nyuuzyou/wikis huggingface dataset:
Congruent to Zero iff Modulo is Divisor

== Theorem ==
Let $a, z \in \R$.
Then $a$ is congruent to $0$ modulo $z$ $a$ is an integer multiple of $z$.
:$\exists k \in \Z: k z = a \iff a \equiv 0 \pmod z$
If $z \in \Z$, then further:
:$z \divides a \iff a \equiv 0 \pmod z$
== Proof ==
Thus by definition of congruence, $a \equiv 0 \pmod z$ and the result is proved.
If $z$ is an integer, then by definition of divisor:
:$z \divides a \iff \exists k \in \Z: a = k z$
Hence the result for integer $z$.
Category:Modulo Arithmetic
Conic Section through Five Points

== Theorem ==
Let $A, B, C, D, E$ be distinct points in the plane such that no $3$ of them are collinear.
Then it is possible to draw a conic section that passes through all $5$ points.
== Proof ==
== Historical Note ==
== Sources ==
* : $5$
* : $5$
Category:Conic Sections
Category:5
Conjugacy Action is not Transitive

== Theorem ==
Let $\struct {G, \circ}$ be a non-trivial group whose identity is $e$.
Let $*: G \times G \to G$ be the conjugacy group action:
:$\forall g, h \in G: g * h = g \circ h \circ g^{-1}$
Then $*$ is not a transitive group action.
== Proof ==
Proof by Counterexample:
For $G$ to be a transitive group action, the orbit of any element of $G$ needs to be the whole of $G$.
Take $h = e$.
Then:
Thus by definition of orbit:
:$\Orb e = \set e$
Only when $G$ is the trivial group, that is, when:
:$G = \set e$
then:
:$\Orb e = G$.
Hence the result by definition of transitive group action.
== Sources ==
* : $\S 3.3$: Group actions and coset decompositions: Examples of group actions: $\text{(v)}$
Category:Conjugacy Action
Category:Transitive Group Actions
Conjugacy Action on Abelian Group is Trivial

== Theorem ==
Let $\struct {G, \circ}$ be an abelian group whose identity is $e$.
Let $*: G \times G \to G$ be the conjugacy group action:
:$\forall g, h \in G: g * h = g \circ h \circ g^{-1}$
Then $*$ is a trivial group action.
== Proof ==
For $G$ to be a trivial group action, the orbit of any element of $G$ is a singleton containing only that element.
Take $h \in G$.
Then:
Thus by definition of orbit:
:$\Orb h = \set h$
Hence the result by definition of trivial group action.
== Sources ==
* : $\S 3.3$: Group actions and coset decompositions: Examples of group actions: $\text{(v)}$
Category:Conjugacy Action
Conjugacy Action on Group Elements is Group Action

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
The conjugacy action on $G$:
: $\forall g, h \in G: g * h = g \circ h \circ g^{-1}$
is a group action on itself.
== Proof ==
We have that:
:$e * x = e \circ x \circ e^{-1} = x$
and so is fulfilled.
 is shown to be fulfilled thus:
== Proof 2 ==
== Also see ==
* Stabilizer of Element under Conjugacy Action is Centralizer
* Orbit of Element under Conjugacy Action is Conjugacy Class
== Sources ==
* : $\S 5.5$. Groups acting on sets: Example $103$
* : $\S 3.3$: Group actions and coset decompositions: Examples of group actions: $\text{(v)}$
* : Chapter $10$: The Orbit-Stabiliser Theorem: Example $10.5$
Category:Conjugacy Action on Group Elements is Group Action
Category:Conjugacy Action
Conjugacy Action on Group Elements is Group Action/Proof 2

== Theorem ==
== Proof ==
Let $X$ be the set of all subgroups of $G$.
By definition, the (left) conjugacy action on subgroups is the group action $*_X : G \times X \to X$ defined as:
:$g *_X X = g \circ X \circ g^{-1}$
By Conjugacy Action on Subgroups is Group Action, the (left) conjugacy action on subgroups $*_X$ is a group action.
By Subset Product Action is Group Action, it follows that the conjugacy action $*: G \times G \to G$ such that:
:$\forall g, h \in G: g * h = g \circ h \circ g^{-1}$
is a group action, as required.
Category:Conjugacy Action on Group Elements is Group Action
Conjugacy Action on Identity

== Theorem ==
Let $G$ be a group whose identity is $e$.
For the conjugacy action:
:$\order {\Orb e} = 1$
and thus:
:$\Stab e = G$
== Proof ==
So the only conjugate of $e$ is $e$ itself.
Thus:
:$\Orb e = \set e$
and so:
:$\order {\Orb e} = 1$
From the Orbit-Stabilizer Theorem, it follows immediately that:
:$\Stab e = G$
== Sources ==
* : $\S 5.5$. Groups acting on sets: Example $103$
Category:Conjugacy Action
Conjugacy Action on Subgroups is Group Action

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $X$ be the set of all subgroups of $G$.
For any $H \le G$ and for any $g \in G$, the conjugacy action:
:$g * H := g \circ H \circ g^{-1}$
is a group action.
== Proof ==
Clearly is fulfilled as $e * H = H$.
 is shown to be fulfilled thus:
== Also see ==
* Stabilizer of Conjugacy Action on Subgroup is Normalizer
* Orbit of Conjugacy Action on Subgroup is Set of Conjugate Subgroups
== Sources ==
* : Chapter $10$: The Orbit-Stabiliser Theorem: Example $10.7$
Category:Conjugacy Action
Conjugacy Action on Subsets is Group Action

== Theorem ==
Let $\powerset G$ be the set of all subgroups of $G$.
For any $S \in \powerset G$ and for any $g \in G$, the conjugacy action:
:$g * S := g \circ S \circ g^{-1}$
is a group action.
== Proof ==
By , $e \in G$, thus:
Thus is seen to be fulfilled.
Then:
Thus is shown to be fulfilled.
== Sources ==
* : $\S 5.5$. Groups acting on sets: Example $105$
Category:Conjugacy Action
Conjugacy Class Equation

== Theorem ==
Let $G$ be a group.
Let $\order G$ denote the order of $G$.
Let $\map Z G$ denote the center of $G$.
Let $x \in G$.
Let $\map {N_G} x$ denote the normalizer of $x$ in $G$.
Let $\index G {\map {N_G} x}$ denote the index of $\map {N_G} x$ in $G$.
Let $m$ be the number of non-singleton conjugacy classes of $G$.
Let $x_j: j \in \set {1, 2, \ldots, m}$ be arbitrary elements of those conjugacy classes.
Then:
:$\ds \order G = \order {\map Z G} + \sum_{j \mathop = 1}^m \index G {\map {N_G} {x_j} }$
== Proof 1 ==
== Proof 2 ==
== Also known as ==
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: Exercise $25.16 \ \text{(c)}$
* 
Category:Conjugacy Class Equation
Category:Conjugacy
Category:Normal Subgroups
Category:Named Theorems
Category:Normalizers
Category:Conjugacy Classes
Conjugacy Class Equation/Proof 1

== Theorem ==
== Proof ==
From Conjugacy Class of Element of Center is Singleton, all elements of $\map Z G$ form their own singleton conjugacy classes.
=== Abelian Group ===
Suppose $G$ is abelian.
Then by definition of abelian group:
:$\map Z G = G$
So there are as many conjugacy classes as there are elements in $\map Z G$ and hence in $G$.
So in this case the result certainly holds.
=== Non-Abelian Group ===
Now suppose $G$ is non-abelian.
Thus:
:$\map Z G \ne G$
and therefore:
:$G \setminus \map Z G \ne \O$
From Conjugacy Class of Element of Center is Singleton, all the non-singleton conjugacy classes of $G$ are in $G \setminus \map Z G$.
From the way the theorem has been worded, there are $m$ of them.
Let us choose one element from each of the non-singleton conjugacy classes and call them $x_1, x_2, \ldots, x_m$.
Thus, these conjugacy classes can be written:
:$\conjclass {x_1}, \conjclass {x_2}, \ldots, \conjclass {x_m}$
So:
:$\ds \order {G \setminus \map Z G} = \sum_{j \mathop = 1}^m \order {\conjclass {x_j} }$
or:
:$\ds \order G - \order {\map Z G} = \sum_{j \mathop = 1}^m \order {\conjclass {x_j} }$
From Size of Conjugacy Class is Index of Normalizer:
:$\order {\conjclass {x_j} } = \index G {\map {N_G} {x_j} }$
and the result follows.
== Sources ==
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 51$
Category:Conjugacy Class Equation
Conjugacy Class Equation/Proof 2

== Theorem ==
== Proof ==
Let the distinct orbits of $G$ under the conjugacy action be:
:$\Orb {x_1}, \Orb {x_2}, \ldots, \Orb {x_s}$
Then from the Partition Equation:
:$\order G = \order {\Orb {x_1} } + \order {\Orb {x_2} } + \cdots + \order {\Orb {x_s} }$
From the Orbit-Stabilizer Theorem:
:$\order {\Orb {x_i} } \divides \order G, i = 1, \ldots, s$
The result follows from the definition of the conjugacy action.
== Sources ==
* : $\S 3.3$: Group actions and coset decompositions
Category:Conjugacy Class Equation
Conjugacy Class of Element of Center is Singleton

== Theorem ==
Let $G$ be a group.
Let $\map Z G$ denote the center of $G$.
The elements of $\map Z G$ form singleton conjugacy classes, and the elements of $G \setminus \map Z G$ belong to multi-element conjugacy classes.
=== Corollary ===
== Proof ==
Let $\conjclass a$ be the conjugacy class of $a$ in $G$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: Exercise $25.16 \ \text{(a)}$
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 51$
* : $\S 48.3$ Conjugacy
* 
Category:Conjugacy Classes
Category:Centers of Groups
Category:Singletons
Conjugacy Class of Element of Center is Singleton/Corollary

== Corollary to Conjugacy Class of Element of Center is Singleton ==
Let $G$ be a group.
Let $\map Z G$ denote the center of $G$.
The number of single-element conjugacy classes of $G$ is the order of $\map Z G$ and divides $\order G$.
== Proof ==
From Conjugacy Class of Element of Center is Singleton, each of the singleton conjugacy classes consists of one of the elements of $\map Z G$.
By Center of Group is Subgroup, $\map Z G$ is a subgroup of $G$.
It follows from Lagrange's Theorem that the number of these divides the order of $G$.
Category:Conjugacy Classes
Category:Normal Subgroups
Category:Singletons
Conjugacy Class of Identity is only Conjugacy Class which is Subgroup

== Theorem ==
Let $G$ be a group.
Let $e$ denote the identity of $G$.
Let $\conjclass g$ denote the conjugacy class of the element $g$.
Then conjugacy class of identity is the only conjugacy class which is a subgroup of $G$:
:$\conjclass g < G \iff g = e$
== Proof ==
=== Necessary Condition ===
Assume $g = e$.
Then by Identity of Group is in Singleton Conjugacy Class, $\conjclass e = \set e$, which is the trivial subgroup.
=== Sufficient Condition ===
Assume $g \neq e$.
Then by Conjugacy Classes are Disjoint, $e \notin \conjclass g$.
Since $\conjclass g$ does not contain the identity element, it could not be a subgroup.
== Sources ==
* 
Category:Conjugacy Classes
Conjugacy Classes of Quaternion Group

== Theorem ==
Let $Q = \Dic 2 = \gen {a, b: a^4 = e, b^2 = a^2, a b a = b}$ be the quaternion group.
The conjugacy classes of $\Dic 2$ are:
:$\set e, \set {a^2}, \set {a, a^3}, \set {b, a^2 b}, \set {a b, a^3 b}$
== Proof ==
From Center of Quaternion Group, we have:
:$\map Z {\Dic 2} = \set {e, a^2}$
Thus from Conjugacy Class of Element of Center is Singleton, $\set e$ and $\set {a^2}$ are two of those conjugacy classes.
By inspection of the Cayley table:
we investigate the remaining $6$ elements in turn, starting with $a$:
So we have a conjugacy class:
:$\set {a, a^3}$
Investigating the remaining $4$ elements in turn, starting with $b$:
So we have a conjugacy class:
:$\set {b, a^2 b}$
Investigating the remaining $2$ elements, starting with $a b$:
We need go no further: the remaining elements $a b$ and $a^3 b$ are in the same conjugacy class:
:$\set {a b, a^3 b}$
Hence the result.
== Sources ==
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 51 \alpha$
Category:Quaternion Group
Category:Examples of Conjugacy Classes
Conjugacy Classes of Symmetric Group

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $S_n$ denote the symmetric group on $n$ letters.
The conjugacy classes of $S_n$ are determined entirely by the cycle type.
That is, the conjugacy class $\conjclass x$ of an element $x$ of $S_n$ consists of all the elements of $S_n$ whose cycle type is the same as the cycle type of $x$.
== Proof ==
Let $\sigma \in S_n$ have cycle type $\tuple {k_1, k_2, \ldots, k_n}$.
Let $\rho$ be conjugate to $\sigma$
From Conjugate Permutations have Same Cycle Type, $\rho$ has the same cycle type $\tuple {k_1, k_2, \ldots, k_n}$ as $\sigma$.
That is, all the elements of the same conjugacy class have the same cycle type.
Let $\sigma, \rho \in S_n$ have the same cycle type $\tuple {k_1, k_2, \ldots, k_n}$.
It is to be demonstrated that $\sigma$ and $\rho$ are in the same conjugacy class.
From Existence and Uniqueness of Cycle Decomposition, $\sigma$ and $\rho$ can each be expressed uniquely as the product of disjoint cycles:
where $\alpha_i$ and $\beta_i$ are $k_i$-cycles.
For each $i$, let the $k_i$-cycles $\alpha_i$ and $\beta_i$ be expressed as:
For all $i, j$ such that $1 \le i \le l$, $1 \le j \le k_i$, let:
:$\tau := \map \tau {\alpha_{i j} } = \beta_{i j}$
Such a $\tau$ is bound to exist in $S_n$, as the underlying set of $S_n$ is the set of all permutations of $\set {1, 2, \ldots, n}$.
Thus from Product of Conjugates equals Conjugate of Products:
:$\tau \alpha_i \tau^{-1} = \beta_i$
Hence:
demonstrating that $\sigma$ and $\rho$ are conjugate.
That is, $\sigma$ and $\rho$ are in the same conjugacy class.
Hence the result.
Category:Symmetric Groups
Category:Conjugacy Classes
Conjugacy is Equivalence Relation

== Theorem ==
Conjugacy of group elements is an equivalence relation.
== Proof ==
Checking each of the criteria for an equivalence relation in turn:
=== Reflexive ===
:$\forall x \in G: e_G \circ x = x \circ e_G \implies x \sim x$
Thus conjugacy of group elements is reflexive.
=== Symmetric ===
Thus conjugacy of group elements is symmetric.
=== Transitive ===
Thus conjugacy of group elements is transitive.
All criteria are satisfied, and so conjugacy of group elements is shown to be an equivalence relation.
== Sources ==
* : Chapter $\text {II}$: Topological Spaces: $\S 10$. Equivalence Relations
* : Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: Exercise $25.16 \ \text{(a)}$
* : $\S 1.9$: Exercise $5.16 \ \text{(i)}$
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 51$
* : $\S 48.1$ Conjugacy
* 
Category:Conjugacy
Category:Examples of Equivalence Relations
Conjugate Lines are Harmonic Conjugates with respect to Tangents from Point of Intersection

== Theorem ==
Let $\CC$ be a circle.
Let $\PP$ and $\QQ$ be conjugate lines $\CC$.
Let $\PP$ and $\QQ$ intersect at $O$.
Let $OS$ and $OT$ be the tangents to $\CC$ from $O$.
Then $\PP$ and $\QQ$ are harmonic conjugates $OS$ and $OT$.
== Proof ==
Let $P$ and $Q$ be the poles of $\PP$ and $\QQ$ $\CC$.
Because $O$ lies on both $\PP$ and $\QQ$, its polar passes through both $P$ and $Q$.
That is, the polar of $O$ is $PQ$ itself.
:
The polar of $O$, by definition, is the chord of contact of $OS$ and $OT$ with $\CC$.
From Intersections of Line joining Conjugate Points with Circle form Harmonic Range, $\tuple {PQ, ST}$ forms a harmonic range.
Hence $\map O {PQ, ST}$ forms a harmonic pencil.
That is, $\PP$ and $\QQ$ are harmonic conjugates $OS$ and $OT$.
== Sources ==
* : Chapter $\text {III}$. The Circle: $8$. Reciprocal property of pole and polar
Category:Conjugate Lines
Conjugate Permutations have Same Cycle Type

== Theorem ==
Let $n \ge 1$ be a natural number.
Let $G$ be a subgroup of the symmetric group on $n$ letters $S_n$.
Let $\sigma, \rho \in G$.
Then $\sigma$ and $\rho$ are conjugate they have the same cycle type.
== Proof ==
Let $\sigma \in G$ have cycle type $\tuple {k_1, k_2, \ldots, k_n}$.
From Existence and Uniqueness of Cycle Decomposition, $\sigma$ can be expressed uniquely as the product of disjoint cycles:
:$\sigma = \alpha_1 \alpha_2 \dotsm \alpha_l$
where $\alpha_i$ is a $k_i$-cycle for some $i\in \set {1, 2, \ldots, l}$.
Let $\tau \in G$ such that:
:$\rho = \tau \sigma \tau^{-1}$
Then:
We have that for $i, j \in \set {1, 2, \ldots, l}$, $\alpha_i$ and $\alpha_j$ are disjoint.
We also have that $\tau$ is a bijection, and so an injection.
It follows that $\alpha_{\map \tau i}$ and $\alpha_{\map \tau j}$ are also disjoint for all $i, j \in \set {1, 2, \ldots, l}$.
Thus:
:the product in $(1)$ is $\rho$ written as the product of disjoint cycles
where:
:$\tau \alpha_i \tau^{-1}$ is a $k_i$-cycle.
Thus $\rho$ has the same cycle type as $\sigma$.
== Sources ==
* : Chapter $2$: The Symmetric Groups: $\S 80 \alpha$
Category:Conjugacy
Category:Permutations
Conjugate Transpose is Involution

== Theorem ==
Let $\mathbf A$ be a complex-valued matrix.
Let $\mathbf A^*$ denote the Hermitian conjugate of $\mathbf A$.
Then the operation of Hermitian conjugate is an involution:
:$\paren {\mathbf A^*}^* = \mathbf A$
== Proof ==
So:
:$\paren {\mathbf A^*}^* = \mathbf A$
Category:Linear Algebra
Category:Involutions
Conjugate of Commuting Elements

== Theorem ==
Let $\struct {S, \circ}$ be a monoid whose identity is $e_S$.
Let $x, y \in S$ such that $x$ and $y$ are both invertible.
Then $x \circ y \circ x^{-1} = y$ $x$ and $y$ commute.
== Proof ==
As $\struct {S, \circ}$ is a monoid, it is by definition a semigroup.
Therefore it is taken for granted that $\circ$ is associative, so we can dispense with parentheses.
We also take for granted the fact that $x$ and $y$ are cancellable from Invertible Element of Monoid is Cancellable.
So:
Category:Monoids
Category:Commutativity
Category:Conjugacy
Conjugate of Cycle

== Theorem ==
Let $n \ge 1$ be a natural number.
Let $S_n$ be the symmetric group on $n$ letters.
Let $\pi, \sigma \in S_n$.
Let $\sigma$ be a cycle of length $k$.
Then the conjugate $\pi \sigma \pi^{-1}$ is a cycle of length $k$.
== Proof ==
Follows directly from Conjugate Permutations have Same Cycle Type.
Category:Cyclic Permutations‎
Category:Conjugacy
Conjugate of Polynomial is Polynomial of Conjugate

== Theorem ==
Let $\map f z = a_n z^n + a_{n - 1} z^{n - 1} + \cdots + a_1 z + a_0$ be a polynomial over complex numbers where $a_0, \ldots, a_n$ are real numbers.
Let $\alpha \in \C$ be a complex number.
Then:
:$\overline {\map f \alpha} = \map f {\overline \alpha}$
where $\overline \alpha$ denotes the complex conjugate of $\alpha$.
== Proof ==
By Power of Complex Conjugate is Complex Conjugate of Power:
:$\overline {\alpha^k} = \paren {\overline \alpha}^k$
for all $k$ between $0$ and $n$.
Then from Product of Complex Conjugates:
:$\overline {a_k \alpha^k} = \overline {a_k} \cdot \overline {\alpha^k}$
But $a_k$ is real.
So by Complex Number equals Conjugate iff Wholly Real:
:$\overline {a_k} = a_k$.
From Sum of Complex Conjugates, it follows that:
:$\overline {\map f \alpha} = a_n \paren {\overline \alpha}^n + a_{n - 1} \paren {\overline \alpha}^{n - 1} + \cdots + a_1 \overline \alpha + a_0$
Hence the result.
== Sources ==
* : $\S 1.2$. The Algebraic Theory: $(1.17)$
Category:Polynomial Theory
Category:Complex Conjugates
Conjugate of Real Polynomial is Polynomial in Conjugate

== Theorem ==
Let $\map P z$ be a polynomial in a complex number $z$.
Let the coefficients of $P$ all be real.
Then:
:$\overline {\map P z} = \map P {\overline z}$
where $\overline z$ denotes the complex conjugate of $z$.
== Proof ==
Let $\map P z$ be expressed as:
:$a_n z^n + a_{n - 1} z^{n - 1} + \cdots + a_1 z + a_0$
Then:
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $133$
Category:Complex Conjugates
Conjugate of Set by Group Product

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $S \subseteq G$.
Let $S^a$ denote the $G$-conjugate of $S$ by $a$ as:
:$S^a := \set {y \in G: \exists x \in S: y = a \circ x \circ a^{-1} } = a \circ S \circ a^{-1}$
Then:
:$\paren {S^a}^b = S^{b \circ a}$
=== Also defined as ===
The concept of set conjugate can be defined in a different way:
Let $S^a$ denote the $G$-conjugate of $S$ by $a$ as:
:$S^a := \set {y \in G: \exists x \in S: y = a^{-1} \circ x \circ a} = a^{-1} \circ S \circ a$
Then:
:$\paren {S^a}^b = S^{a \circ b}$
== Proof ==
$S^a$ is defined as $a \circ S \circ a^{-1}$ from the definition of the conjugate of a set.
From the definition of subset product with a singleton, this can be seen to be the same thing as:
:$S^a = \set a \circ S \circ \set {a^{-1} }$.
Thus we can express $\paren {S^a}^b$ as $b \circ \paren {a \circ S \circ a^{-1} } \circ b^{-1}$, and understand that the refers to subset products.
From Subset Product within Semigroup is Associative (which applies because $\circ$ is associative), it then follows directly that:
=== Proof for Alternative Definition ===
Using the same preliminary argument as above, we then follow:
== Comment ==
This is not always correct in the literature.
For example, defines set conjugate as:
:$S^a := a \circ S \circ a^{-1}$
but then states (without proof) the assertion:
:$\paren {S^a}^b = S^{a \circ b}$
== Sources ==
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 45$
Category:Conjugacy
Conjugate of Set by Identity

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $S \subseteq G$.
Then the conjugate of $S$ by $e$ is $S$:
:$S^e = S$
== Proof ==
== Sources ==
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 45$
Category:Conjugacy
Conjugate of Set with Inverse Closed for Inverses

== Theorem ==
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:
: $\forall x \in \tilde S: x^{-1} \in \tilde S$
== Proof ==
Let $x \in \tilde S$.
That is:
:$\exists s \in \hat S: x = a s a^{-1}$
Then:
Since $s^{-1} \in \hat S$, it follows that $x^{-1} \in \tilde S$.
Category:Conjugacy
Conjugate of Set with Inverse is Closed

== Theorem ==
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:
:$\forall w \in \map W {\tilde S}: \forall a \in G: a w a^{-1} \in \map W {\tilde S}$
== Proof ==
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:
 
As $G$ is a group, all of the $a a_i \in G$.
The result follows.
Category:Conjugacy
Conjugate of Subgroup is Subgroup

== Theorem ==
Let $G$ be a group.
Let $H \le G$ be a subgroup of $G$.
Then the conjugate of $H$ by $a$ is a subgroup of $G$:
:$\forall H \le G, a \in G: H^a \le G$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text{II}$: Groups: Exercise $\text{R}$
* : Chapter $2$: Subgroups and Cosets: $\S 35 \gamma$
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $5$
* : Chapter $7$: Normal subgroups and quotient groups: Exercise $1$
Category:Conjugacy
Category:Subgroups
Category:Conjugate of Subgroup is Subgroup
Conjugate of Subgroup is Subgroup/Proof 1

== Theorem ==
== Proof ==
Let $H \le G$.
First, we show that $x, y \in H^a \implies x \circ y \in H^a$:
Next, we show that $x \in H^a \implies x^{-1} \in H^a$:
Thus by the Two-Step Subgroup Test, $H^a \le G$.
== Sources ==
* : $\S 6.6$. Normal subgroups: Example $124$
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 45$
* : Chapter $7$: Normal subgroups and quotient groups: Proposition $7.2$
Category:Conjugate of Subgroup is Subgroup
Conjugate of Subgroup is Subgroup/Proof 2

== Theorem ==
== Proof ==
Let $*: G \times G / H \to G / H$ be the group action on the (left) coset space:
:$\forall g \in G, \forall g' H \in G / H: g * \paren {g' H} := \paren {g g'} H$
It is established in Action of Group on Coset Space is Group Action that $*$ is a group action.
Then from Stabilizer of Coset under Group Action on Coset Space:
:$\Stab {a H} = a H a^{-1}$
where $\Stab {a H}$ the stabilizer of $a H$ under $*$.
The result follows from Stabilizer is Subgroup.
== Sources ==
* : Chapter $6$: Cosets: Exercise $9$
Category:Conjugate of Subgroup is Subgroup