avewright/proofwiki-math huggingface dataset:
Theorem: Modulus of Sum equals Modulus of Distance implies Quotient is Imaginary

Let $z_1$ and $z_2$ be complex numbers such that:
:$\cmod {z_1 + z_2} = \cmod {z_1 - z_2}$
Then $\dfrac {z_2} {z_1}$ is wholly imaginary.

Proof:
Let $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$.
Then:
{{begin-eqn}}
{{eqn | l = \cmod {z_1 + z_2}
      | r = \cmod {z_1 - z_2}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {x_1 + x_2}^2 + \paren {y_1 + y_2}^2
      | r = \paren {x_1 - x_2}^2 + \paren {y_1 - y_2}^2
      | c = {{Defof|Complex Modulus}}
}}
{{eqn | ll= \leadsto
      | l = {x_1}^2 + 2 x_1 x_2 + {x_2}^2 + {y_1}^2 + 2 y_1 y_2 + {y_2}^2
      | r = {x_1}^2 - 2 x_1 x_2 + {x_1}^2 + {y_1}^2 - 2 y_1 y_2 + {y_1}^2
      | c = Square of Sum, Square of Difference
}}
{{eqn | ll= \leadsto
      | l = 4 x_1 x_2 + 4 y_1 y_2
      | r = 0
      | c = simplifying
}}
{{eqn | ll= \leadsto
      | l = x_1 x_2 + y_1 y_2
      | r = 0
      | c = simplifying
}}
{{end-eqn}}
Now we have:
{{begin-eqn}}
{{eqn | l = \dfrac {z_1} {z_2}
      | r = \frac {x_1 + i y_1} {x_2 + i y_2}
      | c = 
}}
{{eqn | r = \frac {\paren {x_1 + i y_1} \paren {x_2 - i y_2} } { {x_2}^2 + {y_2}^2}
      | c = {{Defof|Complex Division}}
}}
{{eqn | r = \frac {x_1 x_2 + y_1 y_2} { {x_2}^2 + {y_2}^2} + \frac {i \paren {x_2 y_1 - x_1 y_2} } { {x_2}^2 + {y_2}^2}
      | c = {{Defof|Complex Multiplication}}
}}
{{end-eqn}}
But we have:
:$x_1 x_2 + y_1 y_2 = 0$
Thus:
:$\dfrac {z_1} {z_2} = \dfrac {i \paren {x_2 y_1 - x_1 y_2} } { {x_2}^2 + {y_2}^2}$
which is wholly imaginary.
{{qed}}
Theorem: Real Part of Complex Product

Let $z_1$ and $z_2$ be complex numbers.
Then:
:$\map \Re {z_1 z_2} = \map \Re {z_1} \map \Re {z_2} - \map \Im {z_1} \map \Im {z_2}$

Proof:
Let $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$.
By definition of complex multiplication:
:$z_1 z_2 = x_1 x_2 - y_1 y_2 + i \paren {x_1 y_2 + x_2 y_1}$
Then:
{{begin-eqn}}
{{eqn | l = \map \Re {z_1 z_2}
      | r = x_1 x_2 - y_1 y_2
      | c = {{Defof|Real Part}}
}}
{{eqn | r = \map \Re {z_1} \map \Re {z_2} - \map \Im {z_1} \map \Im {z_2}
      | c = {{Defof|Real Part}}
}}
{{end-eqn}}
{{qed}}
Theorem: Imaginary Part of Complex Product

Let $z_1$ and $z_2$ be complex numbers.
Then:
:$\map \Im {z_1 z_2} = \map \Re {z_1} \, \map \Im {z_2} + \map \Im {z_1} \, \map \Re {z_2}$

Proof:
Let $z_1 = x_1 + i y_1$ and $z_2 = x_2 + i y_2$.
By definition of complex multiplication:
:$z_1 z_2 = x_1 x_2 - y_1 y_2 + i \paren {x_1 y_2 + x_2 y_1}$
Then
{{begin-eqn}}
{{eqn | l = \map \Im {z_1 z_2}
      | r = x_1 y_2 + x_2 y_1
      | c = {{Defof|Imaginary Part}}
}}
{{eqn | r = \map \Re {z_1} \, \map \Im {z_2} + \map \Im {z_1} \, \map \Re {z_2}
      | c = {{Defof|Imaginary Part}}
}}
{{end-eqn}}
{{qed}}
Theorem: Equation for Perpendicular Bisector of Two Points

Let $\tuple {x_1, y_1}$ and $\tuple {y_1, y_2}$ be two points in the cartesian plane.
Let $L$ be the perpendicular bisector of the straight line through $z_1$ and $z_2$ in the complex plane.
$L$ can be expressed by the equation:
:$y - \dfrac {y_1 + y_2} 2 = \dfrac {x_1 - x_2} {y_2 - y_1} \paren {x - \dfrac {x_1 + x_2} 2}$

Proof:
Let $M$ be the straight line passing through $z_1$ and $z_2$.
Let $Q$ be the midpoint of $M$.
We have that:
:$Q = \tuple {\dfrac {x_1 + x_2} 2, \dfrac {y_1 + y_2} 2}$
The slope of $M$ is $\dfrac {y_2 - y_1} {x_2 - x_1}$.
As $L$ is perpendicular to the $M$, its slope is $\dfrac {x_1 - x_2} {y_2 - y_1}$.
Thus by Equation of Straight Line in Plane: Point-Slope Form, the equation for $L$ is:
:$y - \dfrac {y_1 + y_2} 2 = \dfrac {x_1 - x_2} {y_2 - y_1} \paren {x - \dfrac {x_1 + x_2} 2}$
{{qed}}
{{improve|I can find no actual page on the web anywhere which gives this result explicitly, so I don't know what the "standard form" may be for this line. Hence I have not tried to simplify it, as any such "simplification" only seems to make it more complicated and less intuitive.}}
Category:Straight Lines
Category:Perpendiculars
3a4hfwwcf3qrt62myok7qyft5t026ex
Theorem: Element of Center in Group whose Order is Power of 2

Let $n \in \Z$ be an integer such that $n \ge 2$.
Let $G$ be a group whose order is $2^n$.
Let $x \in G$ be of order $2^{n - 1}$ in $G$.
Then $x^{2^{n - 2} }$ is an element of the center of $G$.

Proof:
Let $H = \gen x$ be the subgroup of $G$ generated by $x$.
We have that the index of $H$ in $G$ is $2$.
Then by Subgroup of Index 2 is Normal, $H$ is normal in $G$.
Let $y = x^{2^{n - 2} }$.
We have that $y \in H$, by definition of the construction of $H$.
Then the order of $y$ in $G$ is $2$.
As $H$ is normal, we have:
:$\forall g \in G: g y g^{-1} \in H$
By Order of Conjugate Element equals Order of Element, $g y g^{-1}$ is also of order $2$.
We can write $g y g^{-1} = x^m$ for some $m \in \Z$.
By Order of Power of Group Element, we have:
:$2 = \dfrac {2^{n - 1} } {\gcd \set {m, 2^{n - 1} } }$
This leads to:
:$\gcd \set {m, 2^{n - 1} } = 2^{n - 2}$
We have $0 < m < 2^{n - 1}$ and $2^{n - 2} \divides m$.
This forces $m = 2^{n - 2}$ and $g y g^{-1} = y$.
It follows that $g y = y g$.
Hence $y = x^{2^{n - 2} }$ is an element of the center of $G$.
{{qed}}
Theorem: Intersection of Abelian Subgroups is Normal Subgroup of Subgroup Generated by those Subgroups

Let $G$ be a group.
Let $L$ and $M$ be abelian subgroups of $G$.
Let $H = \gen {L, M}$ be the subgroup of $G$ generated by $L$ and $M$.
Then $L \cap M$ is a normal subgroup of $H$.

Proof:
From 
We have that $L$ and $M$ are abelian.
From Intersection of Subgroups is Subgroup, we have that $L \cap M$ is a subgroup of both $L$ and $M$.
From Subgroup of Abelian Group is Normal, we also have that $L \cap M$ is a normal subgroup of both $L$ and $M$.
Thus from Normalizer of Subgroup is Largest Subgroup containing that Subgroup as Normal Subgroup:
:$L \subseteq \map N {L \cap M}$
and:
:$M \subseteq \map N {L \cap M}$
Hence:
:$L \cup M \subseteq \map N {L \cap M}$
But from Normalizer is Subgroup, $\map N {L \cap M}$ is a subgroup of $G$.
So it follows that:
:$H \subseteq \map N {L \cap M}$
Hence the result by Subgroup is Normal Subgroup of Normalizer.
{{qed}}
Theorem: Characterisation of Non-Archimedean Division Ring Norms/Corollary 5

If $\norm {\, \cdot \,}$ is non-Archimedean then:
:$\sup \set {\norm {n \cdot 1_R}: n \in \Z} = 1$
where $n \cdot 1_R = 
\begin{cases}
\underbrace {1_R + 1_R + \dots + 1_R}_{\text {$n$ times} } & : n > 0 \\
0 & : n = 0 \\
\\
-\underbrace {\paren {1_R + 1_R + \dots + 1_R} }_{\text {$-n$ times} } & : n < 0 \\
\end{cases}$

Proof:
By {{Corollary|Characterisation of Non-Archimedean Division Ring Norms|1}}:
:$\sup \set {\norm{n \cdot 1_R}: n \in \N_{> 0}} = 1$
By {{NormAxiomNonArch|1}}:
:$\norm {0 \cdot 1_R} = 0 \le 1$
Let $n < 0$ then:
{{begin-eqn}}
{{eqn | l = \norm {n \cdot 1_R}
      | r = \norm {-\underbrace {\paren {1_R + 1_R + \dots + 1_R} }_{\text {$-n$ times} } }
}}
{{eqn | r = \norm {\underbrace {1_R + 1_R + \dots + 1_R}_{\text {$-n$ times} } }
      | c = Norm of Ring Negative
}}
{{eqn | r = \norm {\paren {-n} \cdot 1_R}
      | c = 
}}
{{eqn | o = \le
      | r = 1
      | c = Characterisation of Non-Archimedean Division Ring Norms
}}
{{end-eqn}}
The result follows.
{{qed}}
Theorem: Subgroup Containing all Squares of Group Elements is Normal

Let $G$ be a group.
Let $H$ be a subgroup of $G$ with the property that:
:$\forall x \in G: x^2 \in H$
Then $H$ is normal in $G$.

Proof:
We have:
{{begin-eqn}}
{{eqn | l = \paren {x h}^2 h^{-1} \paren {x^{-1} }^2
      | r = x h x h h^{-1} x^{-1} x^{-1}
      | c = {{Group-axiom|1}}
}}
{{eqn | r = x h x x^{-1} x^{-1}
      | c = {{Group-axiom|3}}
}}
{{eqn | r = x h x^{-1}
      | c = {{Group-axiom|3}}
}}
{{end-eqn}}
Because $\paren {x h}^2$ and $\paren {x^{-1} }^2$ are in the form $x^2$ for $x \in G$, they are both elements of $H$.
Thus:
:$x h x^{-1} \in H$
and so $H$ is normal in $G$ by definition.
{{qed}}
Theorem: Finite Order Elements of Infinite Abelian Group form Normal Subgroup

Let $G$ be an infinite  abelian group.
Let $H \subseteq G$ be the subset of $G$ defined as:
:$H := \set {x \in G: x \text { is of finite order in } G}$
Then $H$ forms a normal subgroup of $G$.

Proof:
Let $e$ be the identity element of $G$.
From Identity is Only Group Element of Order 1, $\order e = 1$ and so $H \ne \O$.
Let $a \in H$.
Then by Order of Group Element equals Order of Inverse:
:$\order a = \order {a^{-1} }$
and so $a \in H$.
Let $a, b \in H$.
From Order of Product of Abelian Group Elements Divides LCM of Orders of Elements:
:$\order {a b} \divides \lcm \set {\order a, \order b}$
where:
:$\order a$ denotes the order of $a$
:$\divides$ denotes divisibility
:$\lcm$ denotes the lowest common multiple.
Thus $a b$ is also of finite order.
Thus by definition:
:$a b \in H$
By the Two-Step Subgroup Test it follows that $H$ is a subgroup of $G$.
By Subgroup of Abelian Group is Normal, $H$ is normal in $G$.
{{qed}}
Theorem: Commutator is Identity iff Elements Commute

Let $G$ be a group whose identity is $e$.
Let $x, y \in G$.
Let $\sqbrk {x, y}$ denote the commutator of $x$ and $y$.
Then $\sqbrk {x, y} = e$ {{iff}} $x$ and $y$ commute.

Proof:
As $G$ is a group, it is {{afortiori}} a monoid.
Hence Product of Commuting Elements with Inverses applies:
:$x y x^{-1} y^{-1} = e = x^{-1} y^{-1} x y$
{{qed}}
Theorem: Commutator of Quotient Group Elements

Let $G$ be a group.
Let $N$ be a normal subgroup of $G$.
Let $\sqbrk {x, y}$ denote the commutator of $x, y \in G$:
:$\sqbrk {x, y} = x^{-1} y^{-1} x y$
Then:
:$\forall x, y \in G: \sqbrk {x N, y N} = \sqbrk {x, y} N$
where $x N$ and $y N$ are left cosets of $N$, and so elements of the quotient group $G / N$ of $G$ by $N$.

Proof:
{{begin-eqn}}
{{eqn | l = \sqbrk {x N, y N}
      | r = \paren {x N}^{-1} \paren {y N}^{-1} \paren {x N} \paren {y N}
      | c = {{Defof|Commutator of Group Elements}}
}}
{{eqn | r = \paren {x^{-1} N} \paren {y^{-1} N} \paren {x N} \paren {y N}
      | c = Quotient Group is Group: inverse of $x N$ is $x^{-1} N$
}}
{{eqn | r = \paren {x^{-1} y^{-1} N} \paren {x y N}
      | c = {{Defof|Coset Product}}
}}
{{eqn | r = x^{-1} y^{-1} x y N
      | c = {{Defof|Coset Product}}
}}
{{eqn | r = \sqbrk {x, y} N
      | c = {{Defof|Commutator of Group Elements}}
}}
{{end-eqn}}
{{qed}}
Theorem: Quotient Group is Abelian iff All Commutators in Divisor

Let $G$ be a group.
Let $N$ be a normal subgroup of $G$.
Let $G / N$ be the quotient group of $G$ by $N$.
Then the quotient group $G / N$ is abelian {{iff}}:
:$\forall x, y \in G: \sqbrk {x, y} \in N$
where $\sqbrk {x, y}$ denotes the commutator of $x$ and $y$.
That is, {{iff}} $x y x^{-1} y^{-1} \in N$ for all $x, y \in G$.

Proof:
Let $x, y \in G$.
First we establish the following:
{{begin-eqn}}
{{eqn | l = \paren {x N} \paren {y N}
      | r = \paren {y N} \paren {x N}
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \paren {x N} \paren {y N}
      | r = \paren {y N} \paren {x N} N
      | c = Quotient Group is Group: $N$ is the identity of $G / N$
}}
{{eqn | ll= \leadstoandfrom
      | l = \paren {x N}^{-1} \paren {y N}^{-1} \paren {x N} \paren {y N}
      | r = N
      | c = applying $\paren {x N}^{-1} \paren {y N}^{-1}$ to both sides
}}
{{eqn | ll= \leadstoandfrom
      | l = \sqbrk {x N, y N}
      | r = N
      | c = {{Defof|Commutator of Group Elements}}
}}
{{eqn | ll= \leadstoandfrom
      | l = \sqbrk {x, y} N
      | r = N
      | c = Commutator of Quotient Group Elements
}}
{{eqn | n = 1
      | ll= \leadstoandfrom
      | l = \sqbrk {x, y}
      | o = \in
      | r = N
      | c = Commutator of Quotient Group Elements
}}
{{end-eqn}}
{{qed|lemma}}
=== Sufficient Condition ===
Let $G / N$ be abelian.
Then by definition:
:$\forall x, y \in G: \paren {x N} \paren {y N} = \paren {y N} \paren {x N}$
and it follows from $(1)$ that:
:$\forall x, y \in G: \sqbrk {x, y} \in N$
{{qed|lemma}}
=== Necessary Condition ===
Conversely, let:
:$\forall x, y \in G: \sqbrk {x, y} \in N$
Again it follows from $(1)$ that:
:$\forall x, y \in G: \paren {x N} \paren {y N} = \paren {y N} \paren {x N}$
That is, that $G / N$ is abelian.
{{qed}}
Theorem: Quotient Group is Abelian iff All Commutators in Divisor

Let $G$ be a group.
Let $N$ be a normal subgroup of $G$.
Let $G / N$ be the quotient group of $G$ by $N$.
Then the quotient group $G / N$ is abelian {{iff}}:
:$\forall x, y \in G: \sqbrk {x, y} \in N$
where $\sqbrk {x, y}$ denotes the commutator of $x$ and $y$.
That is, {{iff}} $x y x^{-1} y^{-1} \in N$ for all $x, y \in G$.

Proof:
Let $G / N$ be abelian.
Then:
{{begin-eqn}}
{{eqn | q = \forall a N, b N \in G / N
      | l = a N b N
      | r = b N a N
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a b N
      | r = b a N
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a b \paren {b a}^{-1}
      | o = \in
      | r = N
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a b a^{-1} b^{-1}
      | o = \in
      | r = N
      | c = 
}}
{{end-eqn}}
{{qed}}
The argument reverses.
Suppose that $\forall a, b \in G: a b a^{-1} b^{-1} \in N$.
{{begin-eqn}}
{{eqn | q = \forall a, b \in G
      | l = a b a^{-1} b^{-1}
      | o = \in
      | r = N
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a b \paren {b a}^{-1}
      | o = \in
      | r = N
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a b N
      | r = b a N
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a N b N
      | r = b N a N
      | c = 
}}
{{end-eqn}}
Hence the result.
{{Qed}}
Theorem: Sufficient Condition for Quotient Group by Intersection to be Abelian

Let $G$ be a group.
Let $N$ and $K$ be normal subgroups of $G$.
Let the quotient groups $G / N$ and $G / K$ be abelian.
Then the quotient group $G / \paren {N \cap K}$ is also abelian.

Proof:
From Intersection of Normal Subgroups is Normal, we have that $N \cap K$ is normal in $G$.
We are given that $G / N$ and $G / K$ are abelian.
Hence:
{{begin-eqn}}
{{eqn | q = \forall x, y \in G
      | l = \sqbrk {x, y}
      | o = \in
      | r = N
      | c = Quotient Group is Abelian iff All Commutators in Divisor
}}
{{eqn | lo= \land
      | l = \sqbrk {x, y}
      | o = \in
      | r = K
      | c = 
}}
{{eqn | ll= \leadsto
      | q = \forall x, y \in G
      | l = \sqbrk {x, y}
      | o = \in
      | r = N \cap K
      | c = {{Defof|Set Intersection}}
}}
{{eqn | ll= \leadsto
      | l = G / \paren {N \cap K}
      | o = 
      | r = \text {is abelian}
      | c = Quotient Group is Abelian iff All Commutators in Divisor
}}
{{end-eqn}}
{{qed}}
Theorem: Quotient Group by Intersection of Normal Subgroups not necessarily Cyclic if Quotient Groups are

Let $G$ be a group.
Let $N$ and $K$ be normal subgroups of $G$.
Let the quotient groups $G / N$ and $G / K$ be cyclic.
Then the quotient group $G / \paren {N \cap K}$ is not necessarily cyclic.

Proof:
Proof by Counterexample:
Let $G = \set {e, a, b, c}$ be the Klein $4$-group whose identity element is $e$.
Let $N = \set {e, a}$ and $K = \set {e, b}$.
By Subgroups of Klein Four-Group, both $N$ and $K$ are subgroups of $G$.
By Prime Group is Cyclic, both $N$ and $K$ are cyclic.
By Subgroup of Abelian Group is Normal, both $N$ and $K$ are normal in $G$.
Then we have that:
:$N \cap K = e$
and so by Trivial Quotient Group is Quotient Group:
:$G / \paren {N \cap K} \cong G$
But $G$ is the Klein $4$-group, which is not cyclic.
{{qed}}
Theorem: Order of Boolean Group is Power of 2

Let $G$ be a Boolean group.
Let $\order G$ denote the order of $G$.
Then:
:$\order G = 2^n$
where $n \in \Z_{\ge 0}$ is a positive integer.

Proof:
The case where $n = 0$ is clear:
:$\order {\set e} = 1$
and $e^2 = e$.
{{AimForCont}} $\order G = m \times 2^k$ for some odd integer $m$.
Then $m$ itself has an odd prime $p$ as a integer (which may of course equal $m$ if $m$ is itself prime).
Then by Cauchy's Lemma (Group Theory) there exists $g \in G$ such that $\order g = p$.
Hence it is not the case that $g^2 = e$.
Hence $\order G$ has no prime factor which is odd.
The result follows.
{{qed}}
Theorem: Subgroup of Index 2 contains all Squares of Group Elements

Let $G$ be a group.
Let $H$ be a subgroup of $G$ whose index is $2$.
Then:
:$\forall x \in G: x^2 \in H$

Proof:
By Subgroup of Index 2 is Normal, $H$ is normal in $G$.
Hence the quotient group $G / H$ exists.
Then we have:
{{begin-eqn}}
{{eqn | q = \forall x \in G
      | l = \paren {x^2} H
      | r = \paren {x H}^2
      | c = 
}}
{{eqn | r = H
      | c = as $G / H$ is of order $2$
}}
{{end-eqn}}
{{qed}}
Theorem: Subgroup of Index 3 does not necessarily contain all Cubes of Group Elements

Let $G$ be a group.
Let $H$ be a subgroup of $G$ whose index is $3$.
Then it is not necessarily the case that:
:$\forall x \in G: x^3 \in H$

Proof:
Proof by Counterexample:
Consider $S_3$, the symmetric group on $3$ letters.
From Subgroups of Symmetric Group on 3 Letters, the subsets of $S_3$ which form subgroups of $S_3$ are:
{{begin-eqn}}
{{eqn | o = 
      | r = S_3
}}
{{eqn | o = 
      | r = \set e
}}
{{eqn | o = 
      | r = \set {e, \tuple {123}, \tuple {132} }
}}
{{eqn | o = 
      | r = \set {e, \tuple {12} }
}}
{{eqn | o = 
      | r = \set {e, \tuple {13} }
}}
{{eqn | o = 
      | r = \set {e, \tuple {23} }
}}
{{end-eqn}}
Let $H = \set {e, \tuple {12} }$.
The index of $H$ is $3$.
Let $x = \tuple {13}$.
We note that:
:$x \in G$
:$x^3 = \tuple {13}$
But $\tuple {13} \notin \set {e, \tuple {12} }$.
Hence we have:
:$\exists x \in G: x^3 \notin H$
and the result follows.
{{qed}}
Theorem: Center of Non-Abelian Group of Order pq is Trivial

Let $p$ and $q$ be distinct prime numbers.
Let $G$ be a non-abelian group of order $p q$ whose identity is $e$.
Then the center of $G$ is trivial:
:$\map Z G = \set e$

Proof:
From Center of Group is Normal Subgroup, $\map Z G$ is a normal subgroup of $G$.
By Lagrange's Theorem, the order of $\map Z G$ is either $1$, $p$, $q$ or $p q$.
Because $G$ is not abelian, $G \ne \map Z G$.
Hence $\order {\map Z G} \ne p q$.
From Quotient of Group by Center Cyclic implies Abelian:
:$G / \map Z G$ cannot be a cyclic group which is non-trivial.
Then we have:
:$\map C x \subset G$
where $\map C x$ is the centralizer of $x$.
Hence by Prime Group is Cyclic, $G / \map Z G$ cannot be cyclic of order $p$ or $q$.
{{qed}}
Theorem: Number of Elements of Order p in Group of Order pq is Multiple of q

Let $p$ and $q$ be distinct prime numbers.
Let $G$ be a non-abelian group of order $p q$.
Then the number of elements of $G$ of order $p$ is a multiple of $q$.

Proof:
Let $x$ be an element of $G$ of order $p$.
From Center of Non-Abelian Group of Order pq is Trivial:
:$p \notin \map Z G$
where $\map Z G$ denotes the center of $G$.
As $x \notin \map Z G$:
:$\map C x \subsetneq G$
where $\map C x$ is the centralizer of $x$.
From Order of Element divides Order of Centralizer:
:$p \divides \order {\map C x}$
Then from Lagrange's Theorem, it follows directly that:
:$\order {\map C x} = p$
By Number of Conjugates is Number of Cosets of Centralizer:
:$\card {\conjclass x} = \index G {\map {C_G} x}$
where $\conjclass x$ denotes the conjugacy class of $x$.
It follows that:
:$\card {\conjclass x} = q$
and so the element of $G$ of order $p$ come in sets whose cardinality is a divisor of $q$.
{{qed}}
Theorem: Intersection of Normal Subgroup with Center in p-Group

Let $p$ be a prime number
Let $G$ be a $p$-group.
Let $N$ be a non-trivial normal subgroup of $G$.
Let $\map Z G$ denote the center of $G$.
Then:
:$N \cap \map Z G$ is a non-trivial normal subgroup of $G$.

Proof:
First we note that:
:Center of Group is Normal Subgroup
and from Intersection of Normal Subgroups is Normal:
:$N \cap \map Z G$ is normal in $G$.
Suppose $G$ is abelian.
By definition:
:$\map Z G = G$
Then:
:$N \cap \map Z G = N$
which is 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.
From Center of Group of Prime Power Order is Non-Trivial, $\map Z G$ is non-trivial.
From Union of Conjugacy Classes is Normal:
:$N = \ds \bigcup_{x \mathop \in N} \conjclass x$
Let $\conjclass {x_1}, \conjclass {x_2}, \ldots, \conjclass {x_m}$ be the conjugacy classes into which $N$ is partitioned.
From Conjugacy Class of Element of Center is Singleton, all of these will have more than one element.
From Number of Conjugates is Number of Cosets of Centralizer:
:$\order {\conjclass {x_j} } \divides \order G$
for all $j \in \set {1, 2, \ldots, m}$.
{{finish|Needs more thought.}}
Theorem: Normal Subgroup of p-Group of Order p is Subset of Center

Let $p$ be a prime number.
Let $G$ be a $p$-group.
Let $N$ be a normal subgroup of $G$ of order $p$.
Then:
:$N \subseteq \map Z G$
where $\map Z G$ denotes the center of $G$.

Proof:
From Intersection of Normal Subgroup with Center in p-Group:
:$\order {N \cap \map Z G} > 1$
From Intersection of Subgroups is Subgroup, $N \cap \map Z G$ is a subgroup of $N$.
It follows from Lagrange's Theorem that:
:$\order {N \cap \map Z G} = p$
and so:
:$N \cap \map Z G = N$
But from Intersection of Subgroups is Subgroup, $N \cap \map Z G$ is a subgroup of $\map Z G$
That is:
:$N$ is a subgroup of $\map Z G$
and the result follows.
{{qed}}
Theorem: Non-Abelian Group of Order p Cubed has Exactly One Normal Subgroup of Order p

Let $p$ be a prime number.
Let $G$ be a non-abelian group of order $p^3$.
Then $G$ contains exactly one normal subgroup of order $p$.

Proof:
From Center of Group of Prime Power Order is Non-Trivial, $\map Z G$ is not the trivial subgroup.
From Quotient of Group by Center Cyclic implies Abelian, $G / \map G Z$ cannot be cyclic and non-trivial.
Thus $\order {G / \map G Z}$ cannot be $p$ and so must be $p^2$.
Thus $\order {\map G Z} = p$.
Let $N$ be a normal subgroup of $G$ of order $p$.
Then from Normal Subgroup of p-Group of Order p is Subset of Center:
:$N \subseteq \map G Z$
Thus there is no normal subgroup of order $p$ different from $\map G Z$.
Hence the result.
{{qed}}
Theorem: Quaternion Group has Normal Subgroup without Complement

Let $Q$ denote the quaternion group.
There exists a normal subgroup of $Q$ which has no complement.

Proof:
From Subgroups of Quaternion Group:
{{:Subgroups of Quaternion Group}}
From Quaternion Group is Hamiltonian, all these subgroups are normal.
For two subgroups to be complementary, they need to have an intersection which is trivial.
However, apart from $\set e$ itself, all these subgroups contain $a^2$.
Hence none of these subgroups has a complement.
{{qed}}
Theorem: Non-Abelian Order 2p Group has Order p Element

Let $p$ be an odd prime.
Let $G$ be a non-abelian group of order $2 p$.
Then $G$ has at least one element of order $p$.

Proof:
By Lagrange's Theorem, all the elements of $G$ have orders $1$, $2$, $p$ or $2p$.
From Identity is Only Group Element of Order 1, $2 p - 1$ elements of $G$ have orders greater than $1$.
From Cyclic Group is Abelian, $G$ is not the cyclic group $2 p$.
If $g \in G$ was of order $2 p$ then $g$ would generate the cyclic group $C_{2 p}$.
Thus $G$ has no element of order $2 p$.
So all elements of $G$ except the identity have orders $2$ or $p$.
{{AimForCont}} $G$ has no element of order $p$.
Then all elements are of order $2$.
Then $G$ is a Boolean group.
Then by Boolean Group is Abelian, $G$ is abelian.
This contradicts the assertion that $G$ is non-abelian.
Hence the result by Proof by Contradiction.
{{qed}}
Theorem: Fourth Power Modulo 5

Let $n \in \Z$ be an integer.
Then:
:$n^4 \equiv m \pmod 5$
where $m \in \set {0, 1}$.

Proof:
By Congruence of Powers:
:$a \equiv b \pmod 5 \iff a^4 \equiv b^4 \pmod 5$
so it is sufficient to demonstrate the result for $n \in \set {0, 1, 2, 3, 4}$.
Thus:
{{begin-eqn}}
{{eqn | l = 0^4
      | m = 0
      | mo= \equiv
      | r = 0
      | rr= \pmod 5
      | c = 
}}
{{eqn | l = 1^4
      | m = 1
      | mo= \equiv
      | r = 1
      | rr= \pmod 5
      | c = 
}}
{{eqn | l = 2^4
      | m = 16
      | mo= \equiv
      | r = 1
      | rr= \pmod 5
      | c = 
}}
{{eqn | l = 3^4
      | m = 81
      | mo= \equiv
      | r = 1
      | rr= \pmod 5
      | c = 
}}
{{eqn | l = 4^4
      | m = 256
      | mo= \equiv
      | r = 1
      | rr= \pmod 5
      | c = 
}}
{{end-eqn}}
The result follows.
{{qed}}
Category:Fourth Powers
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