avewright/proofwiki-math huggingface dataset:
Theorem: Hermitian Matrix has Real Eigenvalues

Every Hermitian matrix has eigenvalues which are all real numbers.

Proof:
Let $\mathbf A$ be a Hermitian matrix.
Then, by definition:
:$\mathbf A = \mathbf A^\dagger$
where $\mathbf A^\dagger$ denotes the Hermitian conjugate of $\mathbf A$.
Let $\lambda$ be an eigenvalue of $\mathbf A$.
Let $\mathbf v$ be an eigenvector corresponding to the eigenvalue $\lambda$.
Let $\mathbf v$ be represented by a column vector.
By definition of eigenvector:
:$\mathbf{A v} = \lambda \mathbf v$
Left-multiplying both sides by $\mathbf v^\dagger$, we obtain:
{{begin-eqn}}
{{eqn | n = 1
      | l = \mathbf v^\dagger \mathbf {A v}
      | r = \mathbf v^\dagger \lambda \mathbf v
      | c = 
}}
{{eqn | r = \lambda \mathbf v^\dagger \mathbf v
}}
{{end-eqn}}
Firstly, note that $\mathbf v^\dagger$ is a row vector, so $\mathbf v^\dagger \mathbf v$ results in a $1 \times 1$ matrix.
Since $\lambda$ is a scalar, $\mathbf v^\dagger \mathbf A \mathbf v = \lambda \mathbf v^\dagger \mathbf v$ is also a $1 \times 1$ matrix.
Now observe that, using Hermitian Conjugate of Matrix Product:
:$\paren {\mathbf v^\dagger \mathbf{A v} }^\dagger = \mathbf v^\dagger \mathbf A^\dagger \paren {\mathbf v^\dagger}^\dagger$
As $\mathbf A$ is Hermitian, and $\paren {\mathbf v^\dagger}^\dagger = \mathbf v$ by Hermitian Conjugate is Involution, it follows that:
:$\mathbf v^\dagger \mathbf A^\dagger \paren {\mathbf v^\dagger}^\dagger = \mathbf v^\dagger \mathbf{A v}$
That is, $\mathbf v^\dagger \mathbf {A v}$ is also Hermitian.
By Non-Negative Definiteness and $\mathbf v^\dagger \mathbf v = \innerprod {\mathbf v} {\mathbf v}$, $\mathbf v^\dagger \mathbf v$ is Hermitian. 
So both $\mathbf v^\dagger \mathbf {A v}$ and $\mathbf v^\dagger \mathbf v$ are Hermitian $1 \times 1$ matrices. 
Now suppose that we have for some $a,b \in \C$:
:$\mathbf v^\dagger \mathbf {A v} = \begin {bmatrix} a \end {bmatrix}$
:$\mathbf v^\dagger \mathbf v = \begin {bmatrix} b \end {bmatrix}$
Note that $b \ne 0$ as an eigenvector is by definition non-zero.
By definition of Hermitian matrix:
:$\begin {bmatrix} a \end {bmatrix} = \begin {bmatrix} a \end {bmatrix}^\dagger$ and $\begin {bmatrix} b \end {bmatrix} = \begin {bmatrix} b \end {bmatrix}^\dagger$
By definition of Hermitian conjugate:
:$\begin {bmatrix} a \end {bmatrix}^\dagger = \begin {bmatrix} \bar a \end {bmatrix}$ and $\begin {bmatrix} b \end {bmatrix}^\dagger = \begin {bmatrix} \bar b \end {bmatrix}$
where $\bar a$ denotes the complex conjugate of $a$.
So by definition of equality of matrices:
:$a = \bar a$ and $b = \bar b$
By Complex Number equals Conjugate iff Wholly Real:
:$a, b \in \R$, that is, are real.
From $(1)$:
:$\begin {bmatrix} a \end{bmatrix} = \lambda \begin{bmatrix} b \end{bmatrix}$.
Thus:
:$a = \lambda b$
Hence because $b \ne 0$:
:$\lambda = \dfrac a b$
Hence $\lambda$, being a quotient of real numbers, is real.
{{qed}}
Theorem: Hermitian Matrix has Real Eigenvalues

Every Hermitian matrix has eigenvalues which are all real numbers.

Proof:
The result immediately follows from Hermitian Matrix is Hermitian Operator and Eigenvalues of Hermitian Operator are Real.
{{qed}}
Theorem: Hermitian Matrix has Real Eigenvalues

Every Hermitian matrix has eigenvalues which are all real numbers.

Proof:
Let $\mathbf A$ be a Hermitian matrix.
Then, by definition, $\mathbf A = \mathbf A^\dagger$, where $^\dagger$ designates its Hermitian conjugate.
Let $\lambda$ be an eigenvalue of $\mathbf A$.
Let $\mathbf v$ be an eigenvector corresponding to the eigenvalue $\lambda$ of $\mathbf A$.
Denote with $\innerprod \cdot \cdot$ the inner product on $\C$. 
{{begin-eqn}}
{{eqn | l = \lambda \innerprod {\mathbf v} {\mathbf v}
      | r = \innerprod {\lambda \mathbf v} {\mathbf v}
      | c = Linearity in first argument
}}
{{eqn | r = \innerprod {\mathbf A \mathbf v} {\mathbf v}
      | c = {{Defof|Eigenvector of Linear Operator}}: $\lambda \mathbf v = \mathbf A \mathbf v$
}}
{{eqn | r = \innerprod {\mathbf v} {\mathbf A \mathbf v}
      | c = Hermitian Matrix is Hermitian Operator
}}
{{eqn | r = \innerprod {\mathbf v} {\lambda \mathbf v}
      | c = {{Defof|Eigenvector of Linear Operator}}: $\lambda \mathbf v = \mathbf A \mathbf v$
}}
{{eqn | r = \overline {\innerprod {\lambda \mathbf v} {\mathbf v} }
      | c = Conjugate Symmetry
}}
{{eqn | r = \overline {\lambda \innerprod {\mathbf v} {\mathbf v} }
      | c = Linearity in first argument
}}
{{eqn | r = \overline \lambda \innerprod {\mathbf v} {\mathbf v}
      | c = Product of Complex Conjugates and Conjugate Symmetry
}}
{{end-eqn}}
We have that $\mathbf v \ne \mathbf 0$ since it is an eigenvector.
Hence by non-negative definiteness of an inner product:
:$\innerprod {\mathbf v} {\mathbf v} \ne 0$
and we can divide both sides by $\innerprod {\mathbf v} {\mathbf v}$.
Thus:
:$\lambda = \overline \lambda$
By Complex Number equals Conjugate iff Wholly Real, $\lambda$ is a real number.
$\lambda$ was arbitrary, so it follows that every eigenvalue is a real number.
Hence the result.
{{qed}}
Theorem: Permutation of Cosets

Let $G$ be a group and let $H \le G$.
Let $\mathbb S$ be the set of all distinct left cosets of $H$ in $G$.
Then:
:$(1): \quad$ For any $g \in G$, the mapping $\theta_g: \mathbb S \to \mathbb S$ defined by $\map {\theta_g} {x H} = g x H$ is a permutation of $\mathbb S$.
:$(2): \quad$ The mapping $\theta$ defined by $\map \theta g = \theta_g$ is a homomorphism from $G$ into the symmetric group on $\mathbb S$.
:$(3): \quad$ The kernel of $\theta$ is the subgroup $\ds \bigcap_{x \mathop \in G} x H x^{-1}$.

Proof:
First we need to show that $\theta_g$ is well-defined and injective.
{{begin-eqn}}
{{eqn | l = x H
      | r = y H
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = y^{-1} x
      | o = \in
      | r = H
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \paren {g y}^{-1} g x
      | r = y^{-1} x \in H
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = \map {\theta_g} {y H}
      | r = \map {\theta_g} {x H}
      | c = 
}}
{{end-eqn}}
Thus $\theta_g$ is well-defined and injective.
Then we see that $\forall x H \in \mathbb S: \map {\theta_g} {g^{-1} x H} = x H$, so $\theta_g$ is surjective.
Thus $\theta_g$ is a well-defined bijection on $\mathbb S$, and therefore a permutation on $\mathbb S$.
Next we see:
{{begin-eqn}}
{{eqn | l = \map {\theta_{u v} } {x H}
      | r = u v x H
      | c = 
}}
{{eqn | r = \map {\theta_u} {v x H}
      | c = 
}}
{{eqn | r = \map {\theta_u} {\map {\theta_v }{x H} }=
      | c = 
}}
{{end-eqn}}
This shows that $\theta_{u v} = \theta_u \theta_v$, and thus:
:$\map \theta {u v} = \map \theta u \, \map \theta v$
Thus $\theta$ is a homomorphism.
Now to calculate $\map \ker \theta$:
{{begin-eqn}}
{{eqn | l = \map \ker \theta
      | r = \set {g \in G: \theta_g = I_\mathbb S}
      | c = 
}}
{{eqn | r = \set {g \in G: \forall x \in G: \map {\theta_g} {x H} = x H}
      | c = 
}}
{{eqn | r = \set {g \in G: \forall x \in G: g x h = x H}
      | c = 
}}
{{eqn | r = \set {g \in G: \forall x \in G: x^{-1} g x h = H}
      | c = 
}}
{{eqn | r = \set {g \in G: \forall x \in G: x^{-1} g x \in H}
      | c = 
}}
{{eqn | r = \set {g \in G: \forall x \in G: g \in x H x^{-1} }
      | c = 
}}
{{eqn | r = \bigcap_{x \mathop \in G} x H x^{-1}
      | c = 
}}
{{end-eqn}}
as required.
{{Qed}}
Theorem: Cayley's Representation Theorem

Let $S_n$ denote the symmetric group on $n$ letters.
Every finite group is isomorphic to a subgroup of $S_n$ for some $n \in \Z$.

Proof:
Let $G$ be a group and let $a \in G$.
Consider the left regular representation $\lambda_a: G \to G$ defined as:
:$\map {\lambda_a} x = a \cdot x$
From Regular Representations in Group are Permutations we have that $\lambda_a$ is a permutation.
Now let $b \in G$ and consider $\lambda_b: G \to G$ defined as:
:$\map {\lambda_b} x = b \cdot x$
From the Cancellation Laws it follows that $\lambda_a \ne \lambda_b \iff a \ne b$.
Let $H = \set {\lambda_x: x \in G}$.
Consider the mapping $\Phi: G \to H$ defined as:
:$\forall a \in G: \map \Phi a = \lambda_a$
From the above we have that $\Phi$ is a bijection.
Let $a, b \in G$.
From Composition of Regular Representations we have that:
:$\lambda_a \circ \lambda_b = \lambda_{a \cdot b}$
where $\circ$ denotes composition of mappings.
That is, $\Phi$ has the morphism property.
Thus $\Phi$ is seen to be a group isomorphism.
We also have that:
:$\paren {\lambda_a}^{-1} = \lambda_\paren {a^{-1} }$
because:
:$\lambda_a \circ \paren {\lambda_a}^{-1} = \lambda_\paren {a \cdot a^{-1} }$
Hence the set of left regular representations $\set {\lambda_x: x \in G}$ is a group which is isomorphic to $G$.
{{qed}}
Theorem: Cayley's Representation Theorem

Let $S_n$ denote the symmetric group on $n$ letters.
Every finite group is isomorphic to a subgroup of $S_n$ for some $n \in \Z$.

Proof:
We interpret $\struct {G,\cdot}$ as an group category:
{{DefineCategory
| ob = Only one, say $*$
| mor = $a: * \to *$ for all $a \in G$
| comp =  $a \cdot b: * \to *$
| id = $e: * \to *$
}}
where $e$ is the identity element of $G$.
In particular, $G$ is a small category.
By Cayley's Theorem,  $G$ is isomorphic to a subcategory of $\mathbf{Set}$, say $\mathbf D$.
That is, there are functors:
:$F : G \to \mathbf D$
:$H : \mathbf D \to G$
such that:
:$H F = \operatorname{id}_G$
:$F H = \operatorname{id}_{\mathbf D}$
are identity functors.
Let $S := F \ast$.
Then $P := \set { F_\ast a : a \in G }$ is a permutation group on $S$.
Define a mapping $\phi : G \to P$ by:
:$a \mapsto F_\ast a$
Then $\phi : G \to P$ is a gorup isomorphism.
Indeed, if $\map \phi a = \map \phi b$ then:
:$a = \map {\paren {\operatorname{id}_{\mathbf D} }_\ast} a = \paren {H F}_\ast a = H_{F \ast} \map \phi a = H_{F \ast} \map \phi b = \paren {H F}_\ast b = \map {\paren {\operatorname{id}_{\mathbf D} }_\ast} b = b$
Thus $\phi$ is an injection.
{{finish|Details for the last part}}
Theorem: Cayley's Representation Theorem

Let $S_n$ denote the symmetric group on $n$ letters.
Every finite group is isomorphic to a subgroup of $S_n$ for some $n \in \Z$.

Proof:
Let $H = \set e$.
We can apply Permutation of Cosets to $H$ so that:
:$\mathbb S = G$
and:
:$\map \ker \theta = \set e$
The result follows by the First Isomorphism Theorem for Groups.
{{Qed}}
Theorem: Cayley's Representation Theorem

Let $S_n$ denote the symmetric group on $n$ letters.
Every finite group is isomorphic to a subgroup of $S_n$ for some $n \in \Z$.

Proof:
Let $G$ be any arbitrary finite group whose identity is $e_G$.
Let $S$ be the symmetric group on the elements of $G$, where $e_S$ is the identity of $S$.
For any $x \in G$, let $\lambda_x$ be the left regular representation of $G$ with respect to $x$.
From Regular Representations in Group are Permutations, $\forall x \in G: \lambda_x \in S$.
So, we can define a mapping $\theta: G \to S$ as:
:$\forall x \in G: \map \theta x = \lambda_x$
From Composition of Regular Representations, we have:
:$\forall x, y \in G: \lambda_x \circ \lambda_y = \lambda_{x y}$
where in this context $\circ$ denotes composition of mappings.
Thus by definition of $\theta$:
:$\map \theta x \circ \map \theta y = \map \theta {x y}$
demonstrating that $\theta$ is a homomorphism.
Having established that fact, we can now consider $\map \ker \theta$, where $\ker$ denotes the kernel of $\theta$.
Let $x \in G$.
We have that:
{{begin-eqn}}
{{eqn | l = x
      | o = \in
      | r = \map \ker \theta
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map \theta x
      | r = e_S
      | c = {{Defof|Kernel of Group Homomorphism}}
}}
{{eqn | ll= \leadsto
      | l = \lambda_x
      | r = I_G
      | c = where $I_G$ is the identity mapping of $G$
}}
{{eqn | ll= \leadsto
      | l = \map {\lambda_x} {e_G}
      | r = \map {I_G} {e_G}
      | c = {{Defof|Identity Mapping}}
}}
{{eqn | ll= \leadsto
      | l = x
      | r = e_G
      | c = Definitions of $\lambda_x$ and $I_G$
}}
{{end-eqn}}
So $\map \ker \theta$ can contain no element other than $e_G$.
So, since clearly $e_G \in \map \ker \theta$, it follows that:
:$\map \ker \theta = \set {e_G}$
By Kernel is Trivial iff Monomorphism, it follows that $\theta$ is a monomorphism.
By Monomorphism Image is Isomorphic to Domain, we have that:
:$\theta \sqbrk G \cong \Img \theta$
that is, $\theta$ is isomorphic to its image.
Hence the result.
{{qed}}
Theorem: Epimorphism Preserves Modules

Let $\struct {G, +_G, \circ}_R$ be an $R$-module.
Let $\struct {H, +_H, \circ}_R$ be an $R$-algebraic structure.
Let $\phi: G \to H$ be an epimorphism.
Then $H$ is an $R$-module.

Proof:
If $\struct {G, +_G, \circ}_R$ is an $R$-module, then:
$\forall x, y, \in G, \forall \lambda, \mu \in R$:
:$(1): \quad \lambda \circ \paren {x +_G y} = \paren {\lambda \circ x} +_G \paren {\lambda \circ y}$
:$(2): \quad \paren {\lambda +_R \mu} \circ x = \paren {\lambda \circ x} +_G \paren {\mu \circ x}$
:$(3): \quad \paren {\lambda \times_R \mu} \circ x = \lambda \circ \paren {\mu \circ x}$
If $\phi: G \to H$ is an epimorphism, then:
:$\forall x, y \in G: \map \phi {x +_G y} = \map \phi x +_H \map \phi y$
:$\forall x \in S: \forall \lambda \in R: \map \phi {\lambda \circ x} = \lambda \circ \map \phi x$
:$\forall y \in H: \exists x \in G: y = \map \phi x$
As $\phi$ is an epimorphism, we can accurately specify the behaviour of all elements of $H$, as they are the images of elements of $G$.
If $\phi$ were not an epimorphism, that is not surjective, we would have no way of knowing the behaviour of elements of $H$ outside of the image of $G$.
Hence the specification that $\phi$ needs to be an epimorphism.
Now we check the module axioms in turn.
Theorem: Condition for Linear Transformation

Let $G$ be a unitary $R$-module.
Let $H$ be an $R$-module.
Let $\phi: G \to H$ be a mapping.
Then $\phi$ is a linear transformation {{iff}}:
:$\forall x, y \in G: \forall \lambda, \mu \in R: \map \phi {\lambda x + \mu y} = \lambda \map \phi x + \mu \map \phi y$

Proof:
=== Sufficient Condition ===
{{begin-eqn}}
{{eqn | q = \forall x, y \in G: \forall \lambda, \mu \in R
      | l = \map \phi {\lambda x + \mu y}
      | r = \map \phi {\lambda x} + \map \phi {\mu y}
      | c = {{Defof|Linear Transformation}}: condition $(1)$
}}
{{eqn | r = \lambda \map \phi x + \mu \map \phi y
      | c = {{Defof|Linear Transformation}}: condition $(2)$
}}
{{end-eqn}}
{{qed|lemma}}
Theorem: Module of All Mappings is Module

Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {G, +_G, \circ}_R$ be an $R$-module.
Let $S$ be a set.
Let $\struct {G^S, +_G', \circ}_R$ be the module of all mappings from $S$ to $G$.
Then $\struct {G^S, +_G', \circ}_R$ is an $R$-module.

Proof:
To show that $\struct {G^S, +_G', \circ}_R$ is an $R$-module, we verify the following:
$\forall f, g, \in G^S, \forall \lambda, \mu \in R$:
:$(1): \quad \lambda \circ \paren {f +_G' g} = \paren {\lambda \circ f} +_G' \paren {\lambda \circ g}$
:$(2): \quad \paren {\lambda +_R \mu} \circ f = \paren {\lambda \circ f} +_G \paren {\mu \circ f}$
:$(3): \quad \paren {\lambda \times_R \mu} \circ f = \lambda \circ \paren {\mu \circ f}$
Theorem: Kuratowski's Theorem

The following conditions on a graph $\Gamma$ are equivalent:
:$(1): \quad \Gamma$ is planar
:$(2): \quad \Gamma$ contains no subdivision of either the complete graph $K_5$ or the complete bipartite graph $K_{3, 3}$.

Proof:
Let $\Gamma$ be a graph with $V$ vertices, $E$ edges and $F$ faces.
The proof proceeds in two parts: $(1) \implies (2)$ and $(2) \implies (1)$.
Theorem: Fundamental Theorem of Finite Abelian Groups

Every finite abelian group is an internal group direct product of cyclic groups whose orders are prime powers.
The number of terms in the product and the orders of the cyclic groups are uniquely determined by the group.

Proof:
Let $G$ be a finite abelian group.
By means of Abelian Group is Product of Prime-power Order Groups, we factor it uniquely into groups of prime-power order.
Then Abelian Group of Prime-power Order is Product of Cyclic Groups applies to each of these factors.
Hence we conclude $G$ factors into prime-power order cyclic groups.
The factorisation of $G$ into prime-power order factors is already unique.
Therefore, a demonstration of the uniqueness of the secondary factorisation suffices:
Suppose $\order G = p^k$ with $p$ a prime.
Let $G$ have the following two factorisations:
:$G = H_1 \times H_2 \times \cdots \times H_m = K_1 \times K_2 \times \dotsb \times K_n$
where the $H_i$'s and $K_i$'s are non-trivial cyclic subgroups with
:$\order {H_1} \ge \order {H_2} \ge \dotsb \ge \order {H_m}|$
:$\order {K_1} \ge \order {K_2} \ge \dotsb \ge \order {K_n}$
We proceed through complete induction on $k$.
Theorem: Finite Direct Product of Modules is Module

Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {G_1, +_1, \circ_1}_R, \struct {G_2, +_2, \circ_2}_R, \ldots, \struct {G_n, +_n, \circ_n}_R$ be $R$-modules.
Let:
:$\ds G = \prod_{k \mathop = 1}^n G_k$
be their direct product.
Then $G$ is a module.

Proof:
This is a special case of Direct Product of Modules is Module.
Theorem: Subring Module is Module

Let $\struct {R, +, \times}$ be a ring.
Let $\struct {S, +_S, \times_S}$ be a subring of $R$.
Let $\struct {G, +_G, \circ}_R$ be an $R$-module.
Let $\circ_S$ be the restriction of $\circ$ to $S \times G$.
Let $\struct {G, +_G, \circ_S}_S$ be subring module induced by $S$.
Then $\struct {G, +_G, \circ_S}_S$ is an $S$-module.

Proof:
We have that:
:$\forall a, b \in S: a +_S b = a + b$
:$\forall a, b \in S: a \times_S b = a \times b$
:$\forall a \in S: \forall x \in G = a \circ_S x = a \circ x$
as $+_S$, $\times_S$ and $\circ_S$ are restrictions.
Let us verify the module axioms.
Theorem: Z-Module Associated with Abelian Group is Unitary Z-Module

Let $\struct {G, *}$ be an abelian group with identity $e$.
Let $\struct {G, *, \circ}_\Z$ be the $Z$-module associated with $G$.
Then $\struct {G, *, \circ}_\Z$ is a unitary $Z$-module.

Proof:
The notation $*^n x$ can be written as $x^n$.
Let us verify that $\struct {G, *, \circ}_\Z$ is a unitary $\Z$-module by verifying the axioms in turn.
Theorem: Projection on Cartesian Product of Modules

Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {G, +, \circ}_R$ be the cartesian product of a sequence $\sequence {\struct {G_n, +_n, \circ_n}_R}$ of $R$-modules.
Then for each $j \in \closedint 1 n$, the projection $\pr_j$ on the $j$th co-ordinate is an epimorphism from $\struct {G, +, \circ}_R$ onto $\struct {G_j, +_j, \circ_j}_R$.

Proof:
To demonstrate that $\pr_j$ is an epimorphism, we need to show that:
:$(1): \quad \pr_j$ is a surjection
:$(2): \quad \forall x, y \in G: \map {\pr_j} {x + y} = \map {\pr_j} x +_j \map {\pr_j} y$
:$(3): \quad \forall x \in G: \forall \lambda \in R: \map {\pr_j} {\lambda \circ x} = \lambda \circ_j \map {\pr_j} x$
Criteria $(1)$ and $(2)$ are a direct application of Projection is Epimorphism.
Let $x = \tuple {x_1, x_2, \ldots, x_j, \ldots, x_n}$
Then we have:
{{begin-eqn}}
{{eqn | q = \forall x \in G: \forall \lambda \in R
      | l = \map {\pr_j} {\lambda \circ x}
      | r = \map {\pr_j} {\lambda \circ \tuple {x_1, x_2, \ldots, x_j, \ldots, x_n} }
      | c = 
}}
{{eqn | r = \map {\pr_j} {\tuple {\lambda \circ_1 x_1, \lambda \circ_2 x_2, \ldots, \lambda \circ_j x_j, \ldots, \lambda \circ_n x_n} }
      | c = {{Defof|Cartesian Product}}
}}
{{eqn | r = \lambda \circ_j x_j
      | c = {{Defof|Projection (Mapping Theory)}}
}}
{{eqn | r = \lambda \circ_j \map {\pr_j} x
      | c = {{Defof|Projection (Mapping Theory)}}
}}
{{end-eqn}}
{{qed}}
Theorem: Module is Submodule of Itself

Let $\struct {G, +_G, \circ}_R$ be an $R$-module.
Then $\struct {G, +_G, \circ}_R$ is a submodule of itself.

Proof:
Follows directly from Group is Subgroup of Itself.
{{qed}}
Theorem: Null Module Submodule of All

Let $\struct {G, +_G, \circ}_R$ be an $R$-module.
Then the null module:
:$\struct {\set {e_G}, +_G, \circ}_R$
is a submodule of $\struct {G, +_G, \circ}_R$.

Proof:
From Trivial Subgroup is Subgroup, the trivial subgroup is a subgroup of the group $\struct {G, +_G}$.
The result follows from the Submodule Test.
{{qed}}
Theorem: First Sylow Theorem

Let $p$ be a prime number.
Let $G$ be a group such that:
:$\order G = k p^n$
where:
:$\order G$ denotes the order of $G$
:$p$ is not a divisor of $k$.
Then $G$ has at least one Sylow $p$-subgroup.

Proof:
Let $\order G = k p^r$ where $p \nmid k$.
From the First Sylow Theorem, $G$ has a subgroup $S$ of order $p^r$.
From (need to find it), $S$ itself has subgroups of order $p^n$ for all $n \in \set {1, 2, \ldots, r}$.
{{qed}}
Theorem: First Sylow Theorem

Let $p$ be a prime number.
Let $G$ be a group such that:
:$\order G = k p^n$
where:
:$\order G$ denotes the order of $G$
:$p$ is not a divisor of $k$.
Then $G$ has at least one Sylow $p$-subgroup.

Proof:
Let $\order G = m p^n$.
Let $\mathbb S = \set {S \subseteq G: \order S = p^n}$, that is, the set of all of subsets of $G$ which have exactly $p^n$ elements.
Let $N = \order {\mathbb S}$.
Now $N$ is the number of ways $p^n$ elements can be chosen from a set containing $p^n k$ elements.
From Cardinality of Set of Subsets, this is given by:
:$N = \dbinom {p^n m} {p^n} = \dfrac {\paren {p^n m} \paren {p^n m - 1} \cdots \paren {p^n m - i} \cdots \paren {p^n m - p^n + 1} } {\paren {p^n} \paren {p^n - 1} \cdots \paren {p^n - i} \cdots \paren 1}$
For $0 < i < p^n$, the highest power of $p$ which divides $m p^n - i$ is the same as that dividing $p^n - i$.
{{Explain|why the above holds}}
Thus the power of $p$ in the numerator and denominator of $N$ cancel out, except those of $m$.
Thus $m$ and $N$ have the same number of divisors which are power of $p$.
Let $p^r$ be the largest power of $p$ which divides $m$ and $N$.
Now let $G$ act on $\mathbb S$ by the rule:
:$\forall S \in \mathbb S: g * S = g S = \set {x \in G: x = g s: s \in S}$
That is, $g * S$ is the left coset of $S$ by $g$.
From Group Action on Sets with k Elements, this is a group action.
If every orbit under $*$ were divisible by $p^{r - 1}$, then $p^{r + 1} \divides N$.
{{explain|why the above holds}}
But we have stated that $p^r$ is the largest power of $p$ which divides $N$.
Therefore there is at least one orbit under $*$ which can be represented as:
:$\set {S_1, S_2, \ldots, S_k}$
for which $p^{r + 1} \nmid k$.
Let $H = \Stab {S_1}$ denote the stabilizer of $S_1$.
From the Orbit-Stabilizer Theorem, we have that:
:$k = \index G H = \dfrac {\order g} {\order H}$
As $p^r \divides m$, we have:
:$p^{n + r} \divides p^n m$
But:
:$p^n m = \order G = k \order H$
and so:
:$p^{n + r} \divides k \order H$
But we also have that:
:$p^{r + 1} \nmid k$
which means:
:$p^n \divides \order H$
and that:
:$p^n \le \order H$
Now because $H = \Stab {S_1}$, we haveL
:$H g \subseteq S_1$
for all $g \in \S_1$
Thus:
:$\order H = \card {H g} \le \card {S_1} = p^n$
So we have $p^n \le \order H$ and $\order H \le p^n$ which means:
:$\order H = p^n$
and so $H$ is the subgroup of $G$ whose existence we were to demonstrate.
{{qed}}