \exists k, l \in \N: x^k = x^{k + l}$
Now we show that:
:$x^k = x^{k + l} \implies x^{k l} = x^{k l} \circ x^{k l}$
That is, that... | Let $\struct {S, \circ}$ be a [[Definition:Finite Semigroup|finite semigroup]].
For every [[Definition:Element|element]] in $\struct {S, \circ}$, there is a [[Definition:Power of Element of Semigroup|power]] of that [[Definition:Element|element]] which is [[Definition:Idempotent Element|idempotent]].
That is:
:$\for... | From [[Finite Semigroup Equal Elements for Different Powers]], we have:
:$\forall x \in S: \exists m, n \in \N: m \ne n: x^m = x^n$
Let $m > n$.
Let $n = k, m = k + l$.
Then:
: $\forall x \in S: \exists k, l \in \N: x^k = x^{k + l}$
Now we show that:
:$x^k = x^{k + l} \implies x^{k l} = x^{k l} \circ x^{k l}$
Th... | Element has Idempotent Power in Finite Semigroup | https://proofwiki.org/wiki/Element_has_Idempotent_Power_in_Finite_Semigroup | https://proofwiki.org/wiki/Element_has_Idempotent_Power_in_Finite_Semigroup | [
"Semigroups",
"Idempotence"
] | [
"Definition:Finite Semigroup",
"Definition:Element",
"Definition:Power of Element/Semigroup",
"Definition:Element",
"Definition:Idempotence/Element"
] | [
"Finite Semigroup Equal Elements for Different Powers",
"Definition:Idempotence/Element",
"Index Laws/Sum of Indices/Semigroup",
"Principle of Mathematical Induction",
"Definition:Idempotence/Element",
"Index Laws/Sum of Indices/Semigroup",
"Definition:Idempotence/Element"
] |
proofwiki-1016 | Integral Multiple of Ring Element | Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$.
Let $n \cdot x$ be an integral multiple of $x$:
:<nowiki>$n \cdot x = \begin {cases}
0_R & : n = 0 \\
x & : n = 1 \\
\paren {n - 1} \cdot x + x & : n > 1
\end {cases}$</nowiki>
that is:
:$n \cdot x = \underbrace {x + x + \cdots + x}_{\text {$n$ times} }$
For $n... | Proof by induction:
For all $n \in \N$, let $\map P n$ be the proposition:
:$\paren {n \cdot x} \circ x = n \cdot \paren {x \circ x} = x \circ \paren {n \cdot x}$
First we verify $\map P 0$.
When $n = 0$, we have:
{{begin-eqn}}
{{eqn | l = \paren {0 \cdot x} \circ x
| r = 0_R \circ x
| c =
}}
{{eqn | r = 0... | Let $\struct {R, +, \circ}$ be a [[Definition:Ring (Abstract Algebra)|ring]] whose [[Definition:Ring Zero|zero]] is $0_R$.
Let $n \cdot x$ be an [[Definition:Integral Multiple of Ring Element|integral multiple]] of $x$:
:<nowiki>$n \cdot x = \begin {cases}
0_R & : n = 0 \\
x & : n = 1 \\
\paren {n - 1} \cdot x + x & ... | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \N$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\paren {n \cdot x} \circ x = n \cdot \paren {x \circ x} = x \circ \paren {n \cdot x}$
First we verify $\map P 0$.
When $n = 0$, we have:
{{begin-eqn}}
{{eqn | l = \paren {... | Integral Multiple of Ring Element | https://proofwiki.org/wiki/Integral_Multiple_of_Ring_Element | https://proofwiki.org/wiki/Integral_Multiple_of_Ring_Element | [
"Ring Theory",
"Proofs by Induction"
] | [
"Definition:Ring (Abstract Algebra)",
"Definition:Ring Zero",
"Definition:Integral Multiple/Rings and Fields"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-1017 | Power of Conjugate equals Conjugate of Power | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $x, y \in G$ such that $\exists a \in G: x \circ a = a \circ y$.
That is, let $x$ and $y$ be conjugate.
Then:
: $\forall n \in \Z: y^n = \paren {a^{-1} \circ x \circ a}^n = a^{-1} \circ x^n \circ a$
It follows directly that:
: $\exists b \in G: \forall n \i... | Proof by induction:
For all $n \in \N$, let $\map P n$ be the proposition $y^n = a^{-1} \circ x^n \circ a$.
$\map P 0$ is true, as this just says $e = a^{-1} \circ e \circ a$. | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $x, y \in G$ such that $\exists a \in G: x \circ a = a \circ y$.
That is, let $x$ and $y$ be [[Definition:Conjugate of Group Element|conjugate]].
Then:
: $\forall n \in \Z: y^n = \paren {a^{-1} \circ ... | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \N$, let $\map P n$ be the [[Definition:Proposition|proposition]] $y^n = a^{-1} \circ x^n \circ a$.
$\map P 0$ is true, as this just says $e = a^{-1} \circ e \circ a$. | Power of Conjugate equals Conjugate of Power | https://proofwiki.org/wiki/Power_of_Conjugate_equals_Conjugate_of_Power | https://proofwiki.org/wiki/Power_of_Conjugate_equals_Conjugate_of_Power | [
"Conjugacy"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Conjugate (Group Theory)/Element"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction"
] |
proofwiki-1018 | Product of Conjugates equals Conjugate of Products | Let $\struct {G, \circ}$ be a group.
Then:
:$\forall a, x, y \in G: \paren {a \circ x \circ a^{-1} } \circ \paren {a \circ y \circ a^{-1} } = a \circ \paren {x \circ y} \circ a^{-1}$
That is, the product of conjugates is equal to the conjugate of the product. | Follows directly from the group axioms.
{{Qed}} | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Then:
:$\forall a, x, y \in G: \paren {a \circ x \circ a^{-1} } \circ \paren {a \circ y \circ a^{-1} } = a \circ \paren {x \circ y} \circ a^{-1}$
That is, the [[Definition:Product Element|product]] of [[Definition:Conjugate of Group Element|conjugates]] is equ... | Follows directly from the [[Axiom:Group Axioms|group axioms]].
{{Qed}} | Product of Conjugates equals Conjugate of Products | https://proofwiki.org/wiki/Product_of_Conjugates_equals_Conjugate_of_Products | https://proofwiki.org/wiki/Product_of_Conjugates_equals_Conjugate_of_Products | [
"Conjugacy"
] | [
"Definition:Group",
"Definition:Group Product/Product Element",
"Definition:Conjugate (Group Theory)/Element",
"Definition:Conjugate (Group Theory)/Element",
"Definition:Group Product/Product Element"
] | [
"Axiom:Group Axioms"
] |
proofwiki-1019 | Power of Product with Inverse | Let $G$ be a group whose identity is $e$.
Let $a, b \in G: a b = b a^{-1}$.
Then:
: $\forall n \in \Z: a^n b = b a^{-n}$ | Proof by induction:
For all $n \in \Z$, let $\map P n$ be the proposition $a^n b = b a^{-n}$.
$\map P 0$ is trivially true, as $a^0 b = e b = b = b e = b a^{-0}$. | Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity (Abstract Algebra)|identity]] is $e$.
Let $a, b \in G: a b = b a^{-1}$.
Then:
: $\forall n \in \Z: a^n b = b a^{-n}$ | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \Z$, let $\map P n$ be the [[Definition:Proposition|proposition]] $a^n b = b a^{-n}$.
$\map P 0$ is trivially true, as $a^0 b = e b = b = b e = b a^{-0}$. | Power of Product with Inverse | https://proofwiki.org/wiki/Power_of_Product_with_Inverse | https://proofwiki.org/wiki/Power_of_Product_with_Inverse | [
"Group Theory"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction",
"Principle of Mathematical Induction"
] |
proofwiki-1020 | Powers of Elements in Group Direct Product | Let $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$ be group whose identities are $e_G$ and $e_H$.
Let $\struct {G \times H, \circ}$ be the group direct product of $G$ and $H$.
Then:
:$\forall n \in \Z: \forall g \in G, h \in H: \tuple {g, h}^n = \tuple {g^n, h^n}$ | Proof by induction:
For all $n \in \N$, let $\map P n$ be the proposition $\forall g \in G, h \in H: \tuple {g, h}^n = \tuple {g^n, h^n}$. | Let $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$ be [[Definition:Group|group]] whose [[Definition:Identity (Abstract Algebra)|identities]] are $e_G$ and $e_H$.
Let $\struct {G \times H, \circ}$ be the [[Definition:Group Direct Product|group direct product]] of $G$ and $H$.
Then:
:$\forall n \in \Z: \forall g \i... | Proof by [[Principle of Mathematical Induction|induction]]:
For all $n \in \N$, let $\map P n$ be the [[Definition:Proposition|proposition]] $\forall g \in G, h \in H: \tuple {g, h}^n = \tuple {g^n, h^n}$. | Powers of Elements in Group Direct Product | https://proofwiki.org/wiki/Powers_of_Elements_in_Group_Direct_Product | https://proofwiki.org/wiki/Powers_of_Elements_in_Group_Direct_Product | [
"Powers (Abstract Algebra)",
"Group Direct Products"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)",
"Definition:Group Direct Product"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-1021 | General Morphism Property for Semigroups | Let $\struct {S, \circ}$ and $\struct {T, *}$ be semigroups.
Let $\phi: S \to T$ be a homomorphism.
Then:
:$\forall s_k \in S: \map \phi {s_1 \circ s_2 \circ \cdots \circ s_n} = \map \phi {s_1} * \map \phi {s_2} * \cdots * \map \phi {s_n}$
Hence it follows that:
:$\forall n \in \N_{>0}: \forall s \in S: \map \phi {s^n}... | $\forall s_k \in S: \map \phi {s_1 \circ s_2 \circ \cdots \circ s_n} = \map \phi {s_1} * \map \phi {s_2} * \cdots * \map \phi {s_n}$ can be proved by induction.
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$\map \phi {s_1 \circ s_2 \circ \cdots \circ s_n} = \map \phi {s_1} * \map \phi {s_2} * \cdots * \... | Let $\struct {S, \circ}$ and $\struct {T, *}$ be [[Definition:Semigroup|semigroups]].
Let $\phi: S \to T$ be a [[Definition:Semigroup Homomorphism|homomorphism]].
Then:
:$\forall s_k \in S: \map \phi {s_1 \circ s_2 \circ \cdots \circ s_n} = \map \phi {s_1} * \map \phi {s_2} * \cdots * \map \phi {s_n}$
Hence it fol... | $\forall s_k \in S: \map \phi {s_1 \circ s_2 \circ \cdots \circ s_n} = \map \phi {s_1} * \map \phi {s_2} * \cdots * \map \phi {s_n}$ can be proved by [[Principle of Mathematical Induction|induction]].
For all $n \in \N_{>0}$, let $\map P n$ be the [[Definition:Proposition|proposition]]:
:$\map \phi {s_1 \circ s_2 \ci... | General Morphism Property for Semigroups | https://proofwiki.org/wiki/General_Morphism_Property_for_Semigroups | https://proofwiki.org/wiki/General_Morphism_Property_for_Semigroups | [
"Morphism Property",
"Semigroup Homomorphisms"
] | [
"Definition:Semigroup",
"Definition:Semigroup Homomorphism"
] | [
"Principle of Mathematical Induction",
"Definition:Proposition",
"Principle of Mathematical Induction"
] |
proofwiki-1022 | Homomorphism of Power of Group Element | Let $\struct {G, \circ}$ and $\struct {H, \ast}$ be groups.
Let $\phi: S \to T$ be a group homomorphism.
Then:
:$\forall n \in \Z: \forall g \in G: \map \phi {g^n} = \paren {\map \phi g}^n$ | The result for $n \in \N_{>0}$ follows directly from General Morphism Property for Semigroups.
For $n = 0$, we use Homomorphism with Cancellable Codomain Preserves Identity.
For $n < 0$, we use Homomorphism with Identity Preserves Inverses, along with Index Laws for Monoids: Negative Index.
{{qed}} | Let $\struct {G, \circ}$ and $\struct {H, \ast}$ be [[Definition:Group|groups]].
Let $\phi: S \to T$ be a [[Definition:Group Homomorphism|group homomorphism]].
Then:
:$\forall n \in \Z: \forall g \in G: \map \phi {g^n} = \paren {\map \phi g}^n$ | The result for $n \in \N_{>0}$ follows directly from [[General Morphism Property for Semigroups]].
For $n = 0$, we use [[Homomorphism with Cancellable Codomain Preserves Identity]].
For $n < 0$, we use [[Homomorphism with Identity Preserves Inverses]], along with [[Index Laws for Monoids/Negative Index|Index Laws for... | Homomorphism of Power of Group Element | https://proofwiki.org/wiki/Homomorphism_of_Power_of_Group_Element | https://proofwiki.org/wiki/Homomorphism_of_Power_of_Group_Element | [
"Group Homomorphisms"
] | [
"Definition:Group",
"Definition:Group Homomorphism"
] | [
"General Morphism Property for Semigroups",
"Homomorphism with Cancellable Codomain Preserves Identity",
"Homomorphism with Identity Preserves Inverses",
"Index Laws for Monoids/Negative Index"
] |
proofwiki-1023 | Existence of Unique Subsemigroup Generated by Subset | Let $\struct {S, \circ}$ be a semigroup.
Let $\O \subset X \subseteq S$.
Let $\struct {T, \circ}$ be the subsemigroup generated by $X$.
Then $T = \gen X$ exists and is unique. | Let $\mathbb S$ be the set of all subsemigroups of $S$.
From Set of Subsemigroups forms Complete Lattice:
:$\struct {\mathbb S, \subseteq}$ is a complete lattice.
where for every set $\mathbb H$ of subsemigroups of $S$:
:the infimum of $\mathbb H$ necessarily admitted by $\mathbb H$ is $\ds \bigcap \mathbb H$.
Hence th... | Let $\struct {S, \circ}$ be a [[Definition:Semigroup|semigroup]].
Let $\O \subset X \subseteq S$.
Let $\struct {T, \circ}$ be the [[Definition:Generator of Subsemigroup|subsemigroup generated by $X$]].
Then $T = \gen X$ exists and is [[Definition:Unique|unique]]. | Let $\mathbb S$ be the [[Definition:Set|set]] of all [[Definition:Subsemigroup|subsemigroups]] of $S$.
From [[Set of Subsemigroups forms Complete Lattice]]:
:$\struct {\mathbb S, \subseteq}$ is a [[Definition:Complete Lattice|complete lattice]].
where for every [[Definition:Set|set]] $\mathbb H$ of [[Definition:Subse... | Existence of Unique Subsemigroup Generated by Subset/Proof 2 | https://proofwiki.org/wiki/Existence_of_Unique_Subsemigroup_Generated_by_Subset | https://proofwiki.org/wiki/Existence_of_Unique_Subsemigroup_Generated_by_Subset/Proof_2 | [
"Subsemigroups",
"Intersection of Subsemigroups",
"Existence of Unique Subsemigroup Generated by Subset"
] | [
"Definition:Semigroup",
"Definition:Generator of Subsemigroup",
"Definition:Unique"
] | [
"Definition:Set",
"Definition:Subsemigroup",
"Set of Subsemigroups forms Complete Lattice",
"Definition:Complete Lattice",
"Definition:Set",
"Definition:Subsemigroup",
"Definition:Infimum of Set",
"Definition:Infimum of Set"
] |
proofwiki-1024 | Finite Subgroup Test | Let $\struct {G, \circ}$ be a group.
Let $H$ be a non-empty finite subset of $G$.
Then:
:$H$ is a subgroup of $G$
{{iff}}:
:$\forall a, b \in H: a \circ b \in H$
That is, a non-empty finite subset of $G$ is a subgroup {{iff}} it is closed. | Let $H$ be a finite subset of $G$ such that $a, b \in H \implies a \circ b \in H$.
From the Two-Step Subgroup Test, it follows that we only need to show that $a \in H \implies a^{-1} \in H$.
So, let $a \in H$.
First it is straightforward to show by induction that $\set {x \in G: x = a^n: n \in \N} \subseteq H$.
That is... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $H$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] of $G$.
Then:
:$H$ is a [[Definition:Subgroup|subgroup]] of $G$
{{iff}}:
:$\forall a, b \in H: a \circ b \in H$
That is, a [[Definition:Non-Emp... | Let $H$ be a [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] of $G$ such that $a, b \in H \implies a \circ b \in H$.
From the [[Two-Step Subgroup Test]], it follows that we only need to show that $a \in H \implies a^{-1} \in H$.
So, let $a \in H$.
First it is straightforward to show by [[Principle of M... | Finite Subgroup Test/Proof 1 | https://proofwiki.org/wiki/Finite_Subgroup_Test | https://proofwiki.org/wiki/Finite_Subgroup_Test/Proof_1 | [
"Subgroups",
"Finite Subgroup Test"
] | [
"Definition:Group",
"Definition:Non-Empty Set",
"Definition:Finite Set",
"Definition:Subset",
"Definition:Subgroup",
"Definition:Non-Empty Set",
"Definition:Finite Set",
"Definition:Subset",
"Definition:Subgroup",
"Definition:Closure (Abstract Algebra)/Algebraic Structure"
] | [
"Definition:Finite Set",
"Definition:Subset",
"Two-Step Subgroup Test",
"Principle of Mathematical Induction",
"Definition:Finite Set",
"Order of Element Divides Order of Finite Group",
"Definition:Order of Group Element",
"Inverse Element is Power of Order Less 1"
] |
proofwiki-1025 | Finite Subgroup Test | Let $\struct {G, \circ}$ be a group.
Let $H$ be a non-empty finite subset of $G$.
Then:
:$H$ is a subgroup of $G$
{{iff}}:
:$\forall a, b \in H: a \circ b \in H$
That is, a non-empty finite subset of $G$ is a subgroup {{iff}} it is closed. | === Sufficient Condition ===
Let $H$ be a subgroup of $G$.
Then:
:$\forall a, b \in H: a \circ b \in H$
by definition of subgroup.
{{qed|lemma}}
=== Necessary Condition ===
Let $H$ be a non-empty finite subset of $G$ such that:
:$\forall a, b \in H: a \circ b \in H$
Let $x \in H$.
We have by hypothesis that $H$ is clos... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $H$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Finite Set|finite]] [[Definition:Subset|subset]] of $G$.
Then:
:$H$ is a [[Definition:Subgroup|subgroup]] of $G$
{{iff}}:
:$\forall a, b \in H: a \circ b \in H$
That is, a [[Definition:Non-Emp... | === Sufficient Condition ===
Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
Then:
:$\forall a, b \in H: a \circ b \in H$
by definition of [[Definition:Subgroup|subgroup]].
{{qed|lemma}}
=== Necessary Condition ===
Let $H$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Finite Set|finite]] [[Definit... | Finite Subgroup Test/Proof 2 | https://proofwiki.org/wiki/Finite_Subgroup_Test | https://proofwiki.org/wiki/Finite_Subgroup_Test/Proof_2 | [
"Subgroups",
"Finite Subgroup Test"
] | [
"Definition:Group",
"Definition:Non-Empty Set",
"Definition:Finite Set",
"Definition:Subset",
"Definition:Subgroup",
"Definition:Non-Empty Set",
"Definition:Finite Set",
"Definition:Subset",
"Definition:Subgroup",
"Definition:Closure (Abstract Algebra)/Algebraic Structure"
] | [
"Definition:Subgroup",
"Definition:Subgroup",
"Definition:Non-Empty Set",
"Definition:Finite Set",
"Definition:Subset",
"Definition:By Hypothesis",
"Definition:Closure (Abstract Algebra)/Algebraic Structure",
"Definition:Element",
"Definition:Finite Set",
"Powers of Group Elements/Sum of Indices",... |
proofwiki-1026 | Powers of Element form Subgroup | Let $\struct {G, \circ}$ be a group.
Then:
:$\forall a \in G: H = \set {a^n: n \in \Z} \le G$
That is, the subset of $G$ comprising all elements possible as powers of $a \in G$ is a subgroup of $G$. | Clearly $a \in H$, so $H \ne \O$.
Let $x, y \in H$.
{{begin-eqn}}
{{eqn | o =
| r = x, y \in H
| c =
}}
{{eqn | o = \leadsto
| r = \exists m, n \in \Z: x = a^m, y = a^n
| c =
}}
{{eqn | o = \leadsto
| r = x^{-1} y = \paren {a^m}^{-1} a^n
| c =
}}
{{eqn | o = \leadsto
| r = ... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Then:
:$\forall a \in G: H = \set {a^n: n \in \Z} \le G$
That is, the [[Definition:Subset|subset]] of $G$ comprising all [[Definition:Element|elements]] possible as powers of $a \in G$ is a [[Definition:Subgroup|subgroup]] of $G$. | Clearly $a \in H$, so $H \ne \O$.
Let $x, y \in H$.
{{begin-eqn}}
{{eqn | o =
| r = x, y \in H
| c =
}}
{{eqn | o = \leadsto
| r = \exists m, n \in \Z: x = a^m, y = a^n
| c =
}}
{{eqn | o = \leadsto
| r = x^{-1} y = \paren {a^m}^{-1} a^n
| c =
}}
{{eqn | o = \leadsto
| r... | Powers of Element form Subgroup | https://proofwiki.org/wiki/Powers_of_Element_form_Subgroup | https://proofwiki.org/wiki/Powers_of_Element_form_Subgroup | [
"Subgroups"
] | [
"Definition:Group",
"Definition:Subset",
"Definition:Element",
"Definition:Subgroup"
] | [
"One-Step Subgroup Test"
] |
proofwiki-1027 | Existence of Unique Subgroup Generated by Subset | Let $\struct {G, \circ}$ be a group.
Let $\O \subset S \subseteq G$.
Let $\struct {H, \circ}$ be the subgroup generated by $S$.
Then $H = \gen S$ exists and is unique.
Also, $\struct {H, \circ}$ is the intersection of all of the subgroups of $G$ which contain the set $S$:
:$\ds \gen S = \bigcap_i {H_i}: S \subseteq H_i... | === Existence ===
First, we prove that such a subgroup exists.
Let $\mathbb S$ be the set of all subgroups of $G$ which contain $S$.
$\mathbb S \ne \O$ because $G$ is itself a subgroup of $G$, and thus $G \in \mathbb S$.
Let $H$ be the intersection of all the elements of $\mathbb S$.
By Intersection of Subgroups is Sub... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $\O \subset S \subseteq G$.
Let $\struct {H, \circ}$ be the [[Definition:Generator of Subgroup|subgroup generated by $S$]].
Then $H = \gen S$ exists and is [[Definition:Unique|unique]].
Also, $\struct {H, \circ}$ is the [[Definition:Set Intersection|int... | === Existence ===
First, we prove that such a [[Definition:Subgroup|subgroup]] exists.
Let $\mathbb S$ be the [[Definition:Set|set]] of all [[Definition:Subgroup|subgroups]] of $G$ which contain $S$.
$\mathbb S \ne \O$ because [[Group is Subgroup of Itself|$G$ is itself a subgroup]] of $G$, and thus $G \in \mathbb S... | Existence of Unique Subgroup Generated by Subset | https://proofwiki.org/wiki/Existence_of_Unique_Subgroup_Generated_by_Subset | https://proofwiki.org/wiki/Existence_of_Unique_Subgroup_Generated_by_Subset | [
"Group Theory",
"Generated Subgroups"
] | [
"Definition:Group",
"Definition:Generator of Subgroup",
"Definition:Unique",
"Definition:Set Intersection",
"Definition:Subgroup"
] | [
"Definition:Subgroup",
"Definition:Set",
"Definition:Subgroup",
"Group is Subgroup of Itself",
"Definition:Set Intersection",
"Definition:Element",
"Intersection of Subgroups is Subgroup",
"Definition:Subgroup",
"Definition:Subgroup",
"Definition:Subgroup",
"Definition:Subgroup"
] |
proofwiki-1028 | Homomorphism of Generated Group | Let $\struct {G, \circ}$ and $\struct {H, \circ}$ be groups.
Let $\phi: G \to H$ and $\psi: G \to H$ be homomorphisms.
Let $\gen S = G$ be the group generated by $S$.
Let:
:$\forall x \in S: \map \phi x = \map \psi x$
Then:
:$\phi = \psi$ | Let $H = \set {x \in G: \map \phi x = \map \psi x}$.
From Elements of Group with Equal Images under Homomorphisms form Subgroup, $H$ is a subgroup of $G$.
But from the definition of the group generated by $S$, the smallest subgroup that contains $S$ is $G$ itself.
Thus:
:$G = \set {x \in G: \map \phi x = \map \psi x}$
... | Let $\struct {G, \circ}$ and $\struct {H, \circ}$ be [[Definition:Group|groups]].
Let $\phi: G \to H$ and $\psi: G \to H$ be [[Definition:Group Homomorphism|homomorphisms]].
Let $\gen S = G$ be the [[Definition:Generator of Group|group generated by $S$]].
Let:
:$\forall x \in S: \map \phi x = \map \psi x$
Then:
:... | Let $H = \set {x \in G: \map \phi x = \map \psi x}$.
From [[Elements of Group with Equal Images under Homomorphisms form Subgroup]], $H$ is a [[Definition:Subgroup|subgroup]] of $G$.
But from the definition of the [[Definition:Generator of Group|group generated by $S$]], the smallest subgroup that contains $S$ is $G$... | Homomorphism of Generated Group | https://proofwiki.org/wiki/Homomorphism_of_Generated_Group | https://proofwiki.org/wiki/Homomorphism_of_Generated_Group | [
"Group Homomorphisms",
"Generators of Groups"
] | [
"Definition:Group",
"Definition:Group Homomorphism",
"Definition:Generator of Group"
] | [
"Elements of Group with Equal Images under Homomorphisms form Subgroup",
"Definition:Subgroup",
"Definition:Generator of Group"
] |
proofwiki-1029 | Set of Words Generates Group | Let $S \subseteq G$ where $G$ is a group.
Let $\hat S$ be defined as $S \cup S^{-1}$, where $S^{-1}$ is the set of all the inverses of all the elements of $S$.
Then $\gen S = \map W {\hat S}$, where $\map W {\hat S}$ is the set of words of $\hat S$. | Let $H = \gen S$ where $S \subseteq G$.
$H$ must certainly include $\hat S$, because any group containing $s \in S$ must also contain $s^{-1}$.
Thus $\hat S \subseteq H$.
By {{Group-axiom|0}}, $H$ must also contain all products of a finite number of elements of $\hat S$.
Thus $\map W {\hat S} \subseteq H$.
Now we prove... | Let $S \subseteq G$ where $G$ is a [[Definition:Group|group]].
Let $\hat S$ be defined as $S \cup S^{-1}$, where $S^{-1}$ is the [[Definition:Inverse of Subset of Group|set of all the inverses]] of all the elements of $S$.
Then $\gen S = \map W {\hat S}$, where $\map W {\hat S}$ is the [[Definition:Word (Abstract Al... | Let $H = \gen S$ where $S \subseteq G$.
$H$ must certainly include $\hat S$, because any [[Definition:Group|group]] containing $s \in S$ must also contain $s^{-1}$.
Thus $\hat S \subseteq H$.
By {{Group-axiom|0}}, $H$ must also contain all [[Definition:Product Element|products]] of a [[Definition:Finite Set|finite n... | Set of Words Generates Group | https://proofwiki.org/wiki/Set_of_Words_Generates_Group | https://proofwiki.org/wiki/Set_of_Words_Generates_Group | [
"Group Theory",
"Generated Subgroups",
"Set of Words Generates Group"
] | [
"Definition:Group",
"Definition:Inverse of Subset/Group",
"Definition:Word (Abstract Algebra)"
] | [
"Definition:Group",
"Definition:Group Product/Product Element",
"Definition:Finite Set",
"Definition:Element",
"Two-Step Subgroup Test",
"Definition:Group Product/Product Element",
"Definition:Finite Set",
"Definition:Element",
"Definition:Group Product/Product Element",
"Two-Step Subgroup Test",
... |
proofwiki-1030 | Subset Product is Subset of Generator | Let $\struct {G, \circ}$ be a group.
Let $X, Y \subseteq \struct {G, \circ}$.
Then $X \circ Y \subseteq \gen {X, Y}$ where:
:$X \circ Y$ is the Subset Product of $X$ and $Y$ in $G$.
:$\gen {X, Y}$ is the subgroup of $G$ generated by $X$ and $Y$. | It is clear from Set of Words Generates Group that $\map W {\hat X \cup \hat Y} = \gen {X, Y}$.
It is equally clear that $X \circ Y \subseteq \map W {\hat X \cup \hat Y}$.
{{qed}}
Category:Group Theory
Category:Subset Products
3cekksmrfte9xtmhqnc5h4td9k4wcc6 | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $X, Y \subseteq \struct {G, \circ}$.
Then $X \circ Y \subseteq \gen {X, Y}$ where:
:$X \circ Y$ is the [[Definition:Subset Product|Subset Product]] of $X$ and $Y$ in $G$.
:$\gen {X, Y}$ is the [[Definition:Generator of Subgroup|subgroup of $G$ generated b... | It is clear from [[Set of Words Generates Group]] that $\map W {\hat X \cup \hat Y} = \gen {X, Y}$.
It is equally clear that $X \circ Y \subseteq \map W {\hat X \cup \hat Y}$.
{{qed}}
[[Category:Group Theory]]
[[Category:Subset Products]]
3cekksmrfte9xtmhqnc5h4td9k4wcc6 | Subset Product is Subset of Generator | https://proofwiki.org/wiki/Subset_Product_is_Subset_of_Generator | https://proofwiki.org/wiki/Subset_Product_is_Subset_of_Generator | [
"Group Theory",
"Subset Products"
] | [
"Definition:Group",
"Definition:Subset Product",
"Definition:Generator of Subgroup"
] | [
"Set of Words Generates Group",
"Category:Group Theory",
"Category:Subset Products"
] |
proofwiki-1031 | Order of Subset Product with Singleton | Let $\struct {G, \circ}$ be a group.
Let $X, Y \subseteq \struct {G, \circ}$ such that $X$ is a singleton:
:$X = \set x$
Then:
:$\order {X \circ Y} = \order Y = \order {Y \circ X}$
where $\order S$ is defined as the order of $S$. | From Regular Representations of Subset Product, we have that the left regular representation of $\struct {S, \circ}$ with respect to $a$ is:
:$\lambda_x \sqbrk S = \set x \circ S = x \circ S$
The result then follows directly from Regular Representation of Invertible Element is Permutation.
{{qed}} | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $X, Y \subseteq \struct {G, \circ}$ such that $X$ is a [[Definition:Singleton|singleton]]:
:$X = \set x$
Then:
:$\order {X \circ Y} = \order Y = \order {Y \circ X}$
where $\order S$ is defined as the [[Definition:Order of Structure|order of $S$]]. | From [[Regular Representations of Subset Product]], we have that the [[Definition:Left Regular Representation|left regular representation]] of $\struct {S, \circ}$ with respect to $a$ is:
:$\lambda_x \sqbrk S = \set x \circ S = x \circ S$
The result then follows directly from [[Regular Representation of Invertible Ele... | Order of Subset Product with Singleton/Proof 1 | https://proofwiki.org/wiki/Order_of_Subset_Product_with_Singleton | https://proofwiki.org/wiki/Order_of_Subset_Product_with_Singleton/Proof_1 | [
"Subset Products",
"Singletons",
"Order of Subset Product with Singleton"
] | [
"Definition:Group",
"Definition:Singleton",
"Definition:Order of Structure"
] | [
"Regular Representations of Subset Product",
"Definition:Regular Representations/Left Regular Representation",
"Regular Representation of Invertible Element is Permutation"
] |
proofwiki-1032 | Product of Subset with Intersection | Let $\struct {G, \circ}$ be an algebraic structure.
Let $X, Y, Z \subseteq G$.
Then:
:$X \circ \paren {Y \cap Z} \subseteq \paren {X \circ Y} \cap \paren {X \circ Z}$
:$\paren {Y \cap Z} \circ X \subseteq \paren {Y \circ X} \cap \paren {Z \circ X}$
where $X \circ Y$ denotes the subset product of $X$ and $Y$. | Let $x \in X, t \in Y \cap Z$.
By the definition of intersection, $t \in Y$ and $t \in Z$.
Consider $X \circ \paren {Y \cap Z}$.
We have $x \circ t \in X \circ \paren {Y \cap Z}$ by definition of subset product.
As $t \in Y$ and $t \in Z$, we also have $x \circ t \in X \circ Y$ and $x \circ t \in X \circ Z$.
The result... | Let $\struct {G, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]].
Let $X, Y, Z \subseteq G$.
Then:
:$X \circ \paren {Y \cap Z} \subseteq \paren {X \circ Y} \cap \paren {X \circ Z}$
:$\paren {Y \cap Z} \circ X \subseteq \paren {Y \circ X} \cap \paren {Z \circ X}$
where $X \ci... | Let $x \in X, t \in Y \cap Z$.
By the definition of [[Definition:Set Intersection|intersection]], $t \in Y$ and $t \in Z$.
Consider $X \circ \paren {Y \cap Z}$.
We have $x \circ t \in X \circ \paren {Y \cap Z}$ by definition of [[Definition:Subset Product|subset product]].
As $t \in Y$ and $t \in Z$, we also have ... | Product of Subset with Intersection/Proof 1 | https://proofwiki.org/wiki/Product_of_Subset_with_Intersection | https://proofwiki.org/wiki/Product_of_Subset_with_Intersection/Proof_1 | [
"Subset Products",
"Set Intersection",
"Product of Subset with Intersection"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Subset Product"
] | [
"Definition:Set Intersection",
"Definition:Subset Product",
"Definition:Subset Product"
] |
proofwiki-1033 | Product of Subset with Intersection | Let $\struct {G, \circ}$ be an algebraic structure.
Let $X, Y, Z \subseteq G$.
Then:
:$X \circ \paren {Y \cap Z} \subseteq \paren {X \circ Y} \cap \paren {X \circ Z}$
:$\paren {Y \cap Z} \circ X \subseteq \paren {Y \circ X} \cap \paren {Z \circ X}$
where $X \circ Y$ denotes the subset product of $X$ and $Y$. | Consider the relation $\RR \subseteq G \times G$ defined as:
:$\forall g, h \in G: \tuple {g, h} \in \RR \iff \exists g \in X$
Then:
:$\forall S \subseteq G: X \circ S = \RR \sqbrk S$
Then:
{{begin-eqn}}
{{eqn | l = X \circ \paren {Y \cap Z}
| r = \RR \sqbrk {Y \cap Z}
| c =
}}
{{eqn | o = \subseteq
... | Let $\struct {G, \circ}$ be an [[Definition:Algebraic Structure with One Operation|algebraic structure]].
Let $X, Y, Z \subseteq G$.
Then:
:$X \circ \paren {Y \cap Z} \subseteq \paren {X \circ Y} \cap \paren {X \circ Z}$
:$\paren {Y \cap Z} \circ X \subseteq \paren {Y \circ X} \cap \paren {Z \circ X}$
where $X \ci... | Consider the [[Definition:Relation|relation]] $\RR \subseteq G \times G$ defined as:
:$\forall g, h \in G: \tuple {g, h} \in \RR \iff \exists g \in X$
Then:
:$\forall S \subseteq G: X \circ S = \RR \sqbrk S$
Then:
{{begin-eqn}}
{{eqn | l = X \circ \paren {Y \cap Z}
| r = \RR \sqbrk {Y \cap Z}
| c =
}}
... | Product of Subset with Intersection/Proof 2 | https://proofwiki.org/wiki/Product_of_Subset_with_Intersection | https://proofwiki.org/wiki/Product_of_Subset_with_Intersection/Proof_2 | [
"Subset Products",
"Set Intersection",
"Product of Subset with Intersection"
] | [
"Definition:Algebraic Structure/One Operation",
"Definition:Subset Product"
] | [
"Definition:Relation",
"Image of Intersection under Relation",
"Definition:Relation",
"Image of Intersection under Relation"
] |
proofwiki-1034 | Order of Subgroup Product | Let $G$ be a group.
Let $H$ and $K$ be subgroups of $G$.
Then:
:$\order {H K} = \dfrac {\order H \order K} {\order {H \cap K} }$
where:
:$H K$ denotes subset product
:$\order H$ denotes the order of $H$.
{{questionable|The {{RHS}} does not make sense if $\order {H \cap K} {{=}} +\infty$, because it is then $\dfrac {+\i... | From Intersection of Subgroups is Subgroup, we have that $H \cap K \le H$.
Let the number of left cosets of $H \cap K$ in $H$ be $r$.
Then the left coset space of $H \cap K$ in $H$ is:
: $\set {x_1 \paren {H \cap K}, x_2 \paren {H \cap K}, \ldots, x_r \paren {H \cap K} }$
So each element of $H$ is in $x_i \paren {H \ca... | Let $G$ be a [[Definition:Group|group]].
Let $H$ and $K$ be [[Definition:Subgroup|subgroups]] of $G$.
Then:
:$\order {H K} = \dfrac {\order H \order K} {\order {H \cap K} }$
where:
:$H K$ denotes [[Definition:Subset Product|subset product]]
:$\order H$ denotes the [[Definition:Order of Structure|order of $H$]].
{{q... | From [[Intersection of Subgroups is Subgroup]], we have that $H \cap K \le H$.
Let the number of [[Definition:Left Coset|left cosets]] of $H \cap K$ in $H$ be $r$.
Then the [[Definition:Left Coset Space|left coset space]] of $H \cap K$ in $H$ is:
: $\set {x_1 \paren {H \cap K}, x_2 \paren {H \cap K}, \ldots, x_r \pa... | Order of Subgroup Product/Proof 1 | https://proofwiki.org/wiki/Order_of_Subgroup_Product | https://proofwiki.org/wiki/Order_of_Subgroup_Product/Proof_1 | [
"Subgroups",
"Subset Products",
"Order of Subgroup Product"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Subset Product",
"Definition:Order of Structure"
] | [
"Intersection of Subgroups is Subgroup",
"Definition:Coset/Left Coset",
"Definition:Coset Space/Left Coset Space",
"Definition:Coset/Left Coset",
"Definition:Disjoint Sets",
"Left Coset Space forms Partition",
"Left Congruence Class Modulo Subgroup is Left Coset",
"Definition:Contradiction",
"Defini... |
proofwiki-1035 | Order of Subgroup Product | Let $G$ be a group.
Let $H$ and $K$ be subgroups of $G$.
Then:
:$\order {H K} = \dfrac {\order H \order K} {\order {H \cap K} }$
where:
:$H K$ denotes subset product
:$\order H$ denotes the order of $H$.
{{questionable|The {{RHS}} does not make sense if $\order {H \cap K} {{=}} +\infty$, because it is then $\dfrac {+\i... | === Lemma ===
{{:Order of Subgroup Product/Lemma}}{{qed|lemma}}
We have that $H K$ is the union of all left cosets $h K$ with $h \in H$:
:$\ds H K = \bigcup_{h \mathop \in H} h K$
From Left Coset Space forms Partition, unequal $h K$ are disjoint.
From Cosets are Equivalent, each $h K$ contains $\order K$ elements.
From... | Let $G$ be a [[Definition:Group|group]].
Let $H$ and $K$ be [[Definition:Subgroup|subgroups]] of $G$.
Then:
:$\order {H K} = \dfrac {\order H \order K} {\order {H \cap K} }$
where:
:$H K$ denotes [[Definition:Subset Product|subset product]]
:$\order H$ denotes the [[Definition:Order of Structure|order of $H$]].
{{q... | === [[Order of Subgroup Product/Lemma|Lemma]] ===
{{:Order of Subgroup Product/Lemma}}{{qed|lemma}}
We have that $H K$ is the [[Definition:Set Union|union]] of all [[Definition:Left Coset|left cosets]] $h K$ with $h \in H$:
:$\ds H K = \bigcup_{h \mathop \in H} h K$
From [[Left Coset Space forms Partition]], unequal ... | Order of Subgroup Product/Proof 2 | https://proofwiki.org/wiki/Order_of_Subgroup_Product | https://proofwiki.org/wiki/Order_of_Subgroup_Product/Proof_2 | [
"Subgroups",
"Subset Products",
"Order of Subgroup Product"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Subset Product",
"Definition:Order of Structure"
] | [
"Order of Subgroup Product/Lemma",
"Definition:Set Union",
"Definition:Coset/Left Coset",
"Left Coset Space forms Partition",
"Definition:Disjoint Sets",
"Cosets are Equivalent",
"Definition:Element",
"Order of Subgroup Product/Lemma",
"Definition:Coset/Left Coset",
"Definition:Index of Subgroup",... |
proofwiki-1036 | Order of Subgroup Product | Let $G$ be a group.
Let $H$ and $K$ be subgroups of $G$.
Then:
:$\order {H K} = \dfrac {\order H \order K} {\order {H \cap K} }$
where:
:$H K$ denotes subset product
:$\order H$ denotes the order of $H$.
{{questionable|The {{RHS}} does not make sense if $\order {H \cap K} {{=}} +\infty$, because it is then $\dfrac {+\i... | The number of product elements $h k$ that can be formed where $h \in H$ and $k \in K$ is $\order H \order K$, although this may include duplication:
:$h_1 k_1 = h_2 k_2$ may be possible for $h_1, h_2 \in H, k_1, k_2 \in K$.
So, consider the Cartesian product $H \times K$.
From Cardinality of Cartesian Product of Finite... | Let $G$ be a [[Definition:Group|group]].
Let $H$ and $K$ be [[Definition:Subgroup|subgroups]] of $G$.
Then:
:$\order {H K} = \dfrac {\order H \order K} {\order {H \cap K} }$
where:
:$H K$ denotes [[Definition:Subset Product|subset product]]
:$\order H$ denotes the [[Definition:Order of Structure|order of $H$]].
{{q... | The number of [[Definition:Product Element|product elements]] $h k$ that can be formed where $h \in H$ and $k \in K$ is $\order H \order K$, although this may include duplication:
:$h_1 k_1 = h_2 k_2$ may be possible for $h_1, h_2 \in H, k_1, k_2 \in K$.
So, consider the [[Definition:Cartesian Product|Cartesian produc... | Order of Subgroup Product/Proof 3 | https://proofwiki.org/wiki/Order_of_Subgroup_Product | https://proofwiki.org/wiki/Order_of_Subgroup_Product/Proof_3 | [
"Subgroups",
"Subset Products",
"Order of Subgroup Product"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Subset Product",
"Definition:Order of Structure"
] | [
"Definition:Group Product/Product Element",
"Definition:Cartesian Product",
"Cardinality of Cartesian Product of Finite Sets",
"Definition:Relation",
"Definition:Equals",
"Definition:Equivalence Relation",
"Definition:Reflexive Relation",
"Definition:Symmetric Relation",
"Definition:Transitive Relat... |
proofwiki-1037 | Index of Intersection of Subgroups | Let $G$ be a group.
Let $H, K$ be subgroups of finite index of $G$.
Then:
:$\index G {H \cap K} \le \index G H \index G K$
where $\index G H$ denotes the index of $H$ in $G$.
Note that here the symbol $\le$ is being used with its meaning '''less than or equal to'''.
Equality holds {{iff}} $H K = \set {h k: h \in H, k \... | Note that $H \cap K$ is a subgroup of $H$.
From Tower Law for Subgroups, we have:
:$\index G {H \cap K} = \index G H \index H {H \cap K}$
From Index in Subgroup, also:
:$\index H {H \cap K} \le \index G K$
Combining these results yields the desired inequality.
Again from Index in Subgroup, it follows that:
:$\index H {... | Let $G$ be a [[Definition:Group|group]].
Let $H, K$ be [[Definition:Subgroup|subgroups]] of [[Definition:Finite Index|finite index]] of $G$.
Then:
:$\index G {H \cap K} \le \index G H \index G K$
where $\index G H$ denotes the [[Definition:Index of Subgroup|index of $H$ in $G$]].
Note that here the symbol $\le$ i... | Note that $H \cap K$ is a [[Definition:Subgroup|subgroup]] of $H$.
From [[Tower Law for Subgroups]], we have:
:$\index G {H \cap K} = \index G H \index H {H \cap K}$
From [[Index in Subgroup]], also:
:$\index H {H \cap K} \le \index G K$
Combining these results yields the desired inequality.
Again from [[Index in... | Index of Intersection of Subgroups | https://proofwiki.org/wiki/Index_of_Intersection_of_Subgroups | https://proofwiki.org/wiki/Index_of_Intersection_of_Subgroups | [
"Subgroups",
"Index of Subgroups",
"Set Intersection",
"Index of Intersection of Subgroups"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Index of Subgroup/Finite",
"Definition:Index of Subgroup"
] | [
"Definition:Subgroup",
"Tower Law for Subgroups",
"Index in Subgroup",
"Index in Subgroup"
] |
proofwiki-1038 | Intersection of Subgroups of Prime Order | Let $G$ be a group whose identity is $e$.
Let $H$ and $K$ be subsets of $G$ such that:
:$\order H = \order K = p$
:$H \ne K$
:$p$ is prime.
Then:
: $H \cap K = \set e$
That is, the intersection of two unequal subgroups of a group, both of whose order is the same prime, consists solely of the identity. | From Intersection of Subgroups is Subgroup:
:$H \cap K \le G$
and:
:$H \cap K \le H$
where $\le$ denotes subgrouphood.
So:
{{begin-eqn}}
{{eqn | l = H \cap K
| o = \le
| r = H
| c = Intersection of Subgroups is Subgroup
}}
{{eqn | ll= \leadsto
| l = \order {H \cap K}
| o = \divides
|... | Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $H$ and $K$ be [[Definition:Subset|subsets]] of $G$ such that:
:$\order H = \order K = p$
:$H \ne K$
:$p$ is [[Definition:Prime Number|prime]].
Then:
: $H \cap K = \set e$
That is, the [[Definition:Set Intersection|i... | From [[Intersection of Subgroups is Subgroup]]:
:$H \cap K \le G$
and:
:$H \cap K \le H$
where $\le$ denotes [[Definition:Subgroup|subgrouphood]].
So:
{{begin-eqn}}
{{eqn | l = H \cap K
| o = \le
| r = H
| c = [[Intersection of Subgroups is Subgroup]]
}}
{{eqn | ll= \leadsto
| l = \order {H ... | Intersection of Subgroups of Prime Order | https://proofwiki.org/wiki/Intersection_of_Subgroups_of_Prime_Order | https://proofwiki.org/wiki/Intersection_of_Subgroups_of_Prime_Order | [
"Prime Groups",
"Subgroups"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Subset",
"Definition:Prime Number",
"Definition:Set Intersection",
"Definition:Subgroup",
"Definition:Group",
"Definition:Order of Structure",
"Definition:Prime Number",
"Definition:Identity (Abstract Alge... | [
"Intersection of Subgroups is Subgroup",
"Definition:Subgroup",
"Intersection of Subgroups is Subgroup",
"Lagrange's Theorem (Group Theory)",
"Definition:Prime Number",
"Intersection with Subset is Subset"
] |
proofwiki-1039 | Tower Law for Subgroups | Let $\struct {G, \circ}$ be a group.
Let $H$ be a subgroup of $G$ with finite index.
Let $K$ be a subgroup of $H$.
Then:
:$\index G K = \index G H \index H K$
where $\index G H$ denotes the index of $H$ in $G$. | Let $p = \index G H$, $q = \index H K$.
By hypothesis these numbers are finite.
Therefore, there exist $g_1, \ldots, g_p \in G$ such that $G$ is a disjoint union:
$\ds G = \bigsqcup_{i \mathop = 1}^p g_i H$
Similarly, there exist $h_1,\ldots,h_q \in H$ such that $H$ is a disjoint union:
$\ds H = \bigsqcup_{j \mathop = ... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$ with [[Definition:Finite Index|finite index]].
Let $K$ be a [[Definition:Subgroup|subgroup]] of $H$.
Then:
:$\index G K = \index G H \index H K$
where $\index G H$ denotes the [[Definition:Index of Subgro... | Let $p = \index G H$, $q = \index H K$.
By hypothesis these numbers are [[Definition:Finite|finite]].
Therefore, there exist $g_1, \ldots, g_p \in G$ such that $G$ is a [[Definition:Disjoint Union (Set Theory)|disjoint union]]:
$\ds G = \bigsqcup_{i \mathop = 1}^p g_i H$
Similarly, there exist $h_1,\ldots,h_q \in H$... | Tower Law for Subgroups/Proof 1 | https://proofwiki.org/wiki/Tower_Law_for_Subgroups | https://proofwiki.org/wiki/Tower_Law_for_Subgroups/Proof_1 | [
"Subgroups",
"Tower Law for Subgroups",
"Named Theorems",
"Index of Subgroups"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Index of Subgroup/Finite",
"Definition:Subgroup",
"Definition:Index of Subgroup"
] | [
"Definition:Finite",
"Definition:Disjoint Union (Set Theory)",
"Definition:Disjoint Union (Set Theory)",
"Product of Subset with Union",
"Subset Product within Semigroup is Associative/Corollary",
"Definition:Disjoint Union (Set Theory)",
"Definition:Coset",
"Definition:Element",
"Definition:Coset S... |
proofwiki-1040 | Tower Law for Subgroups | Let $\struct {G, \circ}$ be a group.
Let $H$ be a subgroup of $G$ with finite index.
Let $K$ be a subgroup of $H$.
Then:
:$\index G K = \index G H \index H K$
where $\index G H$ denotes the index of $H$ in $G$. | Assume $G$ is finite.
Then:
{{begin-eqn}}
{{eqn | l = \index G H
| r = \frac {\order G} {\order H}
| c = Lagrange's Theorem
}}
{{eqn | l = \index G K
| r = \frac {\order G} {\order K}
| c = Lagrange's Theorem
}}
{{eqn | ll= \leadsto
| l = \index G K
| r = \frac {\order H} {\order K} ... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$ with [[Definition:Finite Index|finite index]].
Let $K$ be a [[Definition:Subgroup|subgroup]] of $H$.
Then:
:$\index G K = \index G H \index H K$
where $\index G H$ denotes the [[Definition:Index of Subgro... | Assume $G$ is [[Definition:Finite Group|finite]].
Then:
{{begin-eqn}}
{{eqn | l = \index G H
| r = \frac {\order G} {\order H}
| c = [[Lagrange's Theorem (Group Theory)|Lagrange's Theorem]]
}}
{{eqn | l = \index G K
| r = \frac {\order G} {\order K}
| c = [[Lagrange's Theorem (Group Theory)|La... | Tower Law for Subgroups/Proof 2 | https://proofwiki.org/wiki/Tower_Law_for_Subgroups | https://proofwiki.org/wiki/Tower_Law_for_Subgroups/Proof_2 | [
"Subgroups",
"Tower Law for Subgroups",
"Named Theorems",
"Index of Subgroups"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Index of Subgroup/Finite",
"Definition:Subgroup",
"Definition:Index of Subgroup"
] | [
"Definition:Finite Group",
"Lagrange's Theorem (Group Theory)",
"Lagrange's Theorem (Group Theory)",
"Lagrange's Theorem (Group Theory)"
] |
proofwiki-1041 | Morphism from Integers to Group | Let $G$ be a group whose identity is $e$.
Let $g \in G$.
Let $\phi: \Z \to G$ be the mapping defined as:
:$\forall n \in \Z: \map \phi n = g^n$.
Then:
:If $g$ has infinite order, then $\phi$ is a group isomorphism from $\struct {\Z, +}$ to $\gen g$.
:If $g$ has finite order such that $\order g = m$, then $\phi$ is a gr... | By Epimorphism from Integers to Cyclic Group, $\phi$ is an epimorphism from $\struct {\Z, +}$ onto $\gen g$.
By Kernel of Group Homomorphism is Subgroup, the kernel $K$ of $G$ is a subgroup of $\struct {\Z, +}$.
Therefore by Subgroup of Integers is Ideal and Principal Ideals of Integers, $\exists m \in \N_{>0}: K = \pa... | Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $g \in G$.
Let $\phi: \Z \to G$ be the [[Definition:Mapping|mapping]] defined as:
:$\forall n \in \Z: \map \phi n = g^n$.
Then:
:If $g$ has [[Definition:Infinite Order Element|infinite order]], then $\phi$ is a [[De... | By [[Epimorphism from Integers to Cyclic Group]], $\phi$ is an [[Definition:Group Epimorphism|epimorphism]] from $\struct {\Z, +}$ onto $\gen g$.
By [[Kernel of Group Homomorphism is Subgroup]], the [[Definition:Kernel of Group Homomorphism|kernel]] $K$ of $G$ is a [[Definition:Subgroup|subgroup]] of $\struct {\Z, +}$... | Morphism from Integers to Group | https://proofwiki.org/wiki/Morphism_from_Integers_to_Group | https://proofwiki.org/wiki/Morphism_from_Integers_to_Group | [
"Group Isomorphisms",
"Group Epimorphisms",
"Integers",
"Ideal Theory"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Mapping",
"Definition:Order of Group Element/Infinite",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Order of Group Element/Finite",
"Definition:Group Epimorphism",
"Definition:Ke... | [
"Epimorphism from Integers to Cyclic Group",
"Definition:Group Epimorphism",
"Kernel of Group Homomorphism is Subgroup",
"Definition:Kernel of Group Homomorphism",
"Definition:Subgroup",
"Subgroup of Integers is Ideal",
"Principal Ideals of Integers",
"Quotient Epimorphism from Integers by Principal I... |
proofwiki-1042 | Identity is Only Group Element of Order 1 | In every group, the identity, and only the identity, has order $1$. | Let $G$ be a group with identity $e$.
Then:
:$e^1 = e$
and:
:$\forall a \in G: a \ne e: a^1 = a \ne e$.
Hence the result.
{{qed}} | In every [[Definition:Group|group]], the [[Definition:Identity Element|identity]], and only the [[Definition:Identity Element|identity]], has [[Definition:Order of Group Element|order]] $1$. | Let $G$ be a [[Definition:Group|group]] with [[Definition:Identity Element|identity]] $e$.
Then:
:$e^1 = e$
and:
:$\forall a \in G: a \ne e: a^1 = a \ne e$.
Hence the result.
{{qed}} | Identity is Only Group Element of Order 1 | https://proofwiki.org/wiki/Identity_is_Only_Group_Element_of_Order_1 | https://proofwiki.org/wiki/Identity_is_Only_Group_Element_of_Order_1 | [
"Identity Elements",
"Order of Group Elements"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Order of Group Element"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity"
] |
proofwiki-1043 | Group Element is Self-Inverse iff Order 2 | Let $\struct {S, \circ}$ be a group whose identity is $e$.
An element $x \in \struct {S, \circ}$ is self-inverse {{iff}}:
:$\order x = 2$ | Let $x \in G: x \ne e$.
{{begin-eqn}}
{{eqn | l = \order x
| r = 2
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x \circ x
| r = e
| c = {{Defof|Order of Group Element}}
}}
{{eqn | ll= \leadstoandfrom
| l = x
| r = x^{-1}
| c = Equivalence of Definitions of Self-Inverse
}}
... | Let $\struct {S, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
An [[Definition:Element|element]] $x \in \struct {S, \circ}$ is [[Definition:Self-Inverse Element|self-inverse]] {{iff}}:
:$\order x = 2$ | Let $x \in G: x \ne e$.
{{begin-eqn}}
{{eqn | l = \order x
| r = 2
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x \circ x
| r = e
| c = {{Defof|Order of Group Element}}
}}
{{eqn | ll= \leadstoandfrom
| l = x
| r = x^{-1}
| c = [[Equivalence of Definitions of Self-Inverse]... | Group Element is Self-Inverse iff Order 2 | https://proofwiki.org/wiki/Group_Element_is_Self-Inverse_iff_Order_2 | https://proofwiki.org/wiki/Group_Element_is_Self-Inverse_iff_Order_2 | [
"Order of Group Elements"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Element",
"Definition:Self-Inverse Element"
] | [
"Equivalence of Definitions of Self-Inverse"
] |
proofwiki-1044 | Powers of Infinite Order Element | Let $G$ be a group whose identity is $e$.
Let $a \in G$ have infinite order in $G$.
Then:
:$\forall m, n \in \Z: m \ne n \implies a^m \ne a^n$ | Let $m, n \in \Z$.
Then:
{{begin-eqn}}
{{eqn | l = a^m
| r = a^n
| c =
}}
{{eqn | ll= \leadsto
| l = a^m \paren {a^n}^{-1}
| r = e
| c =
}}
{{eqn | lo= \land
| l = a^n \paren {a^m}^{-1}
| r = e
| c =
}}
{{eqn | ll= \leadsto
| l = a^{m - n}
| r = a^{n - m} =... | Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $a \in G$ have [[Definition:Infinite Order Element|infinite order]] in $G$.
Then:
:$\forall m, n \in \Z: m \ne n \implies a^m \ne a^n$ | Let $m, n \in \Z$.
Then:
{{begin-eqn}}
{{eqn | l = a^m
| r = a^n
| c =
}}
{{eqn | ll= \leadsto
| l = a^m \paren {a^n}^{-1}
| r = e
| c =
}}
{{eqn | lo= \land
| l = a^n \paren {a^m}^{-1}
| r = e
| c =
}}
{{eqn | ll= \leadsto
| l = a^{m - n}
| r = a^{n - m}... | Powers of Infinite Order Element | https://proofwiki.org/wiki/Powers_of_Infinite_Order_Element | https://proofwiki.org/wiki/Powers_of_Infinite_Order_Element | [
"Order of Group Elements"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Order of Group Element/Infinite"
] | [
"Rule of Transposition"
] |
proofwiki-1045 | Element of Finite Group is of Finite Order | In any finite group, each element has finite order. | Let $G$ be a group whose identity is $e$.
From Element has Idempotent Power in Finite Semigroup, for every element in a ''finite'' semigroup, there is a power of that element which is idempotent.
As $G$, being a group, is also a semigroup, the same applies to $G$.
That is:
:$\forall x \in G: \exists n \in \N_{>0}: x^n ... | In any [[Definition:Finite Group|finite group]], each [[Definition:Element|element]] has [[Definition:Finite Order Element|finite order]]. | Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
From [[Element has Idempotent Power in Finite Semigroup]], for every [[Definition:Element|element]] in a ''finite'' [[Definition:Semigroup|semigroup]], there is a [[Definition:Power of Group Element|power]] of that [[Definit... | Element of Finite Group is of Finite Order/Proof 1 | https://proofwiki.org/wiki/Element_of_Finite_Group_is_of_Finite_Order | https://proofwiki.org/wiki/Element_of_Finite_Group_is_of_Finite_Order/Proof_1 | [
"Finite Groups",
"Order of Group Elements",
"Element of Finite Group is of Finite Order"
] | [
"Definition:Finite Group",
"Definition:Element",
"Definition:Order of Group Element/Finite"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Element has Idempotent Power in Finite Semigroup",
"Definition:Element",
"Definition:Semigroup",
"Definition:Power of Element/Group",
"Definition:Element",
"Definition:Idempotence/Element",
"Definition:Group",
"Defin... |
proofwiki-1046 | Element of Finite Group is of Finite Order | In any finite group, each element has finite order. | Follows as a direct corollary to the result Powers of Infinite Order Element.
{{qed}} | In any [[Definition:Finite Group|finite group]], each [[Definition:Element|element]] has [[Definition:Finite Order Element|finite order]]. | Follows as a direct [[Definition:Corollary|corollary]] to the result [[Powers of Infinite Order Element]].
{{qed}} | Element of Finite Group is of Finite Order/Proof 2 | https://proofwiki.org/wiki/Element_of_Finite_Group_is_of_Finite_Order | https://proofwiki.org/wiki/Element_of_Finite_Group_is_of_Finite_Order/Proof_2 | [
"Finite Groups",
"Order of Group Elements",
"Element of Finite Group is of Finite Order"
] | [
"Definition:Finite Group",
"Definition:Element",
"Definition:Order of Group Element/Finite"
] | [
"Definition:Corollary",
"Powers of Infinite Order Element"
] |
proofwiki-1047 | Inverse Element is Power of Order Less 1 | Let $G$ be a group whose identity is $e$.
Let $g \in G$ be of finite order.
Then:
:$\order g = n \implies g^{n - 1} = g^{-1}$ | {{begin-eqn}}
{{eqn | l = \order g
| r = n
| c =
}}
{{eqn | ll= \leadsto
| l = g^n
| r = e
| c =
}}
{{eqn | ll= \leadsto
| l = g^n g^{-1}
| r = e g^{-1}
| c =
}}
{{eqn | ll= \leadsto
| l = g^{n - 1}
| r = g^{-1}
| c =
}}
{{end-eqn}}
{{qed}} | Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $g \in G$ be of [[Definition:Finite Order Element|finite order]].
Then:
:$\order g = n \implies g^{n - 1} = g^{-1}$ | {{begin-eqn}}
{{eqn | l = \order g
| r = n
| c =
}}
{{eqn | ll= \leadsto
| l = g^n
| r = e
| c =
}}
{{eqn | ll= \leadsto
| l = g^n g^{-1}
| r = e g^{-1}
| c =
}}
{{eqn | ll= \leadsto
| l = g^{n - 1}
| r = g^{-1}
| c =
}}
{{end-eqn}}
{{qed}} | Inverse Element is Power of Order Less 1 | https://proofwiki.org/wiki/Inverse_Element_is_Power_of_Order_Less_1 | https://proofwiki.org/wiki/Inverse_Element_is_Power_of_Order_Less_1 | [
"Inverse Elements",
"Order of Group Elements"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Order of Group Element/Finite"
] | [] |
proofwiki-1048 | Equal Powers of Finite Order Element | :$g^r = g^s \iff k \divides \paren {r - s}$ | === Necessary Condition ===
Suppose that $k \divides \paren {r - s}$.
From the definition of divisor:
:$k \divides \left({r - s}\right) \implies \exists t \in \Z: r - s = k t$
So:
:$g^{r - s} = g^{k t}$
Thus:
{{begin-eqn}}
{{eqn | l = g^r
| r = g^{s + k t}
| c =
}}
{{eqn | r = g^s g^{k t}
| c =
}}
{... | :$g^r = g^s \iff k \divides \paren {r - s}$ | === Necessary Condition ===
Suppose that $k \divides \paren {r - s}$.
From the definition of [[Definition:Divisor of Integer|divisor]]:
:$k \divides \left({r - s}\right) \implies \exists t \in \Z: r - s = k t$
So:
:$g^{r - s} = g^{k t}$
Thus:
{{begin-eqn}}
{{eqn | l = g^r
| r = g^{s + k t}
| c =
}}
{{... | Equal Powers of Finite Order Element | https://proofwiki.org/wiki/Equal_Powers_of_Finite_Order_Element | https://proofwiki.org/wiki/Equal_Powers_of_Finite_Order_Element | [
"Order of Group Elements"
] | [] | [
"Definition:Divisor (Algebra)/Integer"
] |
proofwiki-1049 | Order of Element Divides Order of Finite Group | In a finite group, the order of a group element divides the order of its group:
:$\forall x \in G: \order x \divides \order G$ | Let $G$ be a group.
Let $x \in G$.
By definition, the order of $x$ is the order of the subgroup generated by $x$.
Therefore, by Lagrange's Theorem, $\order x$ is a divisor of $\order G$.
{{qed}} | In a [[Definition:Finite Group|finite group]], the [[Definition:Order of Group Element|order of a group element]] [[Definition:Divisor of Integer|divides]] the [[Definition:Order of Group|order of its group]]:
:$\forall x \in G: \order x \divides \order G$ | Let $G$ be a [[Definition:Group|group]].
Let $x \in G$.
By definition, the [[Definition:Order of Group Element/Definition 2|order of $x$]] is the [[Definition:Order of Group|order]] of the [[Definition:Generated Subgroup|subgroup generated by $x$]].
Therefore, by [[Lagrange's Theorem (Group Theory)|Lagrange's Theore... | Order of Element Divides Order of Finite Group | https://proofwiki.org/wiki/Order_of_Element_Divides_Order_of_Finite_Group | https://proofwiki.org/wiki/Order_of_Element_Divides_Order_of_Finite_Group | [
"Finite Groups",
"Order of Group Elements"
] | [
"Definition:Finite Group",
"Definition:Order of Group Element",
"Definition:Divisor (Algebra)/Integer",
"Definition:Order of Structure"
] | [
"Definition:Group",
"Definition:Order of Group Element/Definition 2",
"Definition:Order of Structure",
"Definition:Generated Subgroup",
"Lagrange's Theorem (Group Theory)",
"Definition:Divisor (Algebra)/Integer"
] |
proofwiki-1050 | Recurrence Relation for Number of Derangements on Finite Set | The number of derangements $D_n$ on a finite set $S$ of cardinality $n$ is:
:$D_n = \paren {n - 1} \paren {D_{n - 1} + D_{n - 2} }$
where $D_1 = 0$, $D_2 = 1$. | Let $\card S = 1$ such that $S = \set s$, say.
Then $\map f s = s$ is the only permutation on $S$.
This by definition is not a derangement.
Thus:
:$D_1 = 0$
Let $\card S = 2$.
Then $S = \set {s, t}$, say.
There are two permutations on $S$:
:$f = \set {\tuple {s, s}, \tuple {t, t} }$
and:
:$g = \set {\tuple {s, t}, \tup... | The number of [[Definition:Derangement|derangements]] $D_n$ on a [[Definition:Finite Set|finite set]] $S$ of [[Definition:Cardinality|cardinality]] $n$ is:
:$D_n = \paren {n - 1} \paren {D_{n - 1} + D_{n - 2} }$
where $D_1 = 0$, $D_2 = 1$. | Let $\card S = 1$ such that $S = \set s$, say.
Then $\map f s = s$ is the only [[Definition:Permutation|permutation]] on $S$.
This by definition is not a [[Definition:Derangement|derangement]].
Thus:
:$D_1 = 0$
Let $\card S = 2$.
Then $S = \set {s, t}$, say.
There are two [[Definition:Permutation|permutations]] ... | Recurrence Relation for Number of Derangements on Finite Set | https://proofwiki.org/wiki/Recurrence_Relation_for_Number_of_Derangements_on_Finite_Set | https://proofwiki.org/wiki/Recurrence_Relation_for_Number_of_Derangements_on_Finite_Set | [
"Number of Derangements on Finite Set",
"Derangements",
"Counting Arguments"
] | [
"Definition:Derangement",
"Definition:Finite Set",
"Definition:Cardinality"
] | [
"Definition:Permutation",
"Definition:Derangement",
"Definition:Permutation",
"Definition:Derangement",
"Definition:Derangement",
"Fundamental Principle of Counting",
"Definition:Derangement",
"Definition:Derangement",
"Definition:Derangement",
"Definition:Derangement",
"Definition:Derangement"
... |
proofwiki-1051 | Element to Power of Group Order is Identity | Let $G$ be a group whose identity is $e$ and whose order is $n$.
Then:
:$\forall g \in G: g^n = e$ | Let $G$ be a group such that $\order G = n$.
Let $g \in G$ and let $\order g = k$.
From Order of Element Divides Order of Finite Group:
:$k \divides n$
So:
:$\exists m \in \Z_{>0}: k m = n$
Thus:
{{begin-eqn}}
{{eqn | l = g^n
| r = \paren {g^k}^m
| c = Powers of Group Elements: Product of Indices
}}
{{eqn |... | Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$ and whose [[Definition:Order of Structure|order]] is $n$.
Then:
:$\forall g \in G: g^n = e$ | Let $G$ be a [[Definition:Group|group]] such that $\order G = n$.
Let $g \in G$ and let $\order g = k$.
From [[Order of Element Divides Order of Finite Group]]:
:$k \divides n$
So:
:$\exists m \in \Z_{>0}: k m = n$
Thus:
{{begin-eqn}}
{{eqn | l = g^n
| r = \paren {g^k}^m
| c = [[Powers of Group Element... | Element to Power of Group Order is Identity | https://proofwiki.org/wiki/Element_to_Power_of_Group_Order_is_Identity | https://proofwiki.org/wiki/Element_to_Power_of_Group_Order_is_Identity | [
"Group Theory"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Order of Structure"
] | [
"Definition:Group",
"Order of Element Divides Order of Finite Group",
"Powers of Group Elements/Product of Indices",
"Power of Identity is Identity"
] |
proofwiki-1052 | Boolean Group is Abelian | Let $G$ be a Boolean group.
Then $G$ is abelian. | By definition of Boolean group, all elements of $G$, other than the identity, have order $2$.
By Group Element is Self-Inverse iff Order 2 and Identity is Self-Inverse, all elements of $G$ are self-inverse.
The result follows directly from All Elements Self-Inverse then Abelian.
{{qed}} | Let $G$ be a [[Definition:Boolean Group|Boolean group]].
Then $G$ is [[Definition:Abelian Group|abelian]]. | By definition of [[Definition:Boolean Group|Boolean group]], all [[Definition:Element|elements]] of $G$, other than the [[Definition:Identity Element|identity]], have [[Definition:Order of Group Element|order]] $2$.
By [[Group Element is Self-Inverse iff Order 2]] and [[Identity is Self-Inverse]], all elements of $G$ ... | Boolean Group is Abelian/Proof 1 | https://proofwiki.org/wiki/Boolean_Group_is_Abelian | https://proofwiki.org/wiki/Boolean_Group_is_Abelian/Proof_1 | [
"Abelian Groups",
"Boolean Groups",
"Boolean Group is Abelian"
] | [
"Definition:Boolean Group",
"Definition:Abelian Group"
] | [
"Definition:Boolean Group",
"Definition:Element",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Order of Group Element",
"Group Element is Self-Inverse iff Order 2",
"Inverse of Identity Element is Itself",
"Definition:Self-Inverse Element",
"All Elements Self-Inverse then A... |
proofwiki-1053 | Boolean Group is Abelian | Let $G$ be a Boolean group.
Then $G$ is abelian. | Let $ a, b \in G$.
By definition of Boolean group:
:$\forall x \in G: x^2 = e$
where $e$ is the identity of $G$.
Then:
{{begin-eqn}}
{{eqn | l = a b
| r = a e b
| c = {{Group-axiom|2}}
}}
{{eqn | r = a \paren {a b}^2 b
| c = as $\forall x \in G: x^2 = e$
}}
{{eqn | r = a \paren {a b} \paren {a b} b
... | Let $G$ be a [[Definition:Boolean Group|Boolean group]].
Then $G$ is [[Definition:Abelian Group|abelian]]. | Let $ a, b \in G$.
By definition of [[Definition:Boolean Group|Boolean group]]:
:$\forall x \in G: x^2 = e$
where $e$ is the [[Definition:Identity Element|identity]] of $G$.
Then:
{{begin-eqn}}
{{eqn | l = a b
| r = a e b
| c = {{Group-axiom|2}}
}}
{{eqn | r = a \paren {a b}^2 b
| c = as $\forall ... | Boolean Group is Abelian/Proof 2 | https://proofwiki.org/wiki/Boolean_Group_is_Abelian | https://proofwiki.org/wiki/Boolean_Group_is_Abelian/Proof_2 | [
"Abelian Groups",
"Boolean Groups",
"Boolean Group is Abelian"
] | [
"Definition:Boolean Group",
"Definition:Abelian Group"
] | [
"Definition:Boolean Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Abelian Group"
] |
proofwiki-1054 | Order of Group Element equals Order of Inverse | Let $G$ be a group whose identity is $e$.
Then:
: $\forall x \in G: \order x = \order {x^{-1} }$
where $\order x$ denotes the order of $x$. | By Powers of Group Elements: Negative Index:
: $\paren {x^k}^{-1} = x^{-k} = \paren {x^{-1} }^k$
Hence:
{{begin-eqn}}
{{eqn | l = x^k
| r = e
| c =
}}
{{eqn | ll= \leadsto
| l = \paren {x^{-1} }^k
| r = e^{-1}
| c =
}}
{{eqn | r = e
| c =
}}
{{eqn | ll= \leadsto
| l = \order... | Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Then:
: $\forall x \in G: \order x = \order {x^{-1} }$
where $\order x$ denotes the [[Definition:Order of Group Element|order]] of $x$. | By [[Powers of Group Elements/Negative Index|Powers of Group Elements: Negative Index]]:
: $\paren {x^k}^{-1} = x^{-k} = \paren {x^{-1} }^k$
Hence:
{{begin-eqn}}
{{eqn | l = x^k
| r = e
| c =
}}
{{eqn | ll= \leadsto
| l = \paren {x^{-1} }^k
| r = e^{-1}
| c =
}}
{{eqn | r = e
| ... | Order of Group Element equals Order of Inverse | https://proofwiki.org/wiki/Order_of_Group_Element_equals_Order_of_Inverse | https://proofwiki.org/wiki/Order_of_Group_Element_equals_Order_of_Inverse | [
"Order of Group Elements",
"Inverse Elements"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Order of Group Element"
] | [
"Powers of Group Elements/Negative Index",
"Definition:Order of Group Element/Infinite"
] |
proofwiki-1055 | Even Order Group has Order 2 Element | Let $G$ be a group whose identity is $e$.
Let $G$ be of even order.
Then:
:$\exists x \in G: \order x = 2$
That is:
:$\exists x \in G: x \ne e: x^2 = e$ | In any group $G$, the identity element $e$ is self-inverse with Identity is Only Group Element of Order 1, and is the only such.
That leaves an odd number of elements.
Each element in $x \in G: \order x > 2$ can be paired off with its inverse, as $\order {x^{-1} } = \order x > 2$ from Order of Group Element equals Orde... | Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $G$ be of even [[Definition:Order of Group|order]].
Then:
:$\exists x \in G: \order x = 2$
That is:
:$\exists x \in G: x \ne e: x^2 = e$ | In any [[Definition:Group|group]] $G$, the [[Identity is Self-Inverse|identity element $e$ is self-inverse]] with [[Identity is Only Group Element of Order 1]], and is the only such.
That leaves an [[Definition:Odd Integer|odd number]] of [[Definition:Element|elements]].
Each element in $x \in G: \order x > 2$ can be... | Even Order Group has Order 2 Element/Proof 1 | https://proofwiki.org/wiki/Even_Order_Group_has_Order_2_Element | https://proofwiki.org/wiki/Even_Order_Group_has_Order_2_Element/Proof_1 | [
"Finite Groups",
"Order of Group Elements",
"Even Order Group has Order 2 Element"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Order of Structure"
] | [
"Definition:Group",
"Inverse of Identity Element is Itself",
"Identity is Only Group Element of Order 1",
"Definition:Odd Integer",
"Definition:Element",
"Definition:Inverse (Abstract Algebra)/Inverse",
"Order of Group Element equals Order of Inverse",
"Definition:Element",
"Definition:Self-Inverse ... |
proofwiki-1056 | Order of Conjugate Element equals Order of Element | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Then
:$\forall a, x \in \struct {G, \circ}: \order {x \circ a \circ x^{-1} } = \order a$
where $\order a$ denotes the order of $a$ in $G$. | Let $\order a = k$.
Then $a^k = e$, and:
:$\forall n \in \N_{>0}: n < k \implies a^n \ne e$
by definition of the order of $a$ in $G$
We have:
{{begin-eqn}}
{{eqn | l = \paren {x \circ a \circ x^{-1} }^k
| r = x \circ a^k \circ x^{-1}
| c = Power of Conjugate equals Conjugate of Power
}}
{{eqn | r = x \circ ... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Then
:$\forall a, x \in \struct {G, \circ}: \order {x \circ a \circ x^{-1} } = \order a$
where $\order a$ denotes the [[Definition:Order of Group Element|order of $a$ in $G$]]. | Let $\order a = k$.
Then $a^k = e$, and:
:$\forall n \in \N_{>0}: n < k \implies a^n \ne e$
by definition of the [[Definition:Order of Group Element|order of $a$ in $G$]]
We have:
{{begin-eqn}}
{{eqn | l = \paren {x \circ a \circ x^{-1} }^k
| r = x \circ a^k \circ x^{-1}
| c = [[Power of Conjugate equal... | Order of Conjugate Element equals Order of Element | https://proofwiki.org/wiki/Order_of_Conjugate_Element_equals_Order_of_Element | https://proofwiki.org/wiki/Order_of_Conjugate_Element_equals_Order_of_Element | [
"Order of Group Elements",
"Conjugacy"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Order of Group Element"
] | [
"Definition:Order of Group Element",
"Power of Conjugate equals Conjugate of Power"
] |
proofwiki-1057 | Order of Homomorphic Image of Group Element | Let $G$ and $H$ be groups whose identities are $e_G$ and $e_H$ respectively.
Let $\phi: G \to H$ be a homomorphism.
Let $g \in G$ be of finite order.
Then:
:$\forall g \in G: \order {\map \phi g} \divides \order g$
where $\divides$ denotes divisibility. | Let $\phi: G \to H$ be a homomorphism.
Let $\order g = n, \order {\map \phi g} = m$.
{{begin-eqn}}
{{eqn | l = \paren {\map \phi g}^n
| r = \map \phi {g^n}
| c = Homomorphism of Power of Group Element
}}
{{eqn | r = \map \phi {e_G}
| c =
}}
{{eqn | r = e_H
| c = Homomorphism to Group Preserves ... | Let $G$ and $H$ be [[Definition:Group|groups]] whose [[Definition:Identity Element|identities]] are $e_G$ and $e_H$ respectively.
Let $\phi: G \to H$ be a [[Definition:Group Homomorphism|homomorphism]].
Let $g \in G$ be of [[Definition:Order of Group Element|finite order]].
Then:
:$\forall g \in G: \order {\map \ph... | Let $\phi: G \to H$ be a [[Definition:Group Homomorphism|homomorphism]].
Let $\order g = n, \order {\map \phi g} = m$.
{{begin-eqn}}
{{eqn | l = \paren {\map \phi g}^n
| r = \map \phi {g^n}
| c = [[Homomorphism of Power of Group Element]]
}}
{{eqn | r = \map \phi {e_G}
| c =
}}
{{eqn | r = e_H
... | Order of Homomorphic Image of Group Element | https://proofwiki.org/wiki/Order_of_Homomorphic_Image_of_Group_Element | https://proofwiki.org/wiki/Order_of_Homomorphic_Image_of_Group_Element | [
"Group Homomorphisms"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Group Homomorphism",
"Definition:Order of Group Element",
"Definition:Divisor (Algebra)/Integer"
] | [
"Definition:Group Homomorphism",
"Homomorphism of Power of Group Element",
"Homomorphism to Group Preserves Identity",
"Element to Power of Multiple of Order is Identity"
] |
proofwiki-1058 | No Group has Two Order 2 Elements | A group can not contain exactly two elements of order $2$. | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Suppose:
: $s, t \in \struct {G, \circ}\: s \ne t, \order s = \order t = 2$
That is. they are self-inverse:
:$s^2 = e = t^2$
The identity is of order $1$.
Hence $s$ nor $t$ is the identity
Hence, as $s \ne t$, then $s \circ t \in G$ is distinct from both $s$ an... | A [[Definition:Group|group]] can not contain exactly two [[Definition:Element|elements]] of [[Definition:Order of Group Element|order]] $2$. | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Suppose:
: $s, t \in \struct {G, \circ}\: s \ne t, \order s = \order t = 2$
That is. they are [[Definition:Self-Inverse Element|self-inverse]]:
:$s^2 = e = t^2$
The [[Definition:Identity Element|identity]]... | No Group has Two Order 2 Elements | https://proofwiki.org/wiki/No_Group_has_Two_Order_2_Elements | https://proofwiki.org/wiki/No_Group_has_Two_Order_2_Elements | [
"Order of Group Elements"
] | [
"Definition:Group",
"Definition:Element",
"Definition:Order of Group Element"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Self-Inverse Element",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Order of Group Element",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Distinct/Plural",... |
proofwiki-1059 | Odd Order Group Element is Square | Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $x \in G$.
Let the order $\order x$ be odd.
Then:
:$\exists y \in G: y^2 = x$ | Let $\order x$ be odd.
Then:
:$\exists n \in \Z_{> 0}: x^{2 n - 1} = e$
from the definition of the order of an element.
Conversely, suppose that
:$\exists n \in \Z_{> 0}: x^{2 n - 1} = e$
Then $\order x$ is a divisor of $2 n - 1$ from Element to Power of Multiple of Order is Identity.
Hence $\order x$ is odd.
So $\ord... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $x \in G$.
Let the [[Definition:Order of Group Element|order]] $\order x$ be [[Definition:Odd Integer|odd]].
Then:
:$\exists y \in G: y^2 = x$ | Let $\order x$ be [[Definition:Odd Integer|odd]].
Then:
:$\exists n \in \Z_{> 0}: x^{2 n - 1} = e$
from the definition of the [[Definition:Order of Group Element|order of an element]].
Conversely, suppose that
:$\exists n \in \Z_{> 0}: x^{2 n - 1} = e$
Then $\order x$ is a [[Definition:Divisor of Integer|divisor]] ... | Odd Order Group Element is Square | https://proofwiki.org/wiki/Odd_Order_Group_Element_is_Square | https://proofwiki.org/wiki/Odd_Order_Group_Element_is_Square | [
"Odd Order Group Element is Square",
"Order of Group Elements"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Order of Group Element",
"Definition:Odd Integer"
] | [
"Definition:Odd Integer",
"Definition:Order of Group Element",
"Definition:Divisor (Algebra)/Integer",
"Element to Power of Multiple of Order is Identity",
"Definition:Odd Integer",
"Definition:Odd Integer"
] |
proofwiki-1060 | Group Isomorphism Preserves Order of Group Element | Let $G$ and $H$ be groups whose identities are $e_G$ and $e_H$.
Let $\phi: G \to H$ be a group isomorphism.
Then:
:$a \in G \implies \order {\map \phi a} = \order a$ | First, suppose $a$ is of finite order.
By definition, $\phi$ is bijective, therefore injective.
The result then follows from Order of Homomorphic Image of Group Element.
{{qed|lemma}}
Now suppose $a$ is of infinite order.
Suppose $\map \phi a$ is of finite order.
Consider the mapping $\phi^{-1}: H \to G$.
Let $b = \map... | Let $G$ and $H$ be [[Definition:Group|groups]] whose [[Definition:Identity Element|identities]] are $e_G$ and $e_H$.
Let $\phi: G \to H$ be a [[Definition:Group Isomorphism|group isomorphism]].
Then:
:$a \in G \implies \order {\map \phi a} = \order a$ | First, suppose $a$ is of [[Definition:Finite Order Element|finite order]].
By definition, $\phi$ is [[Definition:Bijection|bijective]], therefore [[Definition:Injection|injective]].
The result then follows from [[Order of Homomorphic Image of Group Element]].
{{qed|lemma}}
Now suppose $a$ is of [[Definition:Infinit... | Group Isomorphism Preserves Order of Group Element | https://proofwiki.org/wiki/Group_Isomorphism_Preserves_Order_of_Group_Element | https://proofwiki.org/wiki/Group_Isomorphism_Preserves_Order_of_Group_Element | [
"Group Isomorphisms"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism"
] | [
"Definition:Order of Group Element/Finite",
"Definition:Bijection",
"Definition:Injection",
"Order of Homomorphic Image of Group Element",
"Definition:Order of Group Element/Infinite",
"Definition:Order of Group Element/Finite",
"Definition:Order of Group Element/Finite",
"Rule of Transposition"
] |
proofwiki-1061 | Non-Trivial Group has Non-Trivial Cyclic Subgroup | Let $G$ be a group whose identity element is $e$.
Let $g \in G$.
If $g$ has infinite order, then $\gen g$ is an infinite cyclic group.
If $\order g = n$, then $\gen g$ is a cyclic group with $n$ elements.
Thus, every group which is non-trivial has at least one cyclic subgroup which is also non-trivial.
In the case that... | === Infinite Order ===
Suppose that $g$ has infinite order.
We have that $\gen g$ consists of all possible powers of $g$.
So $\gen g$ can contain a finite number of elements only if some of these were equal.
Then we would have:
:$\exists i, j \in \Z, i < j: g^i = g^j$
and so:
:$g^{j - i} = g^{i - i} = g^0 = e$
which wo... | Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity element]] is $e$.
Let $g \in G$.
If $g$ has [[Definition:Infinite Order Element|infinite order]], then $\gen g$ is an [[Definition:Infinite Cyclic Group|infinite cyclic group]].
If $\order g = n$, then $\gen g$ is a [[Definition:Cy... | === Infinite Order ===
Suppose that $g$ has [[Definition:Infinite Order Element|infinite order]].
We have that $\gen g$ consists of all possible [[Powers of Group Elements|powers of $g$]].
So $\gen g$ can contain a [[Definition:Finite Set|finite number of elements]] only if some of these were equal.
Then we would h... | Non-Trivial Group has Non-Trivial Cyclic Subgroup | https://proofwiki.org/wiki/Non-Trivial_Group_has_Non-Trivial_Cyclic_Subgroup | https://proofwiki.org/wiki/Non-Trivial_Group_has_Non-Trivial_Cyclic_Subgroup | [
"Cyclic Groups"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Order of Group Element/Infinite",
"Definition:Infinite Cyclic Group",
"Definition:Cyclic Group",
"Definition:Element",
"Definition:Group",
"Definition:Non-Trivial Group",
"Definition:Cyclic Group",
"Defini... | [
"Definition:Order of Group Element/Infinite",
"Powers of Group Elements",
"Definition:Finite Set",
"Definition:Order of Group Element/Finite",
"Definition:Infinite Cyclic Group"
] |
proofwiki-1062 | Epimorphism from Integers to Cyclic Group | Let $\gen a = \struct {G, \circ}$ be a cyclic group.
Let $f: \Z \to G$ be a mapping defined as:
$\forall n \in \Z: \map f n = a^n$.
Then $f$ is a (group) epimorphism from $\struct {\Z, +}$ onto $\gen a$. | By Powers of Element form Subgroup:
:$\forall n \in \N: a^n \in \gen a$
Hence by the Index Law for Monoids: Negative Index:
:$\forall n \in \Z: a^n \in \gen a$
Also, by Index Law for Monoids: Sum of Indices, $f$ is a homomorphism from $\struct {\Z, +}$ into $\struct {G, \circ}$.
By Homomorphism Preserves Subsemigroups,... | Let $\gen a = \struct {G, \circ}$ be a [[Definition:Cyclic Group|cyclic group]].
Let $f: \Z \to G$ be a [[Definition:Mapping|mapping]] defined as:
$\forall n \in \Z: \map f n = a^n$.
Then $f$ is a [[Definition:Group Epimorphism|(group) epimorphism]] from $\struct {\Z, +}$ onto $\gen a$. | By [[Powers of Element form Subgroup]]:
:$\forall n \in \N: a^n \in \gen a$
Hence by the [[Index Laws for Monoids/Negative Index|Index Law for Monoids: Negative Index]]:
:$\forall n \in \Z: a^n \in \gen a$
Also, by [[Index Laws for Monoids/Sum of Indices|Index Law for Monoids: Sum of Indices]], $f$ is a [[Definition:... | Epimorphism from Integers to Cyclic Group | https://proofwiki.org/wiki/Epimorphism_from_Integers_to_Cyclic_Group | https://proofwiki.org/wiki/Epimorphism_from_Integers_to_Cyclic_Group | [
"Integers",
"Group Epimorphisms",
"Cyclic Groups"
] | [
"Definition:Cyclic Group",
"Definition:Mapping",
"Definition:Group Epimorphism"
] | [
"Powers of Element form Subgroup",
"Index Laws for Monoids/Negative Index",
"Index Laws for Monoids/Sum of Indices",
"Definition:Group Homomorphism",
"Homomorphism Preserves Subsemigroups",
"Definition:Codomain (Set Theory)/Mapping",
"Definition:Subgroup",
"Existence of Unique Subgroup Generated by Su... |
proofwiki-1063 | Cyclic Group is Abelian | Let $G$ be a cyclic group.
Then $G$ is abelian. | Let $G$ be a cyclic group.
All elements of $G$ are of the form $a^n$, where $n \in \Z$.
Let $x, y \in G: x = a^p, y = a^q$.
From Powers of Group Elements: Sum of Indices:
: $x y = a^p a^q = a^{p + q} = a^{q + p} = a^q a^p = y x$
Thus:
: $\forall x, y \in G: x y = y x$
and $G$ is by definition abelian.
{{qed}} | Let $G$ be a [[Definition:Cyclic Group|cyclic group]].
Then $G$ is [[Definition:Abelian Group|abelian]]. | Let $G$ be a [[Definition:Cyclic Group|cyclic group]].
All elements of $G$ are of the form $a^n$, where $n \in \Z$.
Let $x, y \in G: x = a^p, y = a^q$.
From [[Powers of Group Elements/Sum of Indices|Powers of Group Elements: Sum of Indices]]:
: $x y = a^p a^q = a^{p + q} = a^{q + p} = a^q a^p = y x$
Thus:
: $\foral... | Cyclic Group is Abelian/Proof 1 | https://proofwiki.org/wiki/Cyclic_Group_is_Abelian | https://proofwiki.org/wiki/Cyclic_Group_is_Abelian/Proof_1 | [
"Cyclic Groups",
"Abelian Groups",
"Cyclic Group is Abelian"
] | [
"Definition:Cyclic Group",
"Definition:Abelian Group"
] | [
"Definition:Cyclic Group",
"Powers of Group Elements/Sum of Indices",
"Definition:Abelian Group"
] |
proofwiki-1064 | Cyclic Group is Abelian | Let $G$ be a cyclic group.
Then $G$ is abelian. | We have that Integers under Addition form Abelian Group.
The result then follows from combining:
:Epimorphism from Integers to Cyclic Group
:Epimorphism Preserves Commutativity.
{{qed}} | Let $G$ be a [[Definition:Cyclic Group|cyclic group]].
Then $G$ is [[Definition:Abelian Group|abelian]]. | We have that [[Integers under Addition form Abelian Group]].
The result then follows from combining:
:[[Epimorphism from Integers to Cyclic Group]]
:[[Epimorphism Preserves Commutativity]].
{{qed}} | Cyclic Group is Abelian/Proof 2 | https://proofwiki.org/wiki/Cyclic_Group_is_Abelian | https://proofwiki.org/wiki/Cyclic_Group_is_Abelian/Proof_2 | [
"Cyclic Groups",
"Abelian Groups",
"Cyclic Group is Abelian"
] | [
"Definition:Cyclic Group",
"Definition:Abelian Group"
] | [
"Integers under Addition form Abelian Group",
"Epimorphism from Integers to Cyclic Group",
"Epimorphism Preserves Commutativity"
] |
proofwiki-1065 | Cyclic Groups of Same Order are Isomorphic | Two cyclic groups of the same order are isomorphic to each other. | Let $G_1$ and $G_2$ be cyclic groups, both of finite order $k$.
Let $G_1 = \gen a, G_2 = \gen b$.
Then, by the definition of a cyclic group:
:$\order a = \order b = k$
Also, by definition:
:$G_1 = \set {a^0, a^1, \ldots, a^{k - 1} }$
and:
:$G_2 = \set {b^0, b^1, \ldots, b^{k - 1} }$
Let us set up the obvious bijection:... | Two [[Definition:Cyclic Group|cyclic groups]] of the same [[Definition:Order of Structure|order]] are [[Definition:Group Isomorphism|isomorphic]] to each other. | Let $G_1$ and $G_2$ be [[Definition:Cyclic Group|cyclic groups]], both of [[Definition:Finite Group|finite order]] $k$.
Let $G_1 = \gen a, G_2 = \gen b$.
Then, by the definition of a [[Definition:Cyclic Group|cyclic group]]:
:$\order a = \order b = k$
Also, by definition:
:$G_1 = \set {a^0, a^1, \ldots, a^{k - 1} }$... | Cyclic Groups of Same Order are Isomorphic | https://proofwiki.org/wiki/Cyclic_Groups_of_Same_Order_are_Isomorphic | https://proofwiki.org/wiki/Cyclic_Groups_of_Same_Order_are_Isomorphic | [
"Cyclic Groups",
"Group Isomorphisms"
] | [
"Definition:Cyclic Group",
"Definition:Order of Structure",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism"
] | [
"Definition:Cyclic Group",
"Definition:Finite Group",
"Definition:Cyclic Group",
"Definition:Bijection",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Division Theorem",
"Element to Power of Remainder",
"Definition:Group Homomorphism",
"Definition:Bijection",
"Definition:Isomorph... |
proofwiki-1066 | Order of Subgroup of Cyclic Group | Let $C_n = \gen g$ be the cyclic group of order $n$ which is generated by $g$ whose identity is $e$.
Let $a \in C_n: a = g^i$.
Let $H = \gen a$.
Then:
:$\order H = \dfrac n {\gcd \set {n, i} }$
where:
:$\order H$ denotes the order of $H$
:$\gcd \set {n, i}$ denotes the greatest common divisor of $n$ and $i$. | The fact that $H$ is cyclic follows from Subgroup of Cyclic Group is Cyclic.
We need to show that $H$ has $\dfrac n d$ elements.
Let $\order H = k$.
By Non-Trivial Group has Non-Trivial Cyclic Subgroup:
:$k = \order a$
where $\order a$ denotes the order of $a$.
That is:
:$a^k = e$
We have that $a = g^i$.
So:
{{begin-eq... | Let $C_n = \gen g$ be the [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Structure|order]] $n$ which is [[Definition:Generator of Cyclic Group|generated]] by $g$ whose [[Definition:Identity Element|identity]] is $e$.
Let $a \in C_n: a = g^i$.
Let $H = \gen a$.
Then:
:$\order H = \dfrac n {\gcd \s... | The fact that $H$ is [[Definition:Cyclic Group|cyclic]] follows from [[Subgroup of Cyclic Group is Cyclic]].
We need to show that $H$ has $\dfrac n d$ elements.
Let $\order H = k$.
By [[Non-Trivial Group has Non-Trivial Cyclic Subgroup]]:
:$k = \order a$
where $\order a$ denotes the [[Definition:Order of Group Eleme... | Order of Subgroup of Cyclic Group | https://proofwiki.org/wiki/Order_of_Subgroup_of_Cyclic_Group | https://proofwiki.org/wiki/Order_of_Subgroup_of_Cyclic_Group | [
"Cyclic Groups"
] | [
"Definition:Cyclic Group",
"Definition:Order of Structure",
"Definition:Cyclic Group/Generator",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Order of Structure",
"Definition:Greatest Common Divisor/Integers"
] | [
"Definition:Cyclic Group",
"Subgroup of Cyclic Group is Cyclic",
"Non-Trivial Group has Non-Trivial Cyclic Subgroup",
"Definition:Order of Group Element",
"Definition:Divisor (Algebra)/Integer",
"Integers Divided by GCD are Coprime",
"Definition:Coprime/Integers",
"Euclid's Lemma"
] |
proofwiki-1067 | Number of Powers of Cyclic Group Element | Let $G$ be a cyclic group of order $n$, generated by $g$.
Let $d \divides n$.
Then the element $g^{n/d}$ has $d$ distinct powers. | Follows directly from Order of Subgroup of Cyclic Group:
:$\order {\gen {g^{n/d} } } = \dfrac n {\gcd \set {n, n/d} } = d$
Thus from List of Elements in Finite Cyclic Group:
:$\gen {g^{n/d} } = \set {e, g^{n/d}, \paren {g^{n/d} }^2, \ldots, \paren {g^{n/d} }^{d - 1} }$
and the result follows.
{{qed}} | Let $G$ be a [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Structure|order]] $n$, [[Definition:Generator of Cyclic Group|generated]] by $g$.
Let $d \divides n$.
Then the element $g^{n/d}$ has $d$ distinct [[Definition:Power of Group Element|powers]]. | Follows directly from [[Order of Subgroup of Cyclic Group]]:
:$\order {\gen {g^{n/d} } } = \dfrac n {\gcd \set {n, n/d} } = d$
Thus from [[List of Elements in Finite Cyclic Group]]:
:$\gen {g^{n/d} } = \set {e, g^{n/d}, \paren {g^{n/d} }^2, \ldots, \paren {g^{n/d} }^{d - 1} }$
and the result follows.
{{qed}} | Number of Powers of Cyclic Group Element | https://proofwiki.org/wiki/Number_of_Powers_of_Cyclic_Group_Element | https://proofwiki.org/wiki/Number_of_Powers_of_Cyclic_Group_Element | [
"Cyclic Groups"
] | [
"Definition:Cyclic Group",
"Definition:Order of Structure",
"Definition:Cyclic Group/Generator",
"Definition:Power of Element/Group"
] | [
"Order of Subgroup of Cyclic Group",
"List of Elements in Finite Cyclic Group"
] |
proofwiki-1068 | Subgroup of Finite Cyclic Group is Determined by Order | Let $G = \gen g$ be a cyclic group whose order is $n$ and whose identity is $e$.
Let $d \divides n$, where $\divides$ denotes divisibility.
Then there exists exactly one subgroup $G_d = \gen {g^{n / d} }$ of $G$ with $d$ elements. | Let $G$ be generated by $g$, such that $\order g = n$.
From Number of Powers of Cyclic Group Element, $g^{n/d}$ has $d$ distinct powers.
Thus $\gen {g^{n / d} }$ has $d$ elements.
Now suppose $H$ is another subgroup of $G$ of order $d$.
Then by Subgroup of Cyclic Group is Cyclic, $H$ is cyclic.
Let $H = \gen y$ where $... | Let $G = \gen g$ be a [[Definition:Cyclic Group|cyclic group]] whose [[Definition:Order of Structure|order]] is $n$ and whose [[Definition:Identity Element|identity]] is $e$.
Let $d \divides n$, where $\divides$ denotes [[Definition:Divisor of Integer|divisibility]].
Then there exists exactly one [[Definition:Subgro... | Let $G$ be [[Definition:Generated Subgroup|generated]] by $g$, such that $\order g = n$.
From [[Number of Powers of Cyclic Group Element]], $g^{n/d}$ has $d$ distinct [[Definition:Power of Group Element|powers]].
Thus $\gen {g^{n / d} }$ has $d$ [[Definition:Element|elements]].
Now suppose $H$ is another [[Definiti... | Subgroup of Finite Cyclic Group is Determined by Order | https://proofwiki.org/wiki/Subgroup_of_Finite_Cyclic_Group_is_Determined_by_Order | https://proofwiki.org/wiki/Subgroup_of_Finite_Cyclic_Group_is_Determined_by_Order | [
"Finite Cyclic Groups",
"Subgroups"
] | [
"Definition:Cyclic Group",
"Definition:Order of Structure",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Divisor (Algebra)/Integer",
"Definition:Subgroup",
"Definition:Element"
] | [
"Definition:Generated Subgroup",
"Number of Powers of Cyclic Group Element",
"Definition:Power of Element/Group",
"Definition:Element",
"Definition:Subgroup",
"Definition:Order of Structure",
"Subgroup of Cyclic Group is Cyclic",
"Definition:Cyclic Group",
"Equal Powers of Finite Order Element",
"... |
proofwiki-1069 | Cyclic Group Elements whose Powers equal Identity | Let $G$ be a cyclic group whose identity is $e$ and whose order is $n$.
Let $d \divides n$.
Then there exist exactly $d$ elements $x \in G$ satisfying the equation $x^d = e$.
These are the elements of the group $G_d$ generated by $g^{n / d}$:
:$G_d = \gen {g^{n / d} }$ | From the argument in Subgroup of Finite Cyclic Group is Determined by Order, it follows that $x$ satisfies the equation $x^d = e$ {{iff}} $x$ is a power of $g^{n/d}$.
Thus there are $d$ solutions to this equation.
{{qed}} | Let $G$ be a [[Definition:Cyclic Group|cyclic group]] whose [[Definition:Identity Element|identity]] is $e$ and whose [[Definition:Order of Structure|order]] is $n$.
Let $d \divides n$.
Then there exist exactly $d$ elements $x \in G$ satisfying the equation $x^d = e$.
These are the elements of the group $G_d$ [[Def... | From the argument in [[Subgroup of Finite Cyclic Group is Determined by Order]], it follows that $x$ satisfies the equation $x^d = e$ {{iff}} $x$ is a [[Definition:Power of Group Element|power]] of $g^{n/d}$.
Thus there are $d$ solutions to this equation.
{{qed}} | Cyclic Group Elements whose Powers equal Identity | https://proofwiki.org/wiki/Cyclic_Group_Elements_whose_Powers_equal_Identity | https://proofwiki.org/wiki/Cyclic_Group_Elements_whose_Powers_equal_Identity | [
"Cyclic Groups"
] | [
"Definition:Cyclic Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Order of Structure",
"Definition:Cyclic Group/Generator"
] | [
"Subgroup of Finite Cyclic Group is Determined by Order",
"Definition:Power of Element/Group"
] |
proofwiki-1070 | Prime Group is Cyclic | Let $p$ be a prime number.
Let $G$ be a group whose order is $p$.
Then $G$ is cyclic. | Let $a \in G: a \ne e$ where $e$ is the identity of $G$.
From Group of Prime Order p has p-1 Elements of Order p, $a$ has order $p$.
Hence by definition, $a$ generates $G$.
Hence also by definition, $G$ is cyclic.
{{qed}} | Let $p$ be a [[Definition:Prime Number|prime number]].
Let $G$ be a [[Definition:Group|group]] whose [[Definition:Order of Group|order]] is $p$.
Then $G$ is [[Definition:Cyclic Group|cyclic]]. | Let $a \in G: a \ne e$ where $e$ is the [[Definition:Identity Element|identity]] of $G$.
From [[Group of Prime Order p has p-1 Elements of Order p]], $a$ has [[Definition:Order of Group|order]] $p$.
Hence by definition, $a$ [[Definition:Generator of Cyclic Group|generates]] $G$.
Hence also by definition, $G$ is [[De... | Prime Group is Cyclic | https://proofwiki.org/wiki/Prime_Group_is_Cyclic | https://proofwiki.org/wiki/Prime_Group_is_Cyclic | [
"Prime Groups",
"Finite Cyclic Groups"
] | [
"Definition:Prime Number",
"Definition:Group",
"Definition:Order of Structure",
"Definition:Cyclic Group"
] | [
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Group of Prime Order p has p-1 Elements of Order p",
"Definition:Order of Structure",
"Definition:Cyclic Group/Generator",
"Definition:Cyclic Group"
] |
proofwiki-1071 | Group of Order less than 6 is Abelian | All groups with less than $6$ elements are abelian. | Let $G$ be a non-abelian group.
From Non-Abelian Group has Order Greater than 4, the order of $G$ must be at least $5$.
But $5$ is a prime number.
By Prime Group is Cyclic it follows that a group of order $5$ is cyclic.
By Cyclic Group is Abelian this group is abelian.
Hence the result.
{{qed}} | All [[Definition:Group|groups]] with less than $6$ [[Definition:Element|elements]] are [[Definition:Abelian Group|abelian]]. | Let $G$ be a non-[[Definition:Abelian Group|abelian group]].
From [[Non-Abelian Group has Order Greater than 4]], the [[Definition:Order of Structure|order]] of $G$ must be at least $5$.
But $5$ is a [[Definition:Prime Number|prime number]].
By [[Prime Group is Cyclic]] it follows that a [[Definition:Group|group]] o... | Group of Order less than 6 is Abelian | https://proofwiki.org/wiki/Group_of_Order_less_than_6_is_Abelian | https://proofwiki.org/wiki/Group_of_Order_less_than_6_is_Abelian | [
"Abelian Groups"
] | [
"Definition:Group",
"Definition:Element",
"Definition:Abelian Group"
] | [
"Definition:Abelian Group",
"Non-Abelian Group has Order Greater than 4",
"Definition:Order of Structure",
"Definition:Prime Number",
"Prime Group is Cyclic",
"Definition:Group",
"Definition:Order of Structure",
"Definition:Cyclic Group",
"Cyclic Group is Abelian",
"Definition:Abelian Group"
] |
proofwiki-1072 | Power of Generator of Cyclic Group is Generator iff Power is Coprime with Order | Let $C_n$ be the cyclic group of order $n$.
Let $C_n = \gen a$, that is, that $C_n$ is generated by $a$.
Then:
:$C_n = \gen {a^k} \iff k \perp n$
That is, $C_n$ is also generated by $a^k$ {{iff}} $k$ is coprime to $n$. | === Necessary Condition ===
Let $k \perp n$.
Then by Integer Combination of Coprime Integers:
:$\exists u, v \in \Z: 1 = u k + v n$
So $\forall m \in \Z$, we have:
{{begin-eqn}}
{{eqn | l =a^m
| r = a^{m \paren 1}
| c = Integer Multiplication Identity is One
}}
{{eqn | r = a^{m \paren {u k + v n} }
| ... | Let $C_n$ be the [[Definition:Cyclic Group|cyclic group]] of [[Definition:Order of Structure|order]] $n$.
Let $C_n = \gen a$, that is, that $C_n$ is [[Definition:Generator of Cyclic Group|generated]] by $a$.
Then:
:$C_n = \gen {a^k} \iff k \perp n$
That is, $C_n$ is also [[Definition:Generator of Cyclic Group|gener... | === Necessary Condition ===
Let $k \perp n$.
Then by [[Integer Combination of Coprime Integers]]:
:$\exists u, v \in \Z: 1 = u k + v n$
So $\forall m \in \Z$, we have:
{{begin-eqn}}
{{eqn | l =a^m
| r = a^{m \paren 1}
| c = [[Integer Multiplication Identity is One]]
}}
{{eqn | r = a^{m \paren {u k + v n... | Power of Generator of Cyclic Group is Generator iff Power is Coprime with Order | https://proofwiki.org/wiki/Power_of_Generator_of_Cyclic_Group_is_Generator_iff_Power_is_Coprime_with_Order | https://proofwiki.org/wiki/Power_of_Generator_of_Cyclic_Group_is_Generator_iff_Power_is_Coprime_with_Order | [
"Cyclic Groups",
"Coprime Integers",
"Generators of Groups"
] | [
"Definition:Cyclic Group",
"Definition:Order of Structure",
"Definition:Cyclic Group/Generator",
"Definition:Cyclic Group/Generator",
"Definition:Coprime/Integers"
] | [
"Integer Combination of Coprime Integers",
"Integer Multiplication Identity is One",
"Integer Multiplication Distributes over Addition",
"Exponent Combination Laws/Product of Powers",
"Integer Multiplication is Associative",
"Integer Multiplication is Commutative",
"Exponent Combination Laws/Power of Po... |
proofwiki-1073 | Order of Conjugate of Subgroup | Let $G$ be a group.
Let $H$ be a subgroup of $G$ such that $H$ is of finite order.
Then $\order {H^a} = \order H$. | From the definition of Conjugate of Group Subet we have $H^a = a H a^{-1}$.
From Set Equivalence of Regular Representations:
:$\order {a H a^{-1} } = \order {a H} = \order H$
{{qed}} | Let $G$ be a [[Definition:Group|group]].
Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$ such that $H$ is of [[Definition:Order of Structure|finite order]].
Then $\order {H^a} = \order H$. | From the definition of [[Definition:Conjugate of Group Subset|Conjugate of Group Subet]] we have $H^a = a H a^{-1}$.
From [[Set Equivalence of Regular Representations]]:
:$\order {a H a^{-1} } = \order {a H} = \order H$
{{qed}} | Order of Conjugate of Subgroup | https://proofwiki.org/wiki/Order_of_Conjugate_of_Subgroup | https://proofwiki.org/wiki/Order_of_Conjugate_of_Subgroup | [
"Conjugacy"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Order of Structure"
] | [
"Definition:Conjugate (Group Theory)/Subset",
"Set Equivalence of Regular Representations"
] |
proofwiki-1074 | Subgroup of Index 2 is Normal | A subgroup of index $2$ is always normal. | Suppose $H \le G$ such that $\index G H = 2$.
Thus $H$ has two left cosets (and two right cosets) in $G$.
If $g \in H$, then $g H = H = H g$.
If $g \notin H$, then $g H = G \setminus H$ as there are only two cosets and the cosets partition $G$.
For the same reason, $g \notin H \implies H g = G \setminus H$.
That is, $g... | A [[Definition:Subgroup|subgroup]] of [[Definition:Index of Subgroup|index $2$]] is always [[Definition:Normal Subgroup|normal]]. | Suppose $H \le G$ such that $\index G H = 2$.
Thus $H$ has two [[Definition:Left Coset|left cosets]] (and two [[Definition:Right Coset|right cosets]]) in $G$.
If $g \in H$, then $g H = H = H g$.
If $g \notin H$, then $g H = G \setminus H$ as there are only two [[Definition:Coset|cosets]] and the [[Congruence Class M... | Subgroup of Index 2 is Normal | https://proofwiki.org/wiki/Subgroup_of_Index_2_is_Normal | https://proofwiki.org/wiki/Subgroup_of_Index_2_is_Normal | [
"Normal Subgroups"
] | [
"Definition:Subgroup",
"Definition:Index of Subgroup",
"Definition:Normal Subgroup"
] | [
"Definition:Coset/Left Coset",
"Definition:Coset/Right Coset",
"Definition:Coset",
"Congruence Class Modulo Subgroup is Coset",
"Definition:Normal Subgroup"
] |
proofwiki-1075 | Intersection of Normal Subgroups is Normal | Let $G$ be a group.
Let $I$ be an indexing set.
Let $\family {N_i}_{i \mathop \in I}$ be a non-empty indexed family of normal subgroups of $G$.
Then $\ds \bigcap_{i \mathop \in I} N_i$ is a normal subgroup of $G$. | Let $\ds N = \bigcap_{i \mathop \in I} N_i$.
From Intersection of Subgroups is Subgroup, $N \le G$.
Suppose $H \in \set {N_i: i \in I}$.
We have that $N \subseteq H$.
Thus from Subgroup is Superset of Conjugate iff Normal:
:$a N a^{-1} \subseteq a H a^{-1} \subseteq H$
Thus $a N a^{-1}$ is a subset of each one of the s... | Let $G$ be a [[Definition:Group|group]].
Let $I$ be an [[Definition:Indexing Set|indexing set]].
Let $\family {N_i}_{i \mathop \in I}$ be a [[Definition:Non-Empty Set|non-empty]] [[Definition:Indexed Family|indexed family]] of [[Definition:Normal Subgroup|normal subgroups]] of $G$.
Then $\ds \bigcap_{i \mathop \in ... | Let $\ds N = \bigcap_{i \mathop \in I} N_i$.
From [[Intersection of Subgroups is Subgroup]], $N \le G$.
Suppose $H \in \set {N_i: i \in I}$.
We have that $N \subseteq H$.
Thus from [[Subgroup is Superset of Conjugate iff Normal]]:
:$a N a^{-1} \subseteq a H a^{-1} \subseteq H$
Thus $a N a^{-1}$ is a [[Definition:S... | Intersection of Normal Subgroups is Normal | https://proofwiki.org/wiki/Intersection_of_Normal_Subgroups_is_Normal | https://proofwiki.org/wiki/Intersection_of_Normal_Subgroups_is_Normal | [
"Normal Subgroups",
"Set Intersection"
] | [
"Definition:Group",
"Definition:Indexing Set",
"Definition:Non-Empty Set",
"Definition:Indexing Set/Family",
"Definition:Normal Subgroup",
"Definition:Normal Subgroup"
] | [
"Intersection of Subgroups is Subgroup",
"Subgroup is Superset of Conjugate iff Normal",
"Definition:Subset",
"Definition:Subgroup",
"Definition:Set Intersection/Family of Sets",
"Subgroup is Superset of Conjugate iff Normal"
] |
proofwiki-1076 | Union of Conjugacy Classes is Normal | Let $G$ be a group.
Let $H \le G$.
Then $H$ is normal in $G$ {{iff}} $H$ is a union of conjugacy classes of $G$. | {{begin-eqn}}
{{eqn | l = H
| o = \lhd
| r = G
| c = where $\lhd$ denotes that $H$ is normal in $G$
}}
{{eqn | ll= \leadstoandfrom
| q = \forall g \in G
| l = g H g^{-1}
| o = \subseteq
| r = H
| c = {{Defof|Normal Subgroup}}
}}
{{eqn | ll= \leadstoandfrom
| q = \fo... | Let $G$ be a [[Definition:Group|group]].
Let $H \le G$.
Then $H$ is [[Definition:Normal Subgroup|normal]] in $G$ {{iff}} $H$ is a [[Definition:Set Union|union]] of [[Definition:Conjugacy Class|conjugacy classes]] of $G$. | {{begin-eqn}}
{{eqn | l = H
| o = \lhd
| r = G
| c = where $\lhd$ denotes that $H$ is [[Definition:Normal Subgroup|normal]] in $G$
}}
{{eqn | ll= \leadstoandfrom
| q = \forall g \in G
| l = g H g^{-1}
| o = \subseteq
| r = H
| c = {{Defof|Normal Subgroup}}
}}
{{eqn | ll= ... | Union of Conjugacy Classes is Normal | https://proofwiki.org/wiki/Union_of_Conjugacy_Classes_is_Normal | https://proofwiki.org/wiki/Union_of_Conjugacy_Classes_is_Normal | [
"Normal Subgroups"
] | [
"Definition:Group",
"Definition:Normal Subgroup",
"Definition:Set Union",
"Definition:Conjugacy Class"
] | [
"Definition:Normal Subgroup",
"Definition:Conjugacy Class"
] |
proofwiki-1077 | Unique Subgroup of a Given Order is Normal | Let a group $G$ have only one subgroup of a given order.
Then that subgroup is normal. | Let $H \le G$, where $\le$ denotes that $H$ is a subgroup of $G$.
Let $H$ be the only subgroup of $G$ whose order is $\order H$.
Let $g \in G$.
From Conjugate of Subgroup is Subgroup:
:$g H g^{-1} \le G$
From Order of Conjugate of Subgroup:
: $\order {g H g^{-1} } = \order H$
But $H$ is the only subgroup of $G$ of orde... | Let a [[Definition:Group|group]] $G$ have only one [[Definition:Subgroup|subgroup]] of a given [[Definition:Order of Structure|order]].
Then that [[Definition:Subgroup|subgroup]] is [[Definition:Normal Subgroup|normal]]. | Let $H \le G$, where $\le$ denotes that $H$ is a [[Definition:Subgroup|subgroup]] of $G$.
Let $H$ be the only [[Definition:Subgroup|subgroup]] of $G$ whose [[Definition:Order of Structure|order]] is $\order H$.
Let $g \in G$.
From [[Conjugate of Subgroup is Subgroup]]:
:$g H g^{-1} \le G$
From [[Order of Conjugate ... | Unique Subgroup of a Given Order is Normal | https://proofwiki.org/wiki/Unique_Subgroup_of_a_Given_Order_is_Normal | https://proofwiki.org/wiki/Unique_Subgroup_of_a_Given_Order_is_Normal | [
"Normal Subgroups"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Order of Structure",
"Definition:Subgroup",
"Definition:Normal Subgroup"
] | [
"Definition:Subgroup",
"Definition:Subgroup",
"Definition:Order of Structure",
"Conjugate of Subgroup is Subgroup",
"Order of Conjugate of Subgroup",
"Definition:Subgroup",
"Definition:Order of Structure",
"Definition:Subgroup",
"Definition:Order of Structure",
"Subgroup equals Conjugate iff Norma... |
proofwiki-1078 | Group Generated by Normal Intersection is Normal | Let $I$ be an indexing set, and $\set {N_i: i \in I}$ be a set of normal subgroups of the group $G$.
Then $\family {N_i: i \in I}$ is a normal subgroup of $G$. | By definition, $\family {N_i: i \in I}$ is the intersection of all the subgroups of $G$ which contain every $N_i$.
For each $H \le G$, the conjugate $g H g^{-1}$ contains each $g N_i g^{-1}$.
Since each $N_i \lhd G$, it follows that:
:$N_i \subseteq g H g^{-1}$
Thus it follows that:
:$\family {N_i: i \in I} \lhd G$
{{q... | Let $I$ be an [[Definition:Indexing Set|indexing set]], and $\set {N_i: i \in I}$ be a [[Definition:Set|set]] of [[Definition:Normal Subgroup|normal subgroups]] of the [[Definition:Group|group]] $G$.
Then $\family {N_i: i \in I}$ is a [[Definition:Normal Subgroup|normal subgroup]] of $G$. | By definition, $\family {N_i: i \in I}$ is the [[Definition:Set Intersection|intersection]] of all the [[Definition:Subgroup|subgroups]] of $G$ which contain every $N_i$.
For each $H \le G$, the [[Definition:Conjugate of Group Subset|conjugate]] $g H g^{-1}$ contains each $g N_i g^{-1}$.
Since each $N_i \lhd G$, it f... | Group Generated by Normal Intersection is Normal | https://proofwiki.org/wiki/Group_Generated_by_Normal_Intersection_is_Normal | https://proofwiki.org/wiki/Group_Generated_by_Normal_Intersection_is_Normal | [
"Normal Subgroups"
] | [
"Definition:Indexing Set",
"Definition:Set",
"Definition:Normal Subgroup",
"Definition:Group",
"Definition:Normal Subgroup"
] | [
"Definition:Set Intersection",
"Definition:Subgroup",
"Definition:Conjugate (Group Theory)/Subset",
"Category:Normal Subgroups"
] |
proofwiki-1079 | Smallest Normal Subgroup containing Set | Let $S \subseteq G$ where $G$ is a group.
Then there exists a unique smallest normal subgroup of $G$ which contains $S$. | Let $\Bbb S$ be the set of all normal subgroups of $G$ that contain $S$.
Since $S \subseteq G \lhd G$, $\Bbb S \ne \O$.
Let $N = \bigcap \Bbb S$, that is, the intersection of all elements of $\Bbb S$.
By Intersection of Normal Subgroups is Normal, $N \lhd G$.
By the definition of intersection, $S \subseteq N$.
By the m... | Let $S \subseteq G$ where $G$ is a [[Definition:Group|group]].
Then there exists a unique smallest [[Definition:Normal Subgroup|normal subgroup]] of $G$ which contains $S$. | Let $\Bbb S$ be the [[Definition:Set|set]] of all [[Definition:Normal Subgroup|normal subgroups]] of $G$ that contain $S$.
Since $S \subseteq G \lhd G$, $\Bbb S \ne \O$.
Let $N = \bigcap \Bbb S$, that is, the [[Definition:Set Intersection|intersection]] of all [[Definition:Element|elements]] of $\Bbb S$.
By [[Inters... | Smallest Normal Subgroup containing Set | https://proofwiki.org/wiki/Smallest_Normal_Subgroup_containing_Set | https://proofwiki.org/wiki/Smallest_Normal_Subgroup_containing_Set | [
"Normal Subgroups"
] | [
"Definition:Group",
"Definition:Normal Subgroup"
] | [
"Definition:Set",
"Definition:Normal Subgroup",
"Definition:Set Intersection",
"Definition:Element",
"Intersection of Normal Subgroups is Normal",
"Definition:Set Intersection",
"Category:Normal Subgroups"
] |
proofwiki-1080 | Conjugate of Set with Inverse Closed for Inverses | Let $G$ be a group.
Let $S \subseteq G$.
Let $\hat S = S \cup S^{-1}$, where $S^{-1}$ is the set of all the inverses of all the elements of $S$.
Let $\tilde S = \set {a s a^{-1}: s \in \hat S, a \in G}$.
That is, $\tilde S$ is the set containing all the conjugates of the elements of $S$ and all their inverses.
Then:
: ... | Let $x \in \tilde S$.
That is:
:$\exists s \in \hat S: x = a s a^{-1}$
Then:
{{begin-eqn}}
{{eqn | l = x^{-1}
| r = \paren {a s a^{-1} }^{-1}
| c =
}}
{{eqn | r = a s^{-1} a^{-1}
| c = Power of Conjugate equals Conjugate of Power
}}
{{end-eqn}}
Since $s^{-1} \in \hat S$, it follows that $x^{-1} \in \... | Let $G$ be a [[Definition:Group|group]].
Let $S \subseteq G$.
Let $\hat S = S \cup S^{-1}$, where $S^{-1}$ is the [[Definition:Inverse of Subset of Group|set of all the inverses]] of all the elements of $S$.
Let $\tilde S = \set {a s a^{-1}: s \in \hat S, a \in G}$.
That is, $\tilde S$ is the [[Definition:Set|set]]... | Let $x \in \tilde S$.
That is:
:$\exists s \in \hat S: x = a s a^{-1}$
Then:
{{begin-eqn}}
{{eqn | l = x^{-1}
| r = \paren {a s a^{-1} }^{-1}
| c =
}}
{{eqn | r = a s^{-1} a^{-1}
| c = [[Power of Conjugate equals Conjugate of Power]]
}}
{{end-eqn}}
Since $s^{-1} \in \hat S$, it follows that $x^... | Conjugate of Set with Inverse Closed for Inverses | https://proofwiki.org/wiki/Conjugate_of_Set_with_Inverse_Closed_for_Inverses | https://proofwiki.org/wiki/Conjugate_of_Set_with_Inverse_Closed_for_Inverses | [
"Conjugacy"
] | [
"Definition:Group",
"Definition:Inverse of Subset/Group",
"Definition:Set",
"Definition:Conjugate (Group Theory)/Element",
"Definition:Element",
"Definition:Inverse (Abstract Algebra)/Inverse"
] | [
"Power of Conjugate equals Conjugate of Power",
"Category:Conjugacy"
] |
proofwiki-1081 | Conjugate of Set with Inverse is Closed | Let $G$ be a group.
Let $S \subseteq G$.
Let $\hat S = S \cup S$.
Let $\tilde S = \set {a s a^{-1}: s \in \hat S, a \in G}$.
Let $\map W {\tilde S}$ be the set of words of $\tilde S$.
Then:
:$\forall w \in \map W {\tilde S}: \forall a \in G: a w a^{-1} \in \map W {\tilde S}$ | Let $w \in \map W {\tilde S}$.
From the definition of $\map W {\tilde S}$, we have:
:$w = \paren {a_1 s_1 a_1^{-1} } \paren {a_2 s_2 a_2^{-1} } \cdots \paren {a_n s_n a_n^{-1} }, n \in \N^*, a_i \in G, s_1 \in \hat S, 1 \le i \le n$
Thus:
{{begin-eqn}}
{{eqn | l = a w a^{-1}
| r = a \paren {a_1 s_1 a_1^{-1} ... | Let $G$ be a [[Definition:Group|group]].
Let $S \subseteq G$.
Let $\hat S = S \cup S$.
Let $\tilde S = \set {a s a^{-1}: s \in \hat S, a \in G}$.
Let $\map W {\tilde S}$ be the [[Definition:Word (Abstract Algebra)|set of words]] of $\tilde S$.
Then:
:$\forall w \in \map W {\tilde S}: \forall a \in G: a w a^{-1} \... | Let $w \in \map W {\tilde S}$.
From the [[Definition:Word (Abstract Algebra)|definition]] of $\map W {\tilde S}$, we have:
:$w = \paren {a_1 s_1 a_1^{-1} } \paren {a_2 s_2 a_2^{-1} } \cdots \paren {a_n s_n a_n^{-1} }, n \in \N^*, a_i \in G, s_1 \in \hat S, 1 \le i \le n$
Thus:
{{begin-eqn}}
{{eqn | l = a w a^... | Conjugate of Set with Inverse is Closed | https://proofwiki.org/wiki/Conjugate_of_Set_with_Inverse_is_Closed | https://proofwiki.org/wiki/Conjugate_of_Set_with_Inverse_is_Closed | [
"Conjugacy"
] | [
"Definition:Group",
"Definition:Word (Abstract Algebra)"
] | [
"Definition:Word (Abstract Algebra)",
"Definition:Group",
"Category:Conjugacy"
] |
proofwiki-1082 | Generator of Normal Subgroup | Let $G$ be a group.
Let $S \subseteq G$.
Let $\hat S = S \cup S^{-1}$, where $S^{-1}$ is the set of all the inverses of all the elements of $S$.
Let $\tilde S = \set {a s a^{-1}: s \in \hat S, a \in G}$.
Let $\map W {\tilde S}$ be the set of words of $\tilde S$.
Let $N$ be the smallest normal subgroup of $G$ that conta... | Let $N$ be the smallest normal subgroup of $G$ that contains $S$, where $S \subseteq G$.
$N$ must certainly include $\hat S$, because any group containing $s \in S$ must also contain $s^{-1}$.
Therefore, $N$ must be the smallest normal subgroup containing $\hat S$.
Since $N \lhd G$, it follows that $\forall a \in G: \... | Let $G$ be a [[Definition:Group|group]].
Let $S \subseteq G$.
Let $\hat S = S \cup S^{-1}$, where $S^{-1}$ is the [[Definition:Inverse of Subset of Group|set of all the inverses]] of all the elements of $S$.
Let $\tilde S = \set {a s a^{-1}: s \in \hat S, a \in G}$.
Let $\map W {\tilde S}$ be the [[Definition:Word ... | Let $N$ be the smallest [[Definition:Normal Subgroup|normal subgroup]] of $G$ that contains $S$, where $S \subseteq G$.
$N$ must certainly include $\hat S$, because any group containing $s \in S$ must also contain $s^{-1}$.
Therefore, $N$ must be the smallest normal subgroup containing $\hat S$.
Since $N \lhd G$, ... | Generator of Normal Subgroup | https://proofwiki.org/wiki/Generator_of_Normal_Subgroup | https://proofwiki.org/wiki/Generator_of_Normal_Subgroup | [
"Normal Subgroups"
] | [
"Definition:Group",
"Definition:Inverse of Subset/Group",
"Definition:Word (Abstract Algebra)",
"Definition:Normal Subgroup"
] | [
"Definition:Normal Subgroup",
"Axiom:Group Axioms",
"Conjugate of Set with Inverse Closed for Inverses",
"Set of Words Generates Group/Corollary",
"Conjugate of Set with Inverse is Closed",
"Subgroup is Normal iff Contains Conjugate Elements",
"Category:Normal Subgroups"
] |
proofwiki-1083 | Subset Product with Normal Subgroup as Generator | Let $G$ be a group whose identity is $e$.
Let:
:$H$ be a subgroup of $G$
:$N$ be a normal subgroup of $G$.
Then:
:$N \lhd \gen {N, H} = N H = H N \le G$
where:
:$\le$ denotes subgroup
:$\lhd$ denotes normal subgroup
:$\gen {N, H}$ denotes a subgroup generator
:$N H$ denotes subset product. | From Subset Product is Subset of Generator:
:$N H \subseteq \gen {N, H}$
From Subset Product with Normal Subgroup is Subgroup:
:$N H = H N \le G$
Then by the definition of a subgroup generator, $\gen {N, H}$ is the smallest subgroup containing $N H$ and so:
:$\gen {N, H} = N H = H N \le G$
From Normal Subgroup of Subse... | Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let:
:$H$ be a [[Definition:Subgroup|subgroup]] of $G$
:$N$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
Then:
:$N \lhd \gen {N, H} = N H = H N \le G$
where:
:$\le$ denotes [[Definition:Subgroup|subgroup]]
:... | From [[Subset Product is Subset of Generator]]:
:$N H \subseteq \gen {N, H}$
From [[Subset Product with Normal Subgroup is Subgroup]]:
:$N H = H N \le G$
Then by the definition of a [[Definition:Generator of Subgroup|subgroup generator]], $\gen {N, H}$ is the smallest [[Definition:Subgroup|subgroup]] containing $N H$... | Subset Product with Normal Subgroup as Generator | https://proofwiki.org/wiki/Subset_Product_with_Normal_Subgroup_as_Generator | https://proofwiki.org/wiki/Subset_Product_with_Normal_Subgroup_as_Generator | [
"Normal Subgroups",
"Subset Products"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Subgroup",
"Definition:Normal Subgroup",
"Definition:Subgroup",
"Definition:Normal Subgroup",
"Definition:Generator of Subgroup",
"Definition:Subset Product"
] | [
"Subset Product is Subset of Generator",
"Subset Product with Normal Subgroup is Subgroup",
"Definition:Generator of Subgroup",
"Definition:Subgroup",
"Normal Subgroup of Subset Product of Subgroups"
] |
proofwiki-1084 | Subset Product of Normal Subgroups is Normal | Let $\struct {G, \circ}$ be a group.
Let $N$ and $N'$ be normal subgroups of $G$.
Then $N N'$ is also a normal subgroup of $G$. | From Subset Product with Normal Subgroup is Subgroup, we already have that $N N'$ is a subgroup of $G$.
Let $n n' \in N N'$, so that $n \in N, n' \in N'$.
Let $g \in G$.
From Subgroup is Normal iff Contains Conjugate Elements:
:$g n g^{-1}\in N, g n' g^{-1}\in N'$
So:
:$\paren {g n g^{-1} } \paren {g n' g^{-1} } = g n ... | Let $\struct {G, \circ}$ be a [[Definition:Group|group]].
Let $N$ and $N'$ be [[Definition:Normal Subgroup|normal subgroups]] of $G$.
Then $N N'$ is also a [[Definition:Normal Subgroup|normal subgroup]] of $G$. | From [[Subset Product with Normal Subgroup is Subgroup]], we already have that $N N'$ is a [[Definition:Subgroup|subgroup]] of $G$.
Let $n n' \in N N'$, so that $n \in N, n' \in N'$.
Let $g \in G$.
From [[Subgroup is Normal iff Contains Conjugate Elements]]:
:$g n g^{-1}\in N, g n' g^{-1}\in N'$
So:
:$\paren {g n g... | Subset Product of Normal Subgroups is Normal | https://proofwiki.org/wiki/Subset_Product_of_Normal_Subgroups_is_Normal | https://proofwiki.org/wiki/Subset_Product_of_Normal_Subgroups_is_Normal | [
"Normal Subgroups",
"Subset Products"
] | [
"Definition:Group",
"Definition:Normal Subgroup",
"Definition:Normal Subgroup"
] | [
"Subset Product with Normal Subgroup is Subgroup",
"Definition:Subgroup",
"Subgroup is Normal iff Contains Conjugate Elements",
"Definition:Normal Subgroup"
] |
proofwiki-1085 | Prime Group is Simple | Groups of prime order are simple. | Follows directly from Prime Group has no Proper Subgroups: a group of prime order has only itself and the trivial group as subgroups.
From Trivial Subgroup and Group Itself are Normal, these subgroups are normal.
{{qed}} | [[Definition:Group|Groups]] of [[Definition:Prime Number|prime]] [[Definition:Order of Structure|order]] are [[Definition:Simple Group|simple]]. | Follows directly from [[Prime Group has no Proper Subgroups]]: a group of prime order has only itself and the [[Definition:Trivial Group|trivial group]] as [[Definition:Subgroup|subgroups]].
From [[Trivial Subgroup and Group Itself are Normal]], these subgroups are normal.
{{qed}} | Prime Group is Simple | https://proofwiki.org/wiki/Prime_Group_is_Simple | https://proofwiki.org/wiki/Prime_Group_is_Simple | [
"Simple Groups",
"Prime Groups"
] | [
"Definition:Group",
"Definition:Prime Number",
"Definition:Order of Structure",
"Definition:Simple Group"
] | [
"Prime Group has no Proper Subgroups",
"Definition:Trivial Group",
"Definition:Subgroup",
"Trivial Subgroup and Group Itself are Normal"
] |
proofwiki-1086 | Prime Group has no Proper Subgroups | A nontrivial group $G$ has no proper subgroups except the trivial group {{iff}} $G$ is finite and its order is prime. | === Sufficient Condition ===
Suppose $G$ is finite and of prime order $p$.
From Lagrange's Theorem, the order of any subgroup of $G$ must divide the order $p$ of $G$.
From the definition of prime, any subgroups of $p$ can therefore only have order $1$ or $p$.
Hence $G$ can have only itself and the trivial group as subg... | A [[Definition:Non-Trivial Group|nontrivial group]] $G$ has no [[Definition:Proper Subgroup|proper subgroups]] except the [[Definition:Trivial Group|trivial group]] {{iff}} $G$ is [[Definition:Finite Group|finite]] and its [[Definition:Order of Group|order]] is [[Definition:Prime Number|prime]]. | === Sufficient Condition ===
Suppose $G$ is [[Definition:Finite Group|finite]] and of [[Definition:Prime Number|prime]] [[Definition:Order of Group|order]] $p$.
From [[Lagrange's Theorem (Group Theory)|Lagrange's Theorem]], the [[Definition:Order of Group|order]] of any [[Definition:Subgroup|subgroup]] of $G$ must [[... | Prime Group has no Proper Subgroups | https://proofwiki.org/wiki/Prime_Group_has_no_Proper_Subgroups | https://proofwiki.org/wiki/Prime_Group_has_no_Proper_Subgroups | [
"Subgroups",
"Prime Groups"
] | [
"Definition:Non-Trivial Group",
"Definition:Proper Subgroup",
"Definition:Trivial Group",
"Definition:Finite Group",
"Definition:Order of Structure",
"Definition:Prime Number"
] | [
"Definition:Finite Group",
"Definition:Prime Number",
"Definition:Order of Structure",
"Lagrange's Theorem (Group Theory)",
"Definition:Order of Structure",
"Definition:Subgroup",
"Definition:Divisor (Algebra)/Integer",
"Definition:Order of Structure",
"Definition:Prime Number",
"Definition:Subgro... |
proofwiki-1087 | Quotient Group of Cyclic Group | Let $G$ be a cyclic group which is generated by $g$.
Let $H$ be a subgroup of $G$.
Then $g H$ generates $G / H$. | Let $G$ be a cyclic group generated by $g$.
Let $H \le G$.
We need to show that every element of $G / H$ is of the form $\left({g H}\right)^k$ for some $k \in \Z$.
Suppose $x H \in G / H$.
Then, since $G$ is generated by $g$, $x = g^k$ for some $k \in \Z$.
But $\left({g H}\right)^k = \left({g^k}\right) H = x H$.
So $g ... | Let $G$ be a [[Definition:Cyclic Group|cyclic group]] which is [[Definition:Generator of Cyclic Group|generated by $g$]].
Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
Then $g H$ [[Definition:Generator of Cyclic Group|generates]] $G / H$. | Let $G$ be a [[Definition:Cyclic Group|cyclic group]] [[Definition:Generator of Cyclic Group|generated by $g$]].
Let $H \le G$.
We need to show that every element of $G / H$ is of the form $\left({g H}\right)^k$ for some $k \in \Z$.
Suppose $x H \in G / H$.
Then, since $G$ is generated by $g$, $x = g^k$ for some $... | Quotient Group of Cyclic Group/Proof 1 | https://proofwiki.org/wiki/Quotient_Group_of_Cyclic_Group | https://proofwiki.org/wiki/Quotient_Group_of_Cyclic_Group/Proof_1 | [
"Cyclic Groups",
"Quotient Groups",
"Quotient Group of Cyclic Group"
] | [
"Definition:Cyclic Group",
"Definition:Cyclic Group/Generator",
"Definition:Subgroup",
"Definition:Cyclic Group/Generator"
] | [
"Definition:Cyclic Group",
"Definition:Cyclic Group/Generator",
"Definition:Cyclic Group/Generator"
] |
proofwiki-1088 | Quotient Group of Cyclic Group | Let $G$ be a cyclic group which is generated by $g$.
Let $H$ be a subgroup of $G$.
Then $g H$ generates $G / H$. | Let $H$ be a subgroup of the cyclic group $G = \gen g$.
Then by Homomorphism of Powers for Integers:
:$\forall n \in \Z: \map {q_H} {g^n} = \paren {\map {q_H} g}^n = \paren {g H}^n$
As $G = \set {g^n: n \in \Z}$, we conclude that:
:$G / H = q_H \sqbrk G = \set {\paren {g H}^n: n \in \Z}$
Thus, by Epimorphism from Integ... | Let $G$ be a [[Definition:Cyclic Group|cyclic group]] which is [[Definition:Generator of Cyclic Group|generated by $g$]].
Let $H$ be a [[Definition:Subgroup|subgroup]] of $G$.
Then $g H$ [[Definition:Generator of Cyclic Group|generates]] $G / H$. | Let $H$ be a [[Definition:Subgroup|subgroup]] of the [[Definition:Cyclic Group|cyclic group]] $G = \gen g$.
Then by [[Homomorphism of Powers/Integers|Homomorphism of Powers for Integers]]:
:$\forall n \in \Z: \map {q_H} {g^n} = \paren {\map {q_H} g}^n = \paren {g H}^n$
As $G = \set {g^n: n \in \Z}$, we conclude that... | Quotient Group of Cyclic Group/Proof 2 | https://proofwiki.org/wiki/Quotient_Group_of_Cyclic_Group | https://proofwiki.org/wiki/Quotient_Group_of_Cyclic_Group/Proof_2 | [
"Cyclic Groups",
"Quotient Groups",
"Quotient Group of Cyclic Group"
] | [
"Definition:Cyclic Group",
"Definition:Cyclic Group/Generator",
"Definition:Subgroup",
"Definition:Cyclic Group/Generator"
] | [
"Definition:Subgroup",
"Definition:Cyclic Group",
"Homomorphism of Powers/Integers",
"Epimorphism from Integers to Cyclic Group"
] |
proofwiki-1089 | Order of Element in Quotient Group | Let $G$ be a group, and let $H$ be a normal subgroup of $G$.
Let $G / H$ be the quotient group of $G$ by $H$.
The order of $a H \in G / H$ divides the order of $a \in G$. | Let $G$ be a group with normal subgroup $H$.
Let $G / H$ be the quotient of $G$ by $H$.
From Quotient Group Epimorphism is Epimorphism, $G / H$ is a homomorphic image of $G$.
Let $q_H: G \to G / H$ given by $\map f a = a H$ be that quotient epimorphism.
Let $a \in G$ such that $a^n = e$ for some integer $n$.
Then, by t... | Let $G$ be a [[Definition:Group|group]], and let $H$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
Let $G / H$ be the [[Definition:Quotient Group|quotient group]] of $G$ by $H$.
The [[Definition:Order of Group Element|order]] of $a H \in G / H$ [[Definition:Divisor of Integer|divides]] the [[Definition... | Let $G$ be a [[Definition:Group|group]] with [[Definition:Normal Subgroup|normal subgroup]] $H$.
Let $G / H$ be the [[Definition:Quotient Group|quotient of $G$ by $H$]].
From [[Quotient Group Epimorphism is Epimorphism]], $G / H$ is a [[Definition:Homomorphic Image|homomorphic image]] of $G$.
Let $q_H: G \to G / H$ ... | Order of Element in Quotient Group | https://proofwiki.org/wiki/Order_of_Element_in_Quotient_Group | https://proofwiki.org/wiki/Order_of_Element_in_Quotient_Group | [
"Quotient Groups"
] | [
"Definition:Group",
"Definition:Normal Subgroup",
"Definition:Quotient Group",
"Definition:Order of Group Element",
"Definition:Divisor (Algebra)/Integer",
"Definition:Order of Group Element"
] | [
"Definition:Group",
"Definition:Normal Subgroup",
"Definition:Quotient Group",
"Quotient Epimorphism is Epimorphism/Group",
"Definition:Homomorphism (Abstract Algebra)/Image",
"Definition:Quotient Epimorphism",
"Definition:Integer",
"Definition:Morphism Property",
"Definition:By Hypothesis",
"Defi... |
proofwiki-1090 | Intersection of Quotient Groups | Let $N \lhd G$ be a normal subgroup of $G$.
Let:
:$N \le A \le G$
:$N \le B \le G$
For a subgroup $H$ of $G$, let $\alpha$ be the bijection defined as:
:$\map \alpha H = \set {h N: h \in H}$
Then:
:$\map \alpha {A \cap B} = \map \alpha A \cap \map \alpha B$ | From the proof of the Correspondence Theorem:
:$\map \alpha H \subseteq G / N$
Then:
{{begin-eqn}}
{{eqn | l = \map \alpha A \cap \map \alpha B
| r = \set {g N: g \in A \cap B}
| c =
}}
{{eqn | r = \paren {A \cap B} / N
| c =
}}
{{eqn | r = \map \alpha {A \cap B}
| c =
}}
{{end-eqn}}
{{qed}} | Let $N \lhd G$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
Let:
:$N \le A \le G$
:$N \le B \le G$
For a [[Definition:Subgroup|subgroup]] $H$ of $G$, let $\alpha$ be the [[Definition:Bijection|bijection]] defined as:
:$\map \alpha H = \set {h N: h \in H}$
Then:
:$\map \alpha {A \cap B} = \map \alpha ... | From the proof of the [[Correspondence Theorem (Group Theory)|Correspondence Theorem]]:
:$\map \alpha H \subseteq G / N$
Then:
{{begin-eqn}}
{{eqn | l = \map \alpha A \cap \map \alpha B
| r = \set {g N: g \in A \cap B}
| c =
}}
{{eqn | r = \paren {A \cap B} / N
| c =
}}
{{eqn | r = \map \alpha {... | Intersection of Quotient Groups | https://proofwiki.org/wiki/Intersection_of_Quotient_Groups | https://proofwiki.org/wiki/Intersection_of_Quotient_Groups | [
"Quotient Groups"
] | [
"Definition:Normal Subgroup",
"Definition:Subgroup",
"Definition:Bijection"
] | [
"Correspondence Theorem (Group Theory)"
] |
proofwiki-1091 | Generator of Quotient Groups | Let $N \lhd G$ be a normal subgroup of $G$.
Let:
:$N \le A \le G$
:$N \le B \le G$
For a subgroup $H$ of $G$, let $\alpha$ be the bijection defined as:
:$\map \alpha H = \set {h N: h \in H}$
Then:
:$\map \alpha {\gen {A, B} } = \gen {\map \alpha A, \map \alpha B}$
where $\gen {A, B}$ denotes the subgroup of $G$ generat... | From the proof of the Correspondence Theorem:
:$\map \alpha H \subseteq G / N$
Then:
{{begin-eqn}}
{{eqn | l = \map \alpha {\gen {A, B} }
| r = \set {h N: h \in \gen {A, B} }
| c =
}}
{{eqn | r = \set {h N \in \gen {A / N, B / N} }
| c =
}}
{{eqn | r = \gen {\map \alpha A, \map \alpha B}
| c =... | Let $N \lhd G$ be a [[Definition:Normal Subgroup|normal subgroup]] of $G$.
Let:
:$N \le A \le G$
:$N \le B \le G$
For a [[Definition:Subgroup|subgroup]] $H$ of $G$, let $\alpha$ be the [[Definition:Bijection|bijection]] defined as:
:$\map \alpha H = \set {h N: h \in H}$
Then:
:$\map \alpha {\gen {A, B} } = \gen {\m... | From the proof of the [[Correspondence Theorem (Group Theory)|Correspondence Theorem]]:
:$\map \alpha H \subseteq G / N$
Then:
{{begin-eqn}}
{{eqn | l = \map \alpha {\gen {A, B} }
| r = \set {h N: h \in \gen {A, B} }
| c =
}}
{{eqn | r = \set {h N \in \gen {A / N, B / N} }
| c =
}}
{{eqn | r = \... | Generator of Quotient Groups | https://proofwiki.org/wiki/Generator_of_Quotient_Groups | https://proofwiki.org/wiki/Generator_of_Quotient_Groups | [
"Quotient Groups",
"Generated Subgroups"
] | [
"Definition:Normal Subgroup",
"Definition:Subgroup",
"Definition:Bijection",
"Definition:Generated Subgroup"
] | [
"Correspondence Theorem (Group Theory)"
] |
proofwiki-1092 | Quotient of Group by Center Cyclic implies Abelian | Let $G$ be a group.
Let $\map Z G$ be the center of $G$.
Let $G / \map Z G$ be the quotient group of $G$ by $\map Z G$.
Let $G / \map Z G$ be cyclic.
Then $G$ is abelian, so $G = \map Z G$.
That is, the group $G / \map Z G$ cannot be a cyclic group which is non-trivial. | Suppose $G / \map Z G$ is cyclic.
Then by definition:
:$\exists \tau \in G / \map Z G: G / \map Z G = \gen \tau$
Since $\tau$ is a coset by $\map Z G$:
:$\exists t \in G: \tau = t \map Z G$
Thus each coset of $\map Z G$ in $G$ is equal to $\paren {t \map Z G}^i = t^i \map Z G$ for some $i \in \Z$.
Now let $x, y \in G$.... | Let $G$ be a [[Definition:Group|group]].
Let $\map Z G$ be the [[Definition:Center of Group|center]] of $G$.
Let $G / \map Z G$ be the [[Definition:Quotient Group|quotient group]] of $G$ by $\map Z G$.
Let $G / \map Z G$ be [[Definition:Cyclic Group|cyclic]].
Then $G$ is [[Definition:Abelian Group|abelian]], so $G... | Suppose $G / \map Z G$ is [[Definition:Cyclic Group|cyclic]].
Then by definition:
:$\exists \tau \in G / \map Z G: G / \map Z G = \gen \tau$
Since $\tau$ is a [[Definition:Coset|coset]] by $\map Z G$:
:$\exists t \in G: \tau = t \map Z G$
Thus each [[Definition:Coset|coset]] of $\map Z G$ in $G$ is equal to $\paren ... | Quotient of Group by Center Cyclic implies Abelian | https://proofwiki.org/wiki/Quotient_of_Group_by_Center_Cyclic_implies_Abelian | https://proofwiki.org/wiki/Quotient_of_Group_by_Center_Cyclic_implies_Abelian | [
"Quotient Groups",
"Cyclic Groups",
"Abelian Groups",
"Centers of Groups"
] | [
"Definition:Group",
"Definition:Center (Abstract Algebra)/Group",
"Definition:Quotient Group",
"Definition:Cyclic Group",
"Definition:Abelian Group",
"Definition:Cyclic Group",
"Definition:Non-Trivial Group"
] | [
"Definition:Cyclic Group",
"Definition:Coset",
"Definition:Coset",
"Subset Product within Semigroup is Associative",
"Definition:Commutative/Elements",
"Definition:Center (Abstract Algebra)/Group",
"Powers of Group Elements/Sum of Indices",
"Powers of Group Elements/Sum of Indices",
"Definition:Comm... |
proofwiki-1093 | Centralizer is Normal Subgroup of Normalizer | Let $G$ be a group.
Let $H \le G$ be a subgroup of $G$.
Let $\map {C_G} H$ be the centralizer of $H$ in $G$.
Let $\map {N_G} H$ be the normalizer of $H$ in $G$.
Let $\Aut H$ be the automorphism group of $H$.
Then:
:$(1): \quad \map {C_G} H \lhd \map {N_G} H$
:$(2): \quad \map {N_G} H / \map {C_G} H \cong K$
where:
:$\... | In order to invoke the First Isomorphism Theorem for Groups, we must construct a group homomorphism $\phi: \map {N_G} H \to \Aut H$.
Consider the mapping $\phi: x \mapsto \paren {g \mapsto x g x^{-1}}$.
From Inner Automorphism is Automorphism, $g \mapsto x g x^{-1}$ is an automorphism of $G$, so $\phi$ is well-defined.... | Let $G$ be a [[Definition:Group|group]].
Let $H \le G$ be a [[Definition:Subgroup|subgroup]] of $G$.
Let $\map {C_G} H$ be the [[Definition:Centralizer of Subgroup|centralizer]] of $H$ in $G$.
Let $\map {N_G} H$ be the [[Definition:Normalizer|normalizer]] of $H$ in $G$.
Let $\Aut H$ be the [[Definition:Automorphis... | In order to invoke the [[First Isomorphism Theorem for Groups]], we must construct a [[Definition:Group Homomorphism|group homomorphism]] $\phi: \map {N_G} H \to \Aut H$.
Consider the mapping $\phi: x \mapsto \paren {g \mapsto x g x^{-1}}$.
From [[Inner Automorphism is Automorphism]], $g \mapsto x g x^{-1}$ is an [[D... | Centralizer is Normal Subgroup of Normalizer | https://proofwiki.org/wiki/Centralizer_is_Normal_Subgroup_of_Normalizer | https://proofwiki.org/wiki/Centralizer_is_Normal_Subgroup_of_Normalizer | [
"Normal Subgroups",
"Normalizers",
"Quotient Groups"
] | [
"Definition:Group",
"Definition:Subgroup",
"Definition:Centralizer/Subgroup",
"Definition:Normalizer",
"Definition:Automorphism Group/Group",
"Definition:Quotient Group",
"Definition:Subgroup"
] | [
"First Isomorphism Theorem/Groups",
"Definition:Group Homomorphism",
"Inner Automorphism is Automorphism",
"Definition:Group Automorphism",
"Definition:Group Automorphism",
"Definition:Group Homomorphism",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Kernel is Normal Subgroup of Domai... |
proofwiki-1094 | Number of Distinct Conjugate Subsets is Index of Normalizer | Let $G$ be a group.
Let $S$ be a subset of $G$.
Let $\map {N_G} S$ be the normalizer of $S$ in $G$.
Let $\index G {\map {N_G} S}$ be the index of $\map {N_G} S$ in $G$.
The number of distinct subsets of $G$ which are conjugates of $S \subseteq G$ is $\index G {\map {N_G} S}$. | We have that:
:$S^a = S^b \iff S^{a b^{-1}} = S$ (reference to be determined).
That is:
:$S^a = S^b \iff a b^{-1} \in \map {N_G} S$
which is equivalent to:
:$a^{-1} \equiv b^{-1} \pmod {\map {N_G} S}$
Thus we have a bijection between:
:the conjugacy class $\conjclass S$ of subsets of $G$ conjugate to $S$
and:
the left ... | Let $G$ be a [[Definition:Group|group]].
Let $S$ be a [[Definition:Subset|subset]] of $G$.
Let $\map {N_G} S$ be the [[Definition:Normalizer|normalizer]] of $S$ in $G$.
Let $\index G {\map {N_G} S}$ be the [[Definition:Index of Subgroup|index of $\map {N_G} S$ in $G$]].
The number of distinct [[Definition:Subset|s... | We have that:
:$S^a = S^b \iff S^{a b^{-1}} = S$ (reference to be determined).
That is:
:$S^a = S^b \iff a b^{-1} \in \map {N_G} S$
which is equivalent to:
:$a^{-1} \equiv b^{-1} \pmod {\map {N_G} S}$
Thus we have a [[Definition:Bijection|bijection]] between:
:the [[Definition:Conjugacy Class|conjugacy class]] $\conj... | Number of Distinct Conjugate Subsets is Index of Normalizer/Proof 1 | https://proofwiki.org/wiki/Number_of_Distinct_Conjugate_Subsets_is_Index_of_Normalizer | https://proofwiki.org/wiki/Number_of_Distinct_Conjugate_Subsets_is_Index_of_Normalizer/Proof_1 | [
"Normalizers",
"Conjugacy",
"Number of Distinct Conjugate Subsets is Index of Normalizer"
] | [
"Definition:Group",
"Definition:Subset",
"Definition:Normalizer",
"Definition:Index of Subgroup",
"Definition:Subset",
"Definition:Conjugate (Group Theory)/Subset"
] | [
"Definition:Bijection",
"Definition:Conjugacy Class",
"Definition:Subset",
"Definition:Conjugate (Group Theory)/Subset",
"Definition:Coset Space/Left Coset Space",
"Definition:Element"
] |
proofwiki-1095 | Number of Distinct Conjugate Subsets is Index of Normalizer | Let $G$ be a group.
Let $S$ be a subset of $G$.
Let $\map {N_G} S$ be the normalizer of $S$ in $G$.
Let $\index G {\map {N_G} S}$ be the index of $\map {N_G} S$ in $G$.
The number of distinct subsets of $G$ which are conjugates of $S \subseteq G$ is $\index G {\map {N_G} S}$. | Let $G$ act on its power set $\powerset G$ by the rule:
:$\forall g \in G, S \in \powerset G: g * S := S^{g^{-1} } = \set {x \in G: g^{-1} x g \in S}$
That is, the conjugacy action on subsets.
From Conjugacy Action on Subsets is Group Action, $*$ is a group action.
The orbit of $S \in \powerset G$ is the conjugacy cla... | Let $G$ be a [[Definition:Group|group]].
Let $S$ be a [[Definition:Subset|subset]] of $G$.
Let $\map {N_G} S$ be the [[Definition:Normalizer|normalizer]] of $S$ in $G$.
Let $\index G {\map {N_G} S}$ be the [[Definition:Index of Subgroup|index of $\map {N_G} S$ in $G$]].
The number of distinct [[Definition:Subset|s... | Let $G$ [[Definition:Group Action|act on]] its [[Definition:Power Set|power set]] $\powerset G$ by the rule:
:$\forall g \in G, S \in \powerset G: g * S := S^{g^{-1} } = \set {x \in G: g^{-1} x g \in S}$
That is, the [[Definition:Conjugacy Action on Subsets|conjugacy action on subsets]].
From [[Conjugacy Action on Su... | Number of Distinct Conjugate Subsets is Index of Normalizer/Proof 2 | https://proofwiki.org/wiki/Number_of_Distinct_Conjugate_Subsets_is_Index_of_Normalizer | https://proofwiki.org/wiki/Number_of_Distinct_Conjugate_Subsets_is_Index_of_Normalizer/Proof_2 | [
"Normalizers",
"Conjugacy",
"Number of Distinct Conjugate Subsets is Index of Normalizer"
] | [
"Definition:Group",
"Definition:Subset",
"Definition:Normalizer",
"Definition:Index of Subgroup",
"Definition:Subset",
"Definition:Conjugate (Group Theory)/Subset"
] | [
"Definition:Group Action",
"Definition:Power Set",
"Definition:Conjugacy Action/Subsets",
"Conjugacy Action on Subsets is Group Action",
"Definition:Group Action",
"Definition:Orbit (Group Theory)",
"Definition:Conjugacy Class",
"Definition:Set",
"Definition:Distinct",
"Definition:Subset",
"Defi... |
proofwiki-1096 | Subset has 2 Conjugates then Normal Subgroup | Let $G$ be a group.
Let $S$ be a subset of $G$.
Let $S$ have exactly two conjugates in $G$.
Then $G$ has a proper non-trivial normal subgroup. | {{MissingLinks}}
Consider the centralizer $\map {C_G} S$ of $S$ in $G$.
From Centralizer of Group Subset is Subgroup, $\map {C_G} S$ is a subgroup of $G$.
If $\map {C_G} S = G$, then $S$ has no conjugate but itself.
{{explain|Link to that result.}}
So, in order for $S$ to have exactly two conjugates in $G$, it is neces... | Let $G$ be a [[Definition:Group|group]].
Let $S$ be a [[Definition:Subset|subset]] of $G$.
Let $S$ have exactly two [[Definition:Conjugate of Group Subset|conjugates]] in $G$.
Then $G$ has a [[Definition:Proper Subgroup|proper]] [[Definition:Non-Trivial Group|non-trivial]] [[Definition:Normal Subgroup|normal subgro... | {{MissingLinks}}
Consider the [[Definition:Centralizer|centralizer]] $\map {C_G} S$ of $S$ in $G$.
From [[Centralizer of Group Subset is Subgroup]], $\map {C_G} S$ is a [[Definition:Subgroup|subgroup]] of $G$.
If $\map {C_G} S = G$, then $S$ has no [[Definition:Conjugate of Group Subset|conjugate]] but itself.
{{ex... | Subset has 2 Conjugates then Normal Subgroup | https://proofwiki.org/wiki/Subset_has_2_Conjugates_then_Normal_Subgroup | https://proofwiki.org/wiki/Subset_has_2_Conjugates_then_Normal_Subgroup | [
"Conjugacy",
"Normal Subgroups"
] | [
"Definition:Group",
"Definition:Subset",
"Definition:Conjugate (Group Theory)/Subset",
"Definition:Proper Subgroup",
"Definition:Non-Trivial Group",
"Definition:Normal Subgroup"
] | [
"Definition:Centralizer",
"Centralizer of Group Subset is Subgroup",
"Definition:Subgroup",
"Definition:Conjugate (Group Theory)/Subset",
"Definition:Conjugate (Group Theory)/Subset",
"Definition:Proper Subgroup",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Conjugate (Grou... |
proofwiki-1097 | Element of Group Not Conjugate to Proper Subgroup | Let $G$ be a finite group.
Let $H$ be a proper subgroup of $G$.
Then there is at least one element of $G$ not contained in $H$ or in any of its conjugates. | Let $S \subseteq G$ be defined by:
:$S := \set {g \in G: \exists h \in H, a \in G: g = a h a^{-1} }$
It is required to show that $S \ne G$.
Let $\map {N_G} H$ be the normalizer of $H$ in $G$.
By Subgroup is Subgroup of Normalizer, $H \le \map {N_G} H$.
Therefore, by definition of index:
:$\index G {\map {N_G} H} \le \i... | Let $G$ be a [[Definition:Finite Group|finite group]].
Let $H$ be a [[Definition:Proper Subgroup|proper subgroup]] of $G$.
Then there is at least one [[Definition:Element|element]] of $G$ not contained in $H$ or in any of its [[Definition:Conjugate of Group Subset|conjugates]]. | Let $S \subseteq G$ be defined by:
:$S := \set {g \in G: \exists h \in H, a \in G: g = a h a^{-1} }$
It is required to show that $S \ne G$.
Let $\map {N_G} H$ be the [[Definition:Normalizer|normalizer]] of $H$ in $G$.
By [[Subgroup is Subgroup of Normalizer]], $H \le \map {N_G} H$.
Therefore, by definition of [[D... | Element of Group Not Conjugate to Proper Subgroup | https://proofwiki.org/wiki/Element_of_Group_Not_Conjugate_to_Proper_Subgroup | https://proofwiki.org/wiki/Element_of_Group_Not_Conjugate_to_Proper_Subgroup | [
"Conjugacy"
] | [
"Definition:Finite Group",
"Definition:Proper Subgroup",
"Definition:Element",
"Definition:Conjugate (Group Theory)/Subset"
] | [
"Definition:Normalizer",
"Subgroup is Subgroup of Normalizer",
"Definition:Index of Subgroup",
"Definition:Conjugate (Group Theory)/Subset",
"Number of Distinct Conjugate Subsets is Index of Normalizer",
"Definition:Conjugate (Group Theory)/Subset",
"Definition:Conjugate (Group Theory)/Subset",
"Defin... |
proofwiki-1098 | Parity Group is Group | The parity group is in fact a group. | We can completely describe the parity group by showing its Cayley table:
:<nowiki>$\begin{array}{r|rr}
\struct {\set {1, -1}, \times} & 1 & -1\\
\hline
1 & 1 & -1 \\
-1 & -1 & 1 \\
\end{array} \qquad
\begin{array}{r|rr}
\struct {\Z_2, +_2} & \eqclass 0 2 & \eqclass 1 2 \\
\hline
\eqclass 0 2 & \eqclass 0 2 & \eqclass 1... | The [[Definition:Parity Group|parity group]] is in fact a [[Definition:Group|group]]. | We can completely describe the parity group by showing its [[Definition:Cayley Table|Cayley table]]:
:<nowiki>$\begin{array}{r|rr}
\struct {\set {1, -1}, \times} & 1 & -1\\
\hline
1 & 1 & -1 \\
-1 & -1 & 1 \\
\end{array} \qquad
\begin{array}{r|rr}
\struct {\Z_2, +_2} & \eqclass 0 2 & \eqclass 1 2 \\
\hline
\eqclass 0 ... | Parity Group is Group | https://proofwiki.org/wiki/Parity_Group_is_Group | https://proofwiki.org/wiki/Parity_Group_is_Group | [
"Parity Group",
"Groups of Order 2"
] | [
"Definition:Parity Group",
"Definition:Group"
] | [
"Definition:Cayley Table",
"Prime Group is Cyclic",
"Definition:Group",
"Definition:Order of Structure",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Parity Group",
"Definition:Isomorphism (Abstract Algebra)/Group Isomorphism",
"Definition:Cyclic Group",
"Definition:Ord... |
proofwiki-1099 | Inverse of Inner Automorphism | Let $G$ be a group.
Let $x \in G$.
Let $\kappa_x$ be the inner automorphism of $G$ given by $x$.
Then:
:$\paren {\kappa_x}^{-1} = \kappa_{x^{-1} }$ | Let $G$ be a group whose identity is $e$.
Let $x \in G$.
Let $\kappa_x \in \Inn G$.
Then from the definition of inner automorphism:
:$\forall g \in G: \map {\kappa_x} g = x g x^{-1}$
As $G$ is a group:
:$x \in G \implies x^{-1} \in G$
So:
:$\kappa_{x^{-1} } \in \Inn G$
and is defined as:
:$\forall g \in G: \map {\kappa... | Let $G$ be a [[Definition:Group|group]].
Let $x \in G$.
Let $\kappa_x$ be the [[Definition:Inner Automorphism|inner automorphism of $G$ given by $x$]].
Then:
:$\paren {\kappa_x}^{-1} = \kappa_{x^{-1} }$ | Let $G$ be a [[Definition:Group|group]] whose [[Definition:Identity Element|identity]] is $e$.
Let $x \in G$.
Let $\kappa_x \in \Inn G$.
Then from the definition of [[Definition:Inner Automorphism|inner automorphism]]:
:$\forall g \in G: \map {\kappa_x} g = x g x^{-1}$
As $G$ is a [[Definition:Group|group]]:
:$x \... | Inverse of Inner Automorphism | https://proofwiki.org/wiki/Inverse_of_Inner_Automorphism | https://proofwiki.org/wiki/Inverse_of_Inner_Automorphism | [
"Inner Automorphisms"
] | [
"Definition:Group",
"Definition:Inner Automorphism"
] | [
"Definition:Group",
"Definition:Identity (Abstract Algebra)/Two-Sided Identity",
"Definition:Inner Automorphism",
"Definition:Group",
"Definition:Identity Mapping",
"Definition:Group",
"Definition:Group",
"Category:Inner Automorphisms"
] |
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