avewright/proofwiki-math huggingface dataset:
Theorem: Equivalence of Definitions of Equivalent Division Ring Norms/Cauchy Sequence Equivalent implies Open Unit Ball Equivalent

Let $R$ be a division ring.
Let $\norm {\, \cdot \,}_1: R \to \R_{\ge 0}$ and $\norm {\, \cdot \,}_2: R \to \R_{\ge 0}$ be norms on $R$.
Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ satisfy:
:for all sequences $\sequence {x_n}$ in $R$: $\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_1$ {{iff}} $\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_2$
Then $\forall x \in R$:
:$\norm x_1 < 1 \iff \norm x_2 < 1$

Proof:
The contrapositive is proved.
Let there exist $x \in R$ such that $\norm x_1 < 1$ and $\norm x_2 \ge 1$.
Let $\sequence {x_n}$ be the sequence defined by: $\forall n: x_n = x^n$.
By Sequence of Powers of Number less than One in Normed Division Ring then $\sequence {x_n}$ is a null sequence in $\norm {\, \cdot \,}_1$.
By convergent sequence in normed division ring is a Cauchy sequence then $\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_1$.
Let $0_R$ be the zero of $R$ and $1_R$ be the unit of $R$.
By Norm of Unity of Division Ring and the assumption $\norm x_1 < 1$:
:$x \ne 1_R$
Hence:
:$x - 1_R \ne 0_R$
By norm axiom $(\text N 1)$: Positive Definiteness:
:$\norm {x - 1_R}_2 > 0$
Let $\epsilon = \dfrac {\norm {x - 1_R}_2} 2$.
Then $\norm {x - 1_R}_2 > \epsilon$.
We have that $\norm x_2 \ge 1$.
Hence for all $n \in \N$:
{{begin-eqn}}
{{eqn | l = \norm {x_n}_2
      | r = \norm {x^n}_2
      | c = Definition of $x_n$
}}
{{eqn | r = \norm x_2^n
      | c = Norm Axiom $(\text N 2)$: Multiplicativity
}}
{{eqn | o = \ge
      | r = 1
      | c = 
}}
{{end-eqn}}
For all $n \in \N$:
{{begin-eqn}}
{{eqn | l = \norm {x_{n + 1} - x_n}_2
      | r = \norm {x^{n + 1} - x^n}_2
}}
{{eqn | r = \norm {x^n x - x^n}_2
}}
{{eqn | r = \norm {x^n \paren {x - 1_R} }_2
}}
{{eqn | r = \norm {x^n}_2 \norm {x - 1_R}_2
      | c = Norm Axiom $(\text N 2)$: Multiplicativity
}}
{{eqn | o = >
      | r = \epsilon
}}
{{end-eqn}} 
So $\sequence {x_n}$ is not a Cauchy sequence in $\norm {\, \cdot \,}_2$.
The theorem now follows by the Rule of Transposition. 
{{qed}}
Theorem: Equal Order Elements may not be Conjugate

Let $G$ be a group
Let $x, y \in G$ be elements of $G$ such that:
:$\order x = \order y$
where $\order x$ denotes the order of $x$.
Then it is not necessarily the case that $x$ and $y$ are conjugates.

Proof:
Consider the dihedral group $D_4$, whose group presentation is:
{{:Group Presentation of Dihedral Group D4}}
We have that:
:$\order {a^2} = 2$
and:
:$\order b = 2$
but $a^2$ and $b$ are not conjugate to each other.
{{qed}}
Theorem: Group Action of Symmetric Group on Complex Vector Space

Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $S_n$ denote the symmetric group on $n$ letters.
Let $V$ denote a vector space over the complex numbers $\C$.
Let $V$ have a basis:
:$\BB := \set {v_1, v_2, \ldots, v_n}$
Let $*: S_n \times V \to V$ be a group action of $S_n$ on $V$ defined as:
:$\forall \tuple {\rho, v} \in S_n \times V: \rho * v := \lambda_1 v_{\map \rho 1} + \lambda_2 v_{\map \rho 2} + \dotsb + \lambda_n v_{\map \rho n}$
where:
: $v = \lambda_1 v_1 + \lambda_2 v_2 + \dotsb + \lambda_n v_n$
Then $*$ is a group action.

Proof:
Let $\rho, \sigma \in S_n$.
Let $v = \ds \sum_{k \mathop = 1}^n \lambda_k v_k$.
We have:
{{begin-eqn}}
{{eqn | l = \rho * \paren {\sigma * v}
      | r = \rho * \paren {\sigma * \sum_{k \mathop = 1}^n \lambda_k v_k}
      | c = Definition of $v$
}}
{{eqn | r = \rho * \sum_{k \mathop = 1}^n \lambda_k v_{\map \sigma k}
      | c = Definition of $*$
}}
{{eqn | r = \sum_{k \mathop = 1}^n \lambda_k v_{\map \rho {\map \sigma k} }
      | c = Definition of $*$
}}
{{eqn | r = \sum_{k \mathop = 1}^n \lambda_k v_{\map {\rho \mathop \circ \sigma} k}
      | c = {{Defof|Composition of Mappings}}
}}
{{eqn | r = \paren {\rho \circ \sigma} * \sum_{k \mathop = 1}^n \lambda_k v_k
      | c = Definition of $*$
}}
{{eqn | r = \paren {\rho \circ \sigma} * v
      | c = Definition of $v$
}}
{{end-eqn}}
Hence $*$ fulfils {{GroupActionAxiom|1}}.
Let $e$ denote the identity element of $S_n$.
Then:
{{begin-eqn}}
{{eqn | l = e * v
      | r = e * \sum_{k \mathop = 1}^n \lambda_k v_k
      | c = Definition of $v$
}}
{{eqn | r = \sum_{k \mathop = 1}^n \lambda_k v_{\map e k}
      | c = Definition of $*$
}}
{{eqn | r = \sum_{k \mathop = 1}^n \lambda_k v_k
      | c = Definition of $e$
}}
{{eqn | r = v
      | c = Definition of $v$
}}
{{end-eqn}}
Hence $*$ fulfils {{GroupActionAxiom|2}}.
{{qed}}
Theorem: Group Action of Symmetric Group on Complex Vector Space/Orbit

The orbit of an element $v \in V$ is:
:$\ds \Orb v = \set {w \in V: \exists \rho \in S_n: w = \sum_{k \mathop = 1}^n \lambda_k v_{\map \rho k} }$

Proof:
By definition:
{{begin-eqn}}
{{eqn | l = \Orb v
      | r = \set {w \in V: \exists \rho \in S_n: w = \rho * v}
      | c = {{Defof|Orbit (Group Theory)|Orbit}}
}}
{{eqn | r = \set {w \in V: \exists \rho \in S_n: w = \rho * \sum_{k \mathop = 1}^n \lambda_k v_k}
      | c = Definition of $v$
}}
{{eqn | r = \set {w \in V: \exists \rho \in S_n: w = \sum_{k \mathop = 1}^n \lambda_k v_{\map \rho k} }
      | c = Definition of $*$
}}
{{end-eqn}}
{{qed}}
Theorem: Group Action of Symmetric Group on Complex Vector Space/Stabilizer

The stabilizer of an element $v \in V$ is:
:$\ds \Stab v = \set {\rho \in S_n: \sum_{k \mathop = 1}^n \lambda_k v_k = \sum_{k \mathop = 1}^n \lambda_{\map \rho k} v_k}$

Proof:
By definition:
{{begin-eqn}}
{{eqn | l = \Stab v
      | r = \set {\rho \in S_n: \rho * v = v}
      | c = {{Defof|Stabilizer}}
}}
{{eqn | r = \set {\rho \in S_n: \rho * \sum_{k \mathop = 1}^n \lambda_k v_k = \sum_{k \mathop = 1}^n \lambda_k v_k}
      | c = Definition of $v$
}}
{{eqn | r = \set {\rho \in S_n: \sum_{k \mathop = 1}^n \lambda_k v_k = \sum_{k \mathop = 1}^n \lambda_{\map \rho k} v_k}
      | c = Definition of $*$
}}
{{end-eqn}}
{{qed}}
Theorem: Equivalence of Definitions of Equivalent Division Ring Norms/Norm is Power of Other Norm implies Cauchy Sequence Equivalent

Let $\norm {\, \cdot \,}_1$ and $\norm {\, \cdot \,}_2$ satisfy:
:$\exists \alpha \in \R_{> 0}: \forall x \in R: \norm x_1 = \norm x_2^\alpha$
Then for all sequences $\sequence {x_n}$ in $R$:
:$\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_1$ {{iff}} $\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_2$

Proof:
Let $\sequence {x_n}$ be a Cauchy sequence in $\norm {\, \cdot \,}_1$.
Let $\epsilon > 0$ be given.
Since $\sequence {x_n}$ is a Cauchy sequence then:
:$\exists N \in \N: \forall n,m \ge N: \norm {x_n - x_m}_1 < \epsilon^\alpha$
Then:
:$\exists N \in \N: \forall n,m \ge N: \norm {x_n - x_m}_2^\alpha < \epsilon^\alpha$
Hence:
:$\exists N \in \N: \forall n,m \ge N: \norm {x_n - x_m}_2 < \epsilon$
So $\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_2$
It follows that for all sequences $\sequence {x_n}$ in $R$:
:$\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_1 \implies \sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_2$
{{qed|lemma}}
Let $\sequence {x_n}$ be a Cauchy sequence in $\norm {\, \cdot \,}_2$.
Let $\epsilon > 0$ be given.
Since $\sequence {x_n}$ is a Cauchy sequence then:
:$\exists N \in \N: \forall n,m \ge N: \norm {x_n - x_m}_2 \lt \epsilon^{1/\alpha}$
Then:
:$\exists N \in \N: \forall n, m \ge N: \norm {x_n - x_m}_2^\alpha < \epsilon$
Hence:
:$\exists N \in \N: \forall n, m \ge N: \norm {x_n - x_m}_1 < \epsilon$
So $\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_1$
It follows that for all sequences $\sequence {x_n}$ in $R$:
:$\sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_2 \implies \sequence {x_n}$ is a Cauchy sequence in $\norm {\, \cdot \,}_1$
The result follows.
{{qed}}
Theorem: Conjugacy Classes of Symmetric Group

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.
{{qed|lemma}}
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:
{{begin-eqn}}
{{eqn | l = \sigma
      | r = \alpha_1 \alpha_2 \dotsm \alpha_l
}}
{{eqn | l = \rho
      | r = \beta_1 \beta_2 \dotsm \beta_l
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \alpha_i
      | r = \alpha_{i1} \alpha_{i2} \dotsm \alpha_{i k_l}
}}
{{eqn | l = \beta_i
      | r = \beta_{i1} \beta_{i2} \dotsm \beta_{i k_l}
}}
{{end-eqn}}
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:
{{begin-eqn}}
{{eqn | l = \tau \sigma \tau^{-1}
      | r = \tau \alpha_1 \alpha_2 \dotsm \alpha_l \tau^{-1}
}}
{{eqn | r = \tau \alpha_1 \tau^{-1} \tau \alpha_2 \tau^{-1} \dotsm \tau \alpha_l \tau^{-1}
}}
{{eqn | r = \beta_1 \beta_2 \dotsm \beta_l
}}
{{eqn | r = \rho
}}
{{end-eqn}}
demonstrating that $\sigma$ and $\rho$ are conjugate.
That is, $\sigma$ and $\rho$ are in the same conjugacy class.
Hence the result.
{{qed}}
Category:Symmetric Groups
Category:Conjugacy Classes
lqejkraisnceer65vizzoa9d6n6znq1
Theorem: Identity of Group is in Center

Let $G$ be a group.
Let $e$ be the identity of $G$.
Then $e$ is in the center of $G$:
:$e \in \map Z G$

Proof:
From Center is Intersection of Centralizers:
:$\ds \map Z G = \bigcap_{g \mathop \in G} \map {C_G} g$
where $\map {C_G} g$ denotes the centralizer of $g$.
From Centralizer of Group Element is Subgroup, each of $\map {C_G} g$ is a subgroup of $G$.
From Identity of Subgroup:
:$\forall g \in G: e \in \map {C_G} g$
Hence by definition of set intersection:
:$e \in \ds \bigcap_{g \mathop \in G} \map {C_G} g$
whence the result.
{{qed}}
Category:Centers of Groups
Category:Identity Elements
t06u7is8we2jiy0cl6nn7c4a6q7t182
Theorem: Identity of Group is in Singleton Conjugacy Class

Let $G$ be a group.
Let $e$ be the identity of $G$.
Then $e$ is in its own singleton conjugacy class:
:$\conjclass e = \set e$

Proof:
From Identity of Group is in Center:
:$e \in \map Z G$
where $\map Z G$ is the center of $G$.
From Conjugacy Class of Element of Center is Singleton:
:$\conjclass e = \set e$
{{qed}}
Theorem: Finite Group with 2 Conjugacy Classes has 2 Elements

Let $G$ be a finite group.
Let $G$ have exactly $2$ conjugacy classes.
Then $G$ has exactly $2$ elements.

Proof:
Let $G$ be of order $n$.
Let $G$ have exactly $2$ conjugacy classes.
Let $x \in G$ such that $x \ne e$.
Let $\conjclass x$ denote the conjugacy class of $x$.
From Identity of Group is in Singleton Conjugacy Class:
$\conjclass e = \set e$
where $\conjclass e$ denotes the conjugacy class of $e$
The other elements of $G$ are in $\conjclass x$.
Thus:
:$\card {\conjclass x} = n - 1$
where $\card {\, \cdot \,}$ denotes the cardinality of a set.
We have that Size of Conjugacy Class is Index of Normalizer.
From Lagrange's Theorem, the index of any subgroup of $G$ is a divisor of $G$.
Thus:
:$\paren {n - 1} \divides n$
where $\divides$ denotes divisibility.
It follows that $n = 2$.
Hence the result by definition of order of group.
{{qed}}
Theorem: Group of Order 15 has Cyclic Subgroups of Order 3 and Order 5

Let $G$ be a group whose order is $15$.
Then $G$ has
:a cyclic subgroup of order $3$
and:
:a cyclic subgroup of order $5$.

Proof:
Let $G$ be a group of order $15$.
We have that $15 = 3 \times 5$.
Thus from the First Sylow Theorem:
:$G$ has at least one subgroup $H_3$ of order $3$
and:
:$G$ has at least one subgroup $H_5$ of order $5$.
From Prime Group is Cyclic, all such subgroups of order $3$ and order $5$ are cyclic.
{{Qed}}
Theorem: Number of Sylow p-Subgroups in Group of Order 15

Let $G$ be a group whose order is $15$.
Then:
:the number of Sylow $3$-subgroups is in the set $\set {1, 4, 7, \ldots}$
:the number of Sylow $5$-subgroups is in the set $\set {1, 6, 11, \ldots}$

Proof:
Let $G$ be a group of order $15$.
From the Fourth Sylow Theorem:
:the number of Sylow $p$-subgroups is equivalent to $1 \pmod p$
We have that $15 = 3 \times 5$.
Hence the result.
{{Qed}}
Theorem: Direct Product of Sylow p-Subgroups is Sylow p-Subgroup

Let $G_1$ and $G_2$ be groups.
Let $H_1$ and $H_2$ be subgroups of $G_1$ and $G_2$ respectively.
Let $H_1$ be a Sylow $p$-subgroup of $G_1$.
Let $H_2$ be a Sylow $p$-subgroup of $G_2$.
Then $H_1 \times H_2$ is a Sylow $p$-subgroup of $G_1 \times G_2$.

Proof:
By definition of Sylow $p$-subgroup:
:$\order {H_1} = p^r$, where $p^r$ is the highest power of $p$ which is a divisor of $\order {G_1}$.
:$\order {H_2} = p^s$, where $p^s$ is the highest power of $p$ which is a divisor of $\order {G_2}$.
We have that:
:$\order {H_1 \times H_2} = p^{r + s}$
We also have that $p^{r + s}$ is the highest power of $p$ which is a divisor of $\order {G_1 \times G_2}$.
Hence it follows by definition that $H_1 \times H_2$ is a Sylow $p$-subgroup of $G_1 \times G_2$.
{{qed}}
Theorem: Direct Product of Unique Sylow p-Subgroups is Unique Sylow p-Subgroup

Let $G_1$ and $G_2$ be groups.
Let $H_1$ and $H_2$ be subgroups of $G_1$ and $G_2$ respectively.
Let $G_1$ be such that $H_1$ is the unique Sylow $p$-subgroup of $G_1$.
Let $G_2$ be such that $H_2$ is the unique Sylow $p$-subgroup of $G_2$.
Then $H_1 \times H_2$ is the unique Sylow $p$-subgroup of $G_1 \times G_2$.

Proof:
From Direct Product of Sylow p-Subgroups is Sylow p-Subgroup, $H_1 \times H_2$ is a Sylow $p$-subgroup of $G_1 \times G_2$.
By Sylow $p$-Subgroup is Unique iff Normal, each of $H_1$ and $H_2$ are normal in $G_1$ and $G_2$ respectively.
By Direct Product of Normal Subgroups is Normal, $H_1 \times H_2$ is normal in $G_1 \times G_2$.
Again by Sylow $p$-Subgroup is Unique iff Normal, $H_1 \times H_2$ is the unique Sylow $p$-subgroup of $G_1 \times G_2$.
{{qed}}
Theorem: Intersection of Sylow p-Subgroup with Subgroup not necessarily Sylow p-Subgroup

Let $G$ be a group.
Let $P$ be a Sylow $p$-subgroup of $G$.
Let $H$ be a subgroup of $G$.
Then $P \cap H$ is not necessarily a Sylow $p$-subgroup of $H$.

Proof:
We note that from Intersection of Subgroups is Subgroup that $P \cap H$ is a subgroup of $G$ and also of $H$.
Let $G$ be the dihedral group $D_3$, given by its group presentation:
{{:Group Presentation of Dihedral Group D3}}
By definition of Sylow $p$-subgroup, $\gen a$ is a Sylow $3$-subgroup of $G$.
However, $\gen b$ is also a subgroup of $G$, of order $2$.
But:
:$\gen b \cap \gen a = e$
and $e$ is not a Sylow $3$-subgroup of $G$.
{{qed}}
Theorem: Sylow p-Subgroups of Group of Order 2p

Let $p$ be an odd prime.
Let $G$ be a group of order $2 p$.
Then $G$ has exactly one Sylow $p$-subgroup.
This Sylow $p$-subgroup is normal.

Proof:
Let $n_p$ denote the number of Sylow $p$-subgroups of $G$.
From the Fourth Sylow Theorem:
:$n_p \equiv 1 \pmod p$
From the Fifth Sylow Theorem:
:$n_p \divides 2 p$
that is:
:$n_p \in \set {1, 2, p, 2 p}$
But $p$ and $2 p$ are congruent to $0$ modulo $p$
So:
:$n_p \notin \set {p, 2 p}$
Also we have that $p > 2$.
Hence:
:$2 \not \equiv 1 \pmod p$
and so it must be that $n_p = 1$.
It follows from Sylow $p$-Subgroup is Unique iff Normal that this Sylow $p$-subgroup is normal.
{{qed}}
Theorem: Sylow p-Subgroups of Group of Order 2p

Let $p$ be an odd prime.
Let $G$ be a group of order $2 p$.
Then $G$ has exactly one Sylow $p$-subgroup.
This Sylow $p$-subgroup is normal.

Proof:
Let $n_p$ denote the number of Sylow $p$-subgroups of $G$.
From the First Sylow Theorem, there exists at least $1$ Sylow $p$-subgroup of $G$.
Let $P$ be such a Sylow $p$-subgroup of $G$.
The index of $P$ is $2$.
From Subgroup of Index 2 is Normal, $P$ is normal in $G$.
From Sylow $p$-Subgroup is Unique iff Normal it follows that there is only $1$ Sylow $p$-subgroup of $G$.
{{qed}}
Theorem: Sylow p-Subgroups of Group of Order 2p

Let $p$ be an odd prime.
Let $G$ be a group of order $2 p$.
Then $G$ has exactly one Sylow $p$-subgroup.
This Sylow $p$-subgroup is normal.

Proof:
This is a specific instance of Group of Order $p q$ has Normal Sylow $p$-Subgroup, where $q = 2$.
{{qed}}
Theorem: Groups of Order 2p

Let $p$ be a prime number.
Let $G$ be a group.
Let the order of $G$ be $2 p$.
Then $G$ is either:
:the cyclic group $C_{2 p}$
or:
:the dihedral group $D_p$.

Proof:
When $p = 2$, the result follows from Groups of Order 4.
Let $p$ be an odd prime.
From Sylow p-Subgroups of Group of Order 2p, $G$ has exactly $1$ normal subgroup $P$ of order $p$.
$p$ is prime number.
So from Prime Group is Cyclic, $P$ is a cyclic group.
Let $P = \gen x$ for some $x \in G$.
By the First Sylow Theorem there exists at least one subgroup of $G$ of order $2$.
Hence:
:$\exists y \in G: y^2 = e$
It follows that the elements of $G$ are known:
:$G = \set {e, x, \ldots, x^{p - 1}, y, y x, y x^{p - 1} }$
Then:
{{begin-eqn}}
{{eqn | l = y x y^{-1}
      | o = \in
      | r = P
      | c = as $P$ is normal in $G$.
}}
{{eqn | ll= \leadsto
      | l = y x y^{-1}
      | r = x^i
      | c = for some $i \in \Z_{>0}$
}}
{{eqn | ll= \leadsto
      | l = y x y^{-1} x
      | r = x^{i + 1}
      | c = 
}}
{{eqn | n = 1
      | ll= \leadsto
      | l = y x y x = \paren {y x}^2
      | r = x^{i + 1}
      | c = $y^2 = e$, so $y = y^{-1}$
}}
{{end-eqn}}
Thus:
:the even powers of $y x$ are powers of $x$
:the odd powers of $y x$ are of the form $y x^j$ for some $j \in \Z_{>0}$.
By Order of Element Divides Order of Finite Group:
:$\order {y x} \divides 2 p$
where:
:$\order {y x}$ denotes the order of $y x$
:$\divides$ denotes divisibility.
We have that:
:$y x \ne e$
and so by Identity is Only Group Element of Order 1:
:$\order {y x} \ne 1$
Thus:
:$\order {y x} \in \set {2, p, 2 p}$
{{qed|lemma}}
Suppose $i \ne -1$ in $(1)$ above.
Then:
:$\paren {y x}^2 \ne e$
and so:
:$\order {y x} \ne 2$
Because odd powers of $y x$ are of the form $y x^j$:
:$\paren {y x}^p \ne e$
and so:
:$\order {y x} \ne p$
It follows that:
:$\order {y x} = 2 p$
and from Group whose Order equals Order of Element is Cyclic, $G$ is cyclic.
Thus, when $i \ne -1$:
:$G = \gen {y x}$
and so is cyclic.
Thus by Cyclic Group is Abelian:
:$y x = x y$
{{qed|lemma}}
When $i = -1$ in $(1)$ above, we have that:
{{begin-eqn}}
{{eqn | l = y x y^{-1} x
      | r = e
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y x y^{-1}
      | r = x^{-1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = y x
      | r = x^{-1} y
      | c = 
}}
{{end-eqn}}
leading to the group presentation of $G$:
:$G = \gen {x, y: x^p = e = y^2, y x = x^{-1} y}$
which is the Group Presentation of Dihedral Group $D_p$.
{{qed}}
Theorem: Group of Order p q has Normal Sylow p-Subgroup

Let $p$ and $q$ be prime numbers such that $p > q$.
Let $G$ be a group of order $p q$.
Then $G$ has exactly one Sylow $p$-subgroup.
This Sylow $p$-subgroup is normal.

Proof:
Let $n_p$ denote the number of Sylow $p$-subgroups in $G$.
From the Fourth Sylow Theorem:
:$n_p \equiv 1 \pmod p$
From the Fifth Sylow Theorem:
:$n_p \divides p q$
where $\divides$ denotes divisibility.
The divisors of $p q$ are $1$, $p$, $q$ and $p q$.
Of these:
:$p$ and $p q$ are $\equiv 0 \pmod p$
and as $p > q$:
:$q \equiv q \pmod p$
So the only possibility for $n_p$ is $1$.
Let $P$ denote this unique Sylow $p$-subgroup.
From Sylow $p$-Subgroup is Unique iff Normal, $P$ is normal.
{{qed}}
Theorem: Characterisation of Non-Archimedean Division Ring Norms

Let $\struct {R, \norm {\,\cdot\,} }$ be a normed division ring with unity $1_R$.
Then $\norm {\,\cdot\,}$ is non-Archimedean {{iff}}:
:$\forall n \in \N_{>0}: \norm {n \cdot 1_R} \le 1$
where:
:$n \cdot 1_R = \underbrace {1_R + 1_R + \dotsb + 1_R}_{\text {$n$ times} }$

Proof:
=== Necessary Condition ===
{{:Characterisation of Non-Archimedean Division Ring Norms/Necessary Condition}}{{qed|lemma}}
Theorem: Groups of Order 21/Matrix Representation of Non-Abelian Instance

Let $G$ be the group of order $21$ whose group presentation is:
:$\gen {x, y: x^7 = e = y^3, y x y^{-1} = x^2}$
Then $G$ can be instantiated by the following pair of matrices over $\Z_7$:
:$X = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \qquad Y = \begin{pmatrix} 4 & 0 \\ 0 & 2 \end{pmatrix}$

Proof:
We calculate the powers of $X$ and $Y$ in turn:
{{begin-eqn}}
{{eqn | l = X^2
      | r = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}
      | c = 
}}
{{eqn | r = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = X^3
      | r = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}
      | c = 
}}
{{eqn | r = \begin{pmatrix} 1 & 3 \\ 0 & 1 \end{pmatrix}
      | c = 
}}
{{end-eqn}}
and so on to:
{{begin-eqn}}
{{eqn | l = X^7
      | r = \begin{pmatrix} 1 & 6 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}
      | c = 
}}
{{eqn | r = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
      | c = as $7 \equiv 0 \pmod 7$
}}
{{end-eqn}}
Thus we have:
:$X^7 = \mathbf I$
where $\mathbf I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$ is the identity matrix.
Then:
{{begin-eqn}}
{{eqn | l = Y^2
      | r = \begin{pmatrix} 4 & 0 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 4 & 0 \\ 0 & 2 \end{pmatrix}
      | c = 
}}
{{eqn | r = \begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix}
      | c = 
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = Y^3
      | r = \begin{pmatrix} 4 & 0 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix}
      | c = 
}}
{{eqn | r = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
      | c = 
}}
{{end-eqn}}
Thus we have:
:$Y^3 = \mathbf I$
and:
:$Y^{-1} = \begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix}$.
Then:
{{begin-eqn}}
{{eqn | l = Y X Y^{-1}
      | r = \begin{pmatrix} 4 & 0 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix}
      | c = 
}}
{{eqn | r = \begin{pmatrix} 4 & 4 \\ 0 & 2 \end{pmatrix} \begin{pmatrix} 2 & 0 \\ 0 & 4 \end{pmatrix}
      | c = 
}}
{{eqn | r = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix}
      | c = 
}}
{{eqn | r = X^2
      | c = 
}}
{{end-eqn}}
and the result is apparent.
{{qed}}
Theorem: Normal Sylow p-Subgroups in Group of Order 12

Let $G$ be of order $12$.
Then $G$ has either:
:a normal Sylow $2$-subgroup
or:
:a normal Sylow $3$-subgroup.

Proof:
Note that a Sylow $2$-subgroup of $G$ is of order $4$.
From Sylow $3$-Subgroups in Group of Order 12, there are either $1$ or $4$ Sylow $3$-subgroups.
Suppose there is exactly $1$ Sylow $3$-subgroup $P$.
Then from Sylow $p$-Subgroup is Unique iff Normal, $P$ is normal.
{{qed|lemma}}
Suppose there are $4$ Sylow $3$-subgroups $P_1$, $P_2$, $P_3$ and $P_4$.
Each intersection $P_i \cap P_j$ for $i, j \in \set {1, 2, 3, 4}, i \ne j$ is the trivial subgroup of $G$:
:$P_i \cap P_j = \set e$
Thus $G$ contains:
:The identity element $e$
:$8$ elements of order $3$, of which $2$ each are in $P_1$, $P_2$, $P_3$ and $P_4$
:$3$ more elements, which (along with $e$) must form the Sylow $2$-subgroup of order $4$.
This Sylow $2$-subgroup $Q$ must be unique.
Hence by Sylow $p$-Subgroup is Unique iff Normal, $Q$ is normal.
{{qed|lemma}}
Hence the result.
{{qed}}