avewright/proofwiki-math huggingface dataset:
Theorem: 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:
:$\map {N_G} H / \map {C_G} H$ is the quotient group of $\map {N_G} H$ by $\map {C_G} H$
:$K$ is a subgroup of $\Aut H$.

Proof:
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.
As $\phi$ is an group automorphism, $\phi$ is {{afortiori}} a  group 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}}
Theorem: 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 $\index G {\map {N_G} S}$.

Proof:
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 coset space $G / \map {N_G} S$
given by $S^a \to a^{-1} \map {N_G} S$.
Since $G / \map {N_G} S$ has $\index G {\map {N_G} S}$ elements, the result follows.
{{qed}}
Theorem: 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 $\index G {\map {N_G} S}$.

Proof:
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}}
Theorem: 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.

Proof:
{{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}}
Theorem: 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.

Proof:
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}}
Theorem: Parity Group is Group

The parity group is in fact a group.

Proof:
We can completely describe the parity group by showing its Cayley table:
:<nowiki>$\begin{array}{r|rr}
\struct {\set {1, -1}, \times} & 1 & -1\\
\hline
1 & 1 & -1 \\
-1 & -1 & 1 \\
\end{array} \qquad
\begin{array}{r|rr}
\struct {\Z_2, +_2} & \eqclass 0 2 & \eqclass 1 2 \\
\hline
\eqclass 0 2 & \eqclass 0 2 & \eqclass 1 2 \\
\eqclass 1 2 & \eqclass 1 2 & \eqclass 0 2 \\
\end{array}$</nowiki>
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}}
Theorem: 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:
:$\paren {\kappa_x}^{-1} = \kappa_{x^{-1} }$

Proof:
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 | q = \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
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Theorem: Inner Automorphism Group is Isomorphic to Quotient Group with Center

Let $G$ be a group.
Let $\Inn G$ be the inner automorphism group of $G$.
Let $\map Z G$ be the center of $G$.
Let $G / \map Z G$ be the quotient group of $G$ by $\map Z G$.
Then $G / \map Z G \cong \Inn G$.

Proof:
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}}
Theorem: Conjugates of Elements in Centralizer

Let $G$ be a group.
Let $\map {C_G} a$ be the centralizer of $a$ in $G$.
Then $\forall g, h \in G: g a g^{-1} = h a h^{-1}$ {{iff}} $g$ and $h$ belong to the same left coset of $\map {C_G} a$.

Proof:
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}}
Theorem: Number of Conjugates is Number of Cosets of Centralizer

Let $G$ be a group.
Let $\map {C_G} a$ be the centralizer of $a$ in $G$.
Then the number of different conjugates of $a$ in $G$ equals the number of different (left) cosets of $\map {C_G} a$:
:$\card {\conjclass a} = \index G {\map {C_G} a}$
where:
:$\conjclass a$ is the conjugacy class of $a$ in $G$
:$\index G {\map {C_G} a}$ is the index of $\map {C_G} a$ in $G$.
Consequently:
:$\card {\conjclass a} \divides \order G$

Proof:
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}}
Theorem: Size of Conjugacy Class is Index of Normalizer

Let $G$ be a group.
Let $x \in G$.
Let $\conjclass x$ be the conjugacy class of $x$ in $G$.
Let $\map {N_G} x$ be the normalizer of $x$ in $G$.
Let $\index G {\map {N_G} x}$ be the index of $\map {N_G} x$ in $G$.
The number of elements in $\conjclass x$ is $\index G {\map {N_G} x}$.

Proof:
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}}
Theorem: Conjugacy Class of Element of Center is Singleton

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.

Proof:
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
      | q = \forall x \in G
      | l = x a
      | r = a x
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | q = \forall x \in G
      | l = x a x^{-1}
      | r = a
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \conjclass a
      | r = \set a
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Conjugacy Class Equation

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:
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.
{{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:
:$\ds \order {G \setminus \map Z G} = \sum_{j \mathop = 1}^m \order {\conjclass {x_j} }$
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.
{{Qed}}
Theorem: Conjugacy Class Equation

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:
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.
{{Qed}}
Theorem: Group of Order Prime Squared is Abelian

A group whose order is the square of a prime is abelian.

Proof:
Let $G$ be a group of order $p^2$, where $p$ is prime.
Let $\map Z G$ be the center of $G$.
By Center of Group is Subgroup, $\map Z G$ is a subgroup of $G$.
By Lagrange's Theorem:
:$\order {\map Z G} \divides \order G$
It follows that $\order {\map Z G} = 1, p$ or $p^2$.
By Center of Group of Prime Power Order is Non-Trivial:
:$\order {\map Z G} \ne 1$
Suppose $\order {\map Z G} = p$.
Then:
{{begin-eqn}}
{{eqn | l = \order {\map Z G}
      | r = \index G {\map Z G}
      | c = {{Defof|Index of Subgroup}}
}}
{{eqn | r = \order G / \order {\map Z G}
      | c = Lagrange's Theorem
}}
{{eqn | r = p^2 / p
      | c = 
}}
{{eqn | r = p
      | c = 
}}
{{end-eqn}}
So $G / \map Z G$ is non-trivial, and of prime order.
From Prime Group is Cyclic, $G / \map Z G$ is a cyclic group.
But by Quotient of Group by Center Cyclic implies Abelian, that cannot be the case.
Therefore $\order {\map Z G} = p^2$ and therefore $\map Z G = G$.
Therefore $G$ is abelian.
{{qed}}
Theorem: Center of Group of Prime Power Order is Non-Trivial

Let $G$ be a group whose order is the power of a prime.
Then the center of $G$ is non-trivial:
:$\forall G: \order G = p^r: p \in \mathbb P, r \in \N_{>0}: \map Z G \ne \set e$

Proof:
Suppose $G$ is abelian.
By definition of abelian group:
:$\map Z G = G$
and the result is seen to be true as $G$ is itself non-trivial.
From Prime Group is Cyclic and Cyclic Group is Abelian, this will always be the case for $r = 1$.
So, suppose $G$ is non-abelian.
Thus $\map Z G \ne G$ and therefore $G \setminus \map Z G \ne \O$.
Let $\conjclass {x_1}, \conjclass {x_2}, \ldots, \conjclass {x_m}$ be the conjugacy classes into which $G \setminus \map Z G$ is partitioned.
From Conjugacy Class of Element of Center is Singleton, all of these will have more than one element.
From the Conjugacy Class Equation:
:$\ds \order {\map Z G} = \order G - \sum_{j \mathop = 1}^m \order {\conjclass {x_j} }$
From Number of Conjugates is Number of Cosets of Centralizer:
:$\order {\conjclass {x_j} } \divides \order G$
Let $\map {N_G} x$ be the normalizer of $x$ in $G$.
Then:
:$\forall j: 1 \le j \le m: \index G {\map {N_G} {x_j} } > 1 \implies p \divides \index G {\map {N_G} {x_j} }$
Since $p \divides \order G$, it follows that:
:$p \divides \order {\map Z G}$
and the result follows.
{{Qed}}
Theorem: Center of Group of Order Prime Cubed

Let $G$ be a group of order $p^3$, where $p$ is a prime.
Let $\map Z G$ be the center of $G$.
Then $\order {\map Z G} \ne p^2$.

Proof:
{{AimForCont}} $\order {\map Z G} = p^2$.
Then $\order {G / \map Z G} = p$.
From Prime Group is Cyclic it follows that $G / \map Z G$ is cyclic.
From Quotient of Group by Center Cyclic implies Abelian it follows that $G$ is abelian.
by definition of abelian group it follows that $\order {\map Z G} = p^3$.
The result follows by Proof by Contradiction.
{{qed}}
{{expand|This results generalizes naturally to a group of order $p^n$: its center is not of order $p^{n-1}$.}}
Category:Normal Subgroups
Category:Centers of Groups
Category:P-Groups
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Theorem: Prime Power Group has Non-Trivial Proper Normal Subgroup

Let $G$ be a group, whose identity is $e$, such that $\order G = p^n: n > 1, p \in \mathbb P$.
Then $G$ has a proper normal subgroup which is non-trivial.

Proof:
From Center of Group is Normal Subgroup, $\map Z G \lhd G$.
By Center of Group of Prime Power Order is Non-Trivial, $\map Z G$ is non-trivial.
If $\map Z G$ is a proper subgroup, the proof is finished.
Otherwise, $\map Z G = G$.
Then $G$ is abelian by definition.
However, then any $a \in G: a \ne e$ generates a non-trivial normal subgroup $\gen a$, as Subgroup of Abelian Group is Normal.
If $\order a = \order {\gen a} < \order G$, the proof is complete.
Otherwise:
:$\order a = \order G = p^n$
Then $\order {a^p} = p^{n - 1}$ from Subgroup of Finite Cyclic Group is Determined by Order, so $a^p$ generates that non-trivial normal subgroup.
{{qed}}
Theorem: Composition Series of Group of Prime Power Order

Let $G$ be a group whose identity is $e$, and whose order is a prime power:
:$\order G = p^n, p \in \mathbb P, n \ge 1$
Then $G$ has a composition series:
:$\set e = G_0 \subset G_1 \subset \ldots \subset G_n = G$
such that:
:$\order {G_k} = p^k$
where:
:$G_k \lhd G_{k + 1}$
:$G_{k + 1} / G_k$ is cyclic and of order $p$.

Proof:
To be proved by induction on $n$.
Let $P_n$ be the proposition for $\order G = p^n$.
Theorem: Product Rule for Counting

Let it be possible to choose an element $\alpha$ from a given set $S$ in $m$ different ways.
Let it be possible to choose an element $\beta$ from a given set $T$ in $n$ different ways.
Then the ordered pair $\tuple {\alpha, \beta}$ can be chosen from the cartesian product $S \times T$ in $m n$ different ways.

Proof:
{{handwaving}}
The validity of this rule follows directly from the definition of multiplication of integers.
The product $a b$ (for $a, b \in \N_{>0}$) is the number of sequences $\sequence {A, B}$, where $A$ can be any one of $a$ items and $B$ can be any one of $b$ items.
{{qed}}
Theorem: Commutativity of Group Direct Product

Let $\struct {G, \circ_g}$ and $\struct {H, \circ_h}$ be groups.
Let $\struct {G \times H, \circ}$ be the group direct product of $\struct {G, \circ_g}$ and $\struct {H, \circ_h}$, where the operation $\circ$ is defined as:
:$\tuple {g_1, h_1} \circ \tuple {g_2, h_2} = \tuple {g_1 \circ_g g_2, h_1 \circ_h h_2}$
Let $\struct {H \times G, \star}$ be the group direct product of $\struct {H, \circ_h}$ and $\struct {G, \circ_g}$, where the operation $\star$ is defined as:
:$\tuple {h_1, g_1} \star \tuple {h_2, g_2} = \tuple {h_1 \circ_h h_2, g_1 \circ_g g_2}$
The group direct product $\struct {G \times H, \circ}$ is isomorphic to the $\struct {H \times G, \star}$.

Proof:
The mapping $\theta: G \times H \to H \times G$ defined as:
:$\forall g \in G, h \in H: \map \theta {g, h} = \tuple {h, g}$
is to be shown to be a group homomorphism, and that $\theta$ is bijective, as follows:
Theorem: External Direct Product of Abelian Groups is Abelian Group

Let $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$ be groups.
Then the group direct product $\struct {G \times H, \circ}$ is abelian {{iff}} both $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$ are abelian.

Proof:
Let $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$ be groups whose identities are $e_G$ and $e_H$ respectively.
From External Direct Product of Groups is Group, $\struct {G \times H, \circ}$ is indeed a group whose identity is $\tuple {e_G, e_H}$.
Suppose $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$ are both abelian.
Then from External Direct Product Commutativity, $\struct {G \times H, \circ}$ is also abelian.
Now suppose that $\struct {G \times H, \circ}$ is abelian.
Then:
{{begin-eqn}}
{{eqn | l = \tuple {g_1 \circ_1 g_2, e_H}
      | r = \tuple {g_1 \circ_1 g_2, e_H \circ_2 e_H}
      | c = Definition of $e_H$
}}
{{eqn | r = \tuple {g_1, e_H} \circ \tuple {g_2, e_H}
      | c = {{Defof|Group Direct Product}}
}}
{{eqn | r = \tuple {g_2, e_H} \circ \tuple {g_1, e_H}
      | c = as $\struct {G \times H, \circ}$ is abelian
}}
{{eqn | r = \tuple {g_2 \circ_1 g_1, e_H \circ_2 e_H}
      | c = {{Defof|Group Direct Product}}
}}
{{eqn | r = \tuple {g_2 \circ_1 g_1, e_H}
      | c = Definition of $e_H$
}}
{{end-eqn}}
Thus:
: $g_1 \circ_1 g_2 = g_2 \circ_1 g_1$
and $\struct {G, \circ_1}$ is seen to be abelian.
The same argument holds for $\struct {H, \circ_2}$.
{{qed}}
Theorem: Associativity of Group Direct Product

The group direct product $G \times \paren {H \times K}$ is (group) isomorphic to $\paren {G \times H} \times K$.

Proof:
Let $G, H, K$ be groups.
The mapping $\theta: G \times \paren {H \times K} \to \paren {G \times H} \times K$ defined as:
:$\forall g \in G, h \in H, k \in K: \map \theta {\tuple {g, \tuple {h, k} } } = \tuple {\tuple {g, h}, k}$
is shown to be a group isomorphism, as follows:
Theorem: Group Direct Product of Cyclic Groups

Let $G$ and $H$ both be finite cyclic groups with orders $n = \order G$ and $m = \order H$ respectively.
Then:
:The group direct product $G \times H$ is cyclic
{{iff}}:
:$n$ and $m$ are coprime, that is:
::$n \perp m$

Proof:
Let $G$ and $H$ be groups whose identities are $e_G$ and $e_H$ respectively.