hcju/proofwikiqa huggingface dataset:
=== 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 u k + m v n}       | c =  }} {{eqn | r = a^{m u k}       | c = as $a^{m v n} = e$ }} {{eqn | r = \paren {a^k}^{m u}       | c =  }} {{end-eqn}} Thus $a^k$ generate $C_n$. {{qed|lemma}} === Sufficient Condition === Let $C_n = \gen {a^k}$. That is, let $a^k$ generate $C_n$. {{begin-eqn}} {{eqn | lo= \exists u \in \Z:       | l = a       | r = \paren {a^k}^u       | c = as $a$ is an element of the group generated by $a^k$ }} {{eqn | ll= \leadsto       | l = u k       | o = \equiv       | r = 1       | rr= \pmod n       | c =  }} {{eqn | ll= \leadsto       | lo= \exists u, v \in \Z:       | l = 1       | r = u k + v n       | c =  }} {{eqn | ll= \leadsto       | l = k       | o = \perp       | r = n       | c = Integer Combination of Coprime Integers }} {{end-eqn}} {{qed}}
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}}
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, $g H = H g$. The result follows from the definition of normal subgroup. {{qed}}
Let $\displaystyle N = \bigcap_{i \mathop \in I} N_i$. 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 subgroups in $\set {N_i: i \in I}$, and hence in their intersection $N$. That is, $a N a^{-1} \subseteq N$. The result follows by Subgroup is Superset of Conjugate iff Normal. {{qed}}
{{begin-eqn}} {{eqn | l = H       | o = \lhd       | r = G       | c = where $\lhd$ denotes that $H$ is normal in $G$ }} {{eqn | ll= \leadstoandfrom       | l = \forall g \in G: g H g^{-1}       | o = \subseteq       | r = H       | c = {{Defof|Normal Subgroup}} }} {{eqn | ll= \leadstoandfrom       | l = \forall x \in H: \forall g \in G: g x g^{-1}       | o = \in       | r = H       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \forall x \in H: \conjclass x       | o = \subseteq       | r = H       | c = where $\conjclass x$ is the conjugacy class of $x \in G$ }} {{eqn | ll= \leadstoandfrom       | l = H       | r = \bigcup_{x \mathop \in H} \conjclass x       | c =  }} {{end-eqn}} Hence the result. {{qed}}
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 order $\order H$. Hence any subgroup whose order is $\order H$ must in fact ''be'' $H$. That is, $g H g^{-1} = H$. The result follows from Subgroup equals Conjugate iff Normal. {{Qed}}
By definition, $\left \langle {N_i: i \in I} \right \rangle$ 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: : $\left \langle {N_i: i \in I} \right \rangle \lhd G$ {{qed}} Category:Normal Subgroups bae4k1vl8whwigmiqi2pq8hjl1rthii
Let $\Bbb S$ be the set of all normal subgroups of $G$ that contain $S$. $\Bbb S \ne \varnothing$, since $S \subseteq G \lhd G$. Let $N = \bigcap H: H \in \Bbb S$, that is, the intersection of all elements of $\Bbb S$. By Intersection of Normal Subgroups is Normal, $N \lhd G$, and by the definition of intersection, $S \subseteq N$. By the method of construction, $N$ is the smallest such subgroup. By the definition of "smallest", $N$ is unique. {{qed}} Category:Normal Subgroups syea0j00dnpyhtp2qtjk5ukvyhh4kzb
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 $x^{-1} \in \tilde S$. {{Qed}} Category:Conjugacy o56izk8q21jzbzwty9uvh72whozxfbh
Let $w \in W \left({\tilde S}\right)$. From the definition of $W \left({\tilde S}\right)$, we have: $w = \left({a_1 s_1 a_1^{-1}}\right) \left({a_2 s_2 a_2^{-1}}\right) \cdots \left({a_n s_n a_n^{-1}}\right), n \in \N^*, a_i \in G, s_1 \in \hat S, 1 \le i \le n$. Thus: {{begin-eqn}}      {{eqn      | l=a w a^{-1}            | r=a \left({a_1 s_1 a_1^{-1} }\right) \left({a_2 s_2 a_2^{-1} }\right) \cdots \left({a_n s_n a_n^{-1} }\right) a^{-1}            | c= }} {{eqn      | r=a \left({a_1 s_1 a_1^{-1} }\right) a^{-1} a \left({a_2 s_2 a_2^{-1} }\right) a^{-1} \cdots  a \left({a_n s_n a_n^{-1} }\right) a^{-1}            | c= }} {{eqn      | r=\left({\left({a a_1}\right) s_1 \left({a a_1}\right)^{-1} }\right) \left({\left({a a_2}\right) s_2 \left({a a_2}\right)^{-1} }\right) \cdots  \left({\left({a a_n}\right) s_n \left({a a_n}\right)^{-1} }\right)            | c= }} {{end-eqn}} As $G$ is a group, all of the $a a_i \in G$. The result follows. {{qed}} Category:Conjugacy jompv56uogazp88n2mpvhk944cn1tqz
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 $\forall a \in G: \forall s \in \hat S: a s a^{-1} \in N$. Thus, $\forall x \in \tilde S: x \in N$. Thus $\tilde S \subseteq N$. By the closure axiom, $N$ must also contain all products of any finite number of elements of $\tilde S$. Thus $W \left({\tilde S}\right) \subseteq N$. Now we prove that $W \left({\tilde S}\right) \lhd G$. $\tilde S \ne \varnothing$, as $e \left({s s^{-1}}\right) e^{-1} = e \in \tilde S$. By Conjugate of Set with Inverse Closed for Inverses, $\tilde S$ is closed under taking inverses. So from Set of Words Generates Group: Corollary: : $W \left({\tilde S}\right) \le G$ From Conjugate of Set with Inverse is Closed: : $\forall w \in W \left({\tilde S}\right): \forall a \in G: a w a^{-1} \in W \left({\tilde S}\right)$ From Subgroup is Normal iff Contains Conjugate Elements: : $W \left({\tilde S}\right) \lhd G$. The result follows by the definition of the minimality of $N$. {{qed}} Category:Normal Subgroups tmf3it242yg20gtqnpz2aoa5sswhrjt
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 Subset Product of Subgroups we have that: :$N \lhd N H$ Hence the result. {{qed}}
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: :$\paren {g n g^{-1} } \paren {g n' g^{-1} } = g n n' g^{-1} \in N N'$ So $N N'$ is normal. {{qed}}
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}}
=== 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 subgroups. {{qed|lemma}} === Necessary Condition === Suppose $G$ is ''not'' finite and prime. Let the identity of $G$ be $e$. Let $h \in G$ be an element of $G$ such that $h \ne e$. Then $H = \gen h$ is a cyclic subgroup of $G$. If $H \ne G$ then $H$ is a non-trivial proper subgroup of $G$, and the proof is complete. Otherwise, $H = G$ is a cyclic group and there are two possibilities: :$(1): \quad G$ is infinite :$(2): \quad G$ is finite (and of non-prime order). First, suppose $G$ is infinite. From Infinite Cyclic Group is Isomorphic to Integers, $G$ is isomorphic to $\struct {\Z, +}$. From Subgroups of Additive Group of Integers, $\struct {\Z, +}$ has proper subgroups, for example: $\gen 2$. Because $G \cong \struct {\Z, +}$, then so does $G$ have proper subgroups, and the proof is complete. Suppose $G$ is finite, and of order $n$ where $n$ is not prime . Then: : $\exists d \in \N: d \divides n, 1 < d < n$ From Subgroup of Finite Cyclic Group is Determined by Order, $G$ has a proper subgroup of order $d$ and again, the proof is complete. {{qed}}
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 $g H$ generates $G / H$. {{Qed}}
Let $H$ be a subgroup of the cyclic group $G = \left \langle {g} \right \rangle$. Then by Homomorphism of Powers for Integers: :$\forall n \in \Z: q_H \left({g^n}\right) = \left({q_H \left({g}\right)}\right)^n = \left({g H}\right)^n$ As $G = \left\{{g^n: n \in \Z}\right\}$, we conclude that: :$G / H = q_H \left({G}\right) = \left\{{\left({g H}\right)^n: n \in \Z}\right\}$ Thus, by Epimorphism from Integers to Cyclic Group, $g H$ generates $G / H$. {{Qed}}
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 the morphism property of $q_H$: {{begin-eqn}} {{eqn | l = \map {q_H} {a^n}       | r = \paren {\map {q_H} a}^n       | c =  }} {{eqn | r = \paren {a H}^n       | c =  }} {{eqn | r = a^n H       | c = {{Defof|Coset Product}} }} {{eqn | r = e H       | c = by hypothesis }} {{eqn | r = H       | c =  }} {{end-eqn}} Hence $\order H$ divides $n$. {{qed}}
Let $G / H$ be abelian. Then: {{begin-eqn}} {{eqn | lo= \forall a H, b H \in G / H:       | l = a H b H       | r = b H a H       | c =  }} {{eqn | ll= \leadsto       | l = a b H       | r = b a H       | c =  }} {{eqn | ll= \leadsto       | l = a b \paren {b a}^{-1}       | o = \in       | r = H       | c =  }} {{eqn | ll= \leadsto       | l = a b a^{-1} b^{-1}       | o = \in       | r = H       | c =  }} {{end-eqn}} The argument reverses. Suppose that $\forall a, b \in G: a b a^{-1} b^{-1} \in H$. {{begin-eqn}} {{eqn | lo= \forall a, b \in G:       | l = a b a^{-1} b^{-1}       | o = \in       | r = H       | c =  }} {{eqn | ll= \leadsto       | l = a b \paren {b a}^{-1}       | o = \in       | r = H       | c =  }} {{eqn | ll= \leadsto       | l = a b H       | r = b a H       | c =  }} {{eqn | ll= \leadsto       | l = a H b H       | r = b H a H       | c =  }} {{end-eqn}} {{Qed}}
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}}
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 =  }} {{end-eqn}} {{qed}}
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 $x, y \in G$. Suppose $x \in t^m \map Z G, y \in t^n \map Z G$. Then $x = t^m z_1, y = t^n z_2$ for some $z_1, z_2 \in \map Z G$. Thus: {{begin-eqn}} {{eqn | l = x y       | r = t^m z_1 t^n z_2       | c =  }} {{eqn | r = t^m t^n z_1 z_2       | c = $z_1$ commutes with all $t \in G$ since it is in the center }} {{eqn | r = t^{m + n} z_1 z_2       | c =  }} {{eqn | r = t^{n + m} z_2 z_1       | c =  }} {{eqn | r = t^n t^m z_2 z_1       | c =  }} {{eqn | r = t^n z_2 t^m z_1       | c = $z_2$ commutes with all $t \in G$ since it is in the center }} {{eqn | r = y x       | c =  }} {{end-eqn}} This holds for all $x, y \in G$, and thus $G$ is abelian. Thus by Group equals Center iff Abelian $\map Z G = G$. Therefore Quotient of Group by Itself it follows that $G / \map Z G$ is the trivial group. {{qed}}
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. To see that $\phi$ is a homomorphism, notice that for any $x, y \in \map {N_G} H$: {{begin-eqn}} {{eqn | l = \map \phi x \map \phi y       | r = \paren {g \mapsto x g x^{-1} } \circ \paren {g \mapsto y g y^{-1} }       | c = where $\circ$ denote composition of maps }} {{eqn | r = g \mapsto x \paren {y g y^{-1} } x^{-1} }} {{eqn | r = g \mapsto \paren {x y} g \paren {x y}^{-1}       | c = Inverse of Group Product }} {{eqn | r = \map \phi {x y} }} {{end-eqn}} Hence $\phi$ is a homomorphism. Now we prove that $\ker \phi = \map {C_G} H$. Note that for $x \in \map {N_G} H$: {{begin-eqn}} {{eqn | l = x       | o = \in       | r = \ker \phi }} {{eqn | ll = \leadstoandfrom       | l = g       | r = x g x^{-1}       | rr = \forall g \in H       | c = $g \mapsto g$ is the identity of $\Aut H$ }} {{eqn | ll = \leadstoandfrom       | l = g x       | r = x g       | rr = \forall g \in H }} {{eqn | ll = \leadstoandfrom       | l = x       | o = \in       | r = \map {C_G} H       | c = {{Defof|Centralizer of Subgroup}} }} {{end-eqn}} Hence $\ker \phi = \map {C_G} H$. By Kernel is Normal Subgroup of Domain: :$\map {C_G} H \lhd \map {N_G} H$ By First Isomorphism Theorem for Groups: :$\map {N_G} H / \map {C_G} H \cong \Img \phi$ By Image of Group Homomorphism is Subgroup: :$\Img \phi \le \Aut H$ Hence the result. {{qed}}
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 class of $S$. That is, $\Orb S$ is the set of distinct subsets of $G$ which are conjugate to $S$ The stabilizer of $S$ is its normalizer $\map {N_G} S$. From the Orbit-Stabilizer Theorem: :$\order {\Orb x} = \index G {\Stab x} = \dfrac {\order G} {\order {\Stab x} }$ from which the result follows directly. {{Qed}}
{{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 necessary for $\map {C_G} S$ to be a proper subgroup. Let $e$ be the identity of $G$. If $\map {C_G} S = \set e$, then for there to be exactly two conjugates of $S$:  :$\forall a \ne b \in G \setminus \set e: b x b^{-1} = a x a^{-1}$ But: {{begin-eqn}} {{eqn | l = b x b^{-1}       | r = a x a^{-1}       | c =  }} {{eqn | ll= \leadsto       | l = \paren {a^{-1} b} x b^{-1}       | r = x a^{-1}       | c =  }} {{eqn | ll= \leadsto       | l = \paren {a^{-1} b} x \paren {a^{-1} b}^{-1}       | r = x       | c =  }} {{eqn | ll= \leadsto       | l = a^{-1} b       | o = \in       | r = \map {C_G} S       | c =  }} {{end-eqn}} This implies either that $\map {C_G} S$ is actually nontrivial, or that $a^{-1}b = e \iff a = b$, a contradiction. Thus $\map {C_G} S$ is a nontrivial proper subgroup of $G$. We have that there are exactly $2$ conjugacy classes of $S$. These are in one-to-one correspondence with cosets of $S$. Thus the index $\index G {\map {C_G} S}$ of the centralizer is: :$\index G {\map {C_G} S} = 2$ From Subgroup of Index 2 is Normal: :$\map {C_G} S$ is a proper nontrivial normal subgroup of $G$. {{qed}}
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: :$\index G {\map {N_G} H} \le \index G H$ Each of the conjugates of $H$ has $\order H$ elements. By Number of Distinct Conjugate Subsets is Index of Normalizer, there are $\index G {\map {N_G} H}$ conjugates. Therefore: :$\card S \le \order H \index G {\map {N_G} H}$ with equality {{iff}} all conjugates of $H$ are disjoint. Combining this with the earlier inequality, we find that: :$\card S \le \order H \index G H = \order G$ Now if $H$ is normal in $G$, then by Normalizer of Subgroup is Largest Subgroup containing that Subgroup as Normal Subgroup: :$\index G {\map {N_G} H} < \index G H$ for $\map {N_G} H = G$ in that case, and $H$ is a proper subgroup. If $H$ is not normal in $G$, then it has multiple conjugates. By Conjugate of Subgroup is Subgroup, they all are subgroups of $G$. In particular, then, the identity of $G$ is common to all of them. Hence they are not disjoint, and: :$\card S < \order H \index G {\map {N_G} H}$ In either case, it follows that: :$\card S < \order G$ as desired. {{qed}}
We can completely describe the parity group by showing its Cayley table: :$\begin{array}{r|rr} \struct {\set {1, -1}, \times} & 1 & -1\\ \hline 1 & 1 & -1 \\ -1 & -1 & 1 \\ \end{array} \qquad \begin{array}{r|rr} \struct {\Z_2, +_2} & \eqclass 0 2 & \eqclass 1 2 \\ \hline \eqclass 0 2 & \eqclass 0 2 & \eqclass 1 2 \\ \eqclass 1 2 & \eqclass 1 2 & \eqclass 0 2 \\ \end{array}$ From Prime Group is Cyclic, there is only one group of order $2$, up to isomorphism. Thus both instantiations of the parity group are isomorphic to $C_2$, the cyclic group of order $2$. {{qed}}
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: :$\forall g \in G: \map {\kappa_{x^{-1} } } g = x^{-1} g \paren {x^{-1} }^{-1} = x^{-1} g x$ Now we need to show that: :$\kappa_x \circ \kappa_{x^{-1} } = I_G = \kappa_{x^{-1} } \circ \kappa_x$ where $I_G: G \to G$ is the identity mapping. So: {{begin-eqn}} {{eqn | ll= \forall g \in G:       | l = \map {\kappa_x \circ \kappa_{x^{-1} } } g       | r = \map {\kappa_x} {\map {\kappa_{x^{-1} } } g}       | c = {{Defof|Composition of Mappings}} }} {{eqn | r = \map {\kappa_x } {x^{-1} g x}       | c = Definition of $\kappa_{x^{-1} }$ }} {{eqn | r = x \paren {x^{-1} g x} x^{-1}       | c = Definition of $\kappa_x$ }} {{eqn | r = g       | c = Group Properties }} {{eqn | r = x^{-1} \paren {x g x^{-1} } x       | c = Group Properties }} {{eqn | r = \map {\kappa_{x^{-1} } } {x g x^{-1} }       | c = Definition of $\kappa_{x^{-1} }$ }} {{eqn | r = \map {\kappa_{x^{-1} } } {\map {\kappa_x} g}       | c = Definition of $\kappa_x$ }} {{eqn | r = \map {\kappa_{x^{-1} } \circ \kappa_x} g       | c = {{Defof|Composition of Mappings}} }} {{end-eqn}} Thus: :$\forall g \in G: \map {\kappa_x \circ \kappa_{x^{-1} } } g = \map {I_G} g = \map {\kappa_{x^{-1} } \circ \kappa_x} g$ Hence the result. {{qed}} Category:Inner Automorphisms se1c1eh10u9lnbhu4qlivuoylagn5kf
Let $G$ be a group. Let the mapping $\kappa: G \to \Inn G$ be defined as: :$\map \kappa a = \kappa_a$ where $\kappa_a$ is the inner automorphism of $G$ given by $a$. From Kernel of Inner Automorphism Group is Center: :$\map \ker \kappa = \map Z G$ and also that: :$\Img \kappa = \Inn G$ From the First Isomorphism Theorem: :$\Img \kappa \cong G / \map \ker \kappa$ Thus, as $\map \ker \kappa = \map Z G$ and $\Img \kappa = \Inn G$: :$G / \map Z G \cong \Inn G$ {{qed}}
The centralizer of $a$ in $G$ is defined as: :$\map {C_G} a = \set {x \in G: x \circ a = a \circ x}$ Let $g, h \in G$. Then: {{begin-eqn}} {{eqn | l = g a g^{-1}       | r = h a h^{-1}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = g^{-1} \paren {g a g^{-1} } h       | r = g^{-1} \paren {h a h^{-1} } h       | c =  }} {{eqn | ll= \leadstoandfrom       | l = a g^{-1} h       | r = g^{-1} h a       | c =  }} {{eqn | ll= \leadstoandfrom       | l = g^{-1} h       | o = \in       | r = \map {C_G} a       | c = {{Defof|Centralizer of Group Element}} }} {{end-eqn}} By Elements in Same Left Coset iff Product with Inverse in Subgroup: :$g$ and $h$ belong to the same left coset of $\map {C_G} a$ {{iff}} $g^{-1} h \in \map {C_G} a$. The result follows. {{Qed}}
Let $x, y \in \conjclass a$. By definition of $\conjclass a$: :$x a x^{-1} = y a y^{-1}$ By Conjugates of Elements in Centralizer, this is the case {{iff}} $x$ and $y$ belong to the same left coset of $\map {C_G} a$. Hence: :$\card {\conjclass a} = \index G {\map {C_G} a}$ It follows from Lagrange's Theorem that: :$\card {\conjclass a} \divides \order G$ {{qed}}
The number of elements in $\conjclass x$ is the number of conjugates of the set $\set x$. From Number of Distinct Conjugate Subsets is Index of Normalizer, the number of distinct subsets of a $G$ which are conjugates of $S \subseteq G$ is $\index G {\map {N_G} S}$. The result follows. {{qed}}
Let $\conjclass a$ be the conjugacy class of $a$ in $G$. {{begin-eqn}} {{eqn | l = a       | o = \in       | r = \map Z G       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \forall x \in G: x a       | r = a x       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \forall x \in G: x a x^{-1}       | r = a       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \conjclass a       | r = \set a       | c =  }} {{end-eqn}} {{qed}}
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 from Group equals Center iff Abelian we have: :$\map Z G = G$ 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. {{qed|lemma}} === 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: :$\displaystyle \order {G \setminus \map Z G} = \sum_{j \mathop = 1}^m \order {\conjclass {x_j} }$ or: :$\displaystyle \order G - \order {\map Z G} = \sum_{j \mathop = 1}^m \order {\conjclass {x_j} }$ From Size of Conjugacy Class is Index of Normalizer: :$\order {\conjclass {x_j} } = \index G {\map {N_G} {x_j} }$ and the result follows. {{Qed}}