avewright/proofwiki-math huggingface dataset:
Theorem: Subgroup of Order p in Group of Order 2p is Normal/Corollary

Let $G$ be non-abelian.
Every element of $G \setminus K$ is of order $2$, and:
:$\forall b \in G \setminus K: b a b^{-1} = a^{-1}$

Proof:
By Lagrange's Theorem, the elements of $G \setminus K$ can be of order $1$, $2$, $p$ or $2 p$.
$1$ is not possible because Identity is Only Group Element of Order 1.
Then we have that $G$ is non-abelian.
Hence from Cyclic Group is Abelian, $G$ is not cyclic.
Thus $\order b \ne 2 p$.
It remains to investigate $2$ and $p$.
Let $b \in G \setminus K$.
By Subgroup of Index 2 contains all Squares of Group Elements:
:$b^2 \in K$
{{AimForCont}} $b$ has order $p$.
Then by Intersection of Subgroups of Prime Order:
:$K \cap \gen b = e$
which contradicts $b^2 \in K$.
Thus by Proof by Contradiction $\order b \ne p$
Hence it must follow that $\order b = 2$.
{{qed|lemma}}
We have that:
:$b a \in G \setminus K$
Thus:
:$\paren {b a}^2 = e$
{{begin-eqn}}
{{eqn | l = b a
      | o = \in
      | r = G \setminus K
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {b a}^2
      | r = e
      | c = 
}}
{{eqn | ll= \leadsto
      | l = b a b
      | r = \paren {b a}^2 a^{-1}
      | c = 
}}
{{eqn | r = a^{-1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = b a b^{-1}
      | r = a^{-1}
      | c = as $b = b^{-1}$
}}
{{end-eqn}}
{{qed}}
Theorem: Subgroup of Order p in Group of Order 2p is Normal

Let $p$ be an odd prime.
Let $G$ be a group of order $2 p$.
Let $a \in G$ be of order $p$.
Let $K = \gen a$ be the subgroup of $G$ generated by $a$.
Then $K$ is normal in $G$.

Proof:
The result Non-Abelian Order 2p Group has Order p Element demonstrates that such an element $a$ exists in $G$.
By definition of generator of cyclic group, $K$ is the cyclic group $C_p$ of order $p$.
By Lagrange's Theorem, the index of $K$ is:
:$\index G K = \dfrac {\order G} {\order K} = \dfrac {2 p} p = 2$
The result follows from Subgroup of Index 2 is Normal.
{{qed}}
Theorem: Inner Automorphisms form Subgroup of Automorphism Group

Let $G$ be a group.
Then the set $\Inn G$ of all inner automorphisms of $G$ forms a normal subgroup of the automorphism group $\Aut G$ of $G$:
:$\Inn G \le \Aut G$

Proof:
Let $G$ be a group whose identity is $e$.
Let $\kappa_x: G \to G$ be the inner automorphism defined as:
:$\forall g \in G: \map {\kappa_x} g = x g x^{-1}$
We see that:
:$\Inn G \ne \O$
as $\kappa_x$ is defined for all $x \in G$.
We show that:
:$\kappa_x, \kappa_y \in \Inn G: \kappa_x \circ \paren {\kappa_y}^{-1} \in \Inn G$
So:
{{begin-eqn}}
{{eqn | q = \forall g \in G
      | l = \map {\paren {\kappa_x \circ \paren {\kappa_y}^{-1} } } g
      | r = \map {\kappa_x} {\map {\paren {\kappa_y}^{-1} } g}
      | c = 
}}
{{eqn | r = \map {\kappa_x} {\map {\kappa_{y^{-1} } } g}
      | c = Inverse of Inner Automorphism
}}
{{eqn | r = \map {\kappa_x} {y^{-1} g y}
      | c = 
}}
{{eqn | r = x y^{-1} g y x^{-1}
      | c = 
}}
{{eqn | r = \paren {x y^{-1} } g \paren {x y^{-1} }^{-1}
      | c = Inverse of Product
}}
{{eqn | r = \map {\kappa_{x y^{-1} } } g
      | c = 
}}
{{end-eqn}}
As $x y^{-1} \in G$, it follows that:
:$\kappa_{x y^{-1} } \in \Inn G$
By the One-Step Subgroup Test:
:$\Inn G \le \Aut G$
{{qed}}
Theorem: Property of Group Automorphism which Fixes Identity Only

Let $G$ be a finite group whose identity is $e$.
Let $\phi: G \to G$ be a group automorphism.
Let $\phi$ have the property that:
:$\forall g \in G \setminus \set e: \map \phi t \ne t$
That is, the only fixed element of $\phi$ is $e$.
Then:
:$\forall x, y \in G: x^{-1} \, \map \phi x = y^{-1} \, \map \phi y \implies x = y$

Proof:
{{begin-eqn}}
{{eqn | l = x^{-1} \, \map \phi x
      | r = y^{-1} \, \map \phi y
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map \phi x
      | r = x y^{-1} \, \map \phi y
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map \phi x \paren {\map \phi y}^{-1}
      | r = x y^{-1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map \phi x \, \map \phi {y^{-1} }
      | r = x y^{-1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \map \phi {x y^{-1} }
      | r = x y^{-1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x y^{-1}
      | r = e
      | c = Definition of $\phi$
}}
{{eqn | ll= \leadsto
      | l = x
      | r = y
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Property of Group Automorphism which Fixes Identity Only/Corollary 1

:$\forall x \in G: \exists g \in G: x = g^{-1} \, \map \phi g$

Proof:
Let $\psi: G \to G$ be the mapping:
:$\forall x \in G: \map \psi x = x^{-1} \, \map \phi x$
From Property of Group Automorphism which Fixes Identity Only:
:$\forall x, y \in G: x^{-1} \, \map \phi x = y^{-1} \, \map \phi y \implies x = y$
That is, $\psi$ is an injection.
From Injection from Finite Set to Itself is Surjection, it follows that $\psi$ is a surjection.
That is:
:$\forall x \in G: \exists g \in G: x = \map \psi g$
which is a restatement of the hypothesis.
{{qed}}
Theorem: Automorphism Group of Cyclic Group is Abelian

Let $G$ be a cyclic group.
Let $\Aut G$ denote the automorphism group of $G$.
Then $\Aut G$ is abelian.

Proof:
Let $G = \gen g$
Let $\phi, \psi \in \Aut G$.
As $G$ is cyclic:
{{begin-eqn}}
{{eqn | q = \exists a \in \Z
      | l = \map \phi g 
      | r = g^a
}}
{{eqn | q = \exists b \in \Z
      | l = \map \psi g 
      | r = g^b
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = \map {\phi \circ \psi} g
      | r = \paren {g^a}^b
      | c = 
}}
{{eqn | r = g^{a b}
      | c = 
}}
{{eqn | r = g^{b a}
      | c = 
}}
{{eqn | r = \paren {g^b}^a
      | c = 
}}
{{eqn | r = \map {\psi \circ \phi} g
      | c = 
}}
{{end-eqn}}
Thus in particular, $\phi \circ \psi$ and $\psi \circ \phi$ are equal on the generator $g$.
Since $g$ generates $G$, they must be equal as automorphisms.
{{qed}}
Theorem: Order of Automorphism Group of Prime Group

Let $p$ be a prime number.
Let $G$ be a group of order $p$.
Let $\Aut G$ denote the automorphism group of $G$.
Then:
:$\order {\Aut G} = p - 1$
where $\order {\, \cdot \,}$ denotes the order of a group.

Proof:
From Prime Group is Cyclic we have that $G$ is a cyclic group.
From Order of Automorphism Group of Cyclic Group:
:$\order {\Aut G} = \map \phi p$
where $\map \phi n$ denotes the Euler $\phi$ function.
The result follows from Euler Phi Function of Prime.
{{qed}}
Theorem: Square Order 2 Matrices over Real Numbers form Ring with Unity

Let $S$ denote the set of square matrices of order $2$ whose entries are the set of real numbers.
Then $S$ forms a non-commutative '''ring with unity''' whose unity is the matrix $\begin {pmatrix} 1 & 0 \\ 0 & 1 \end {pmatrix}$.

Proof:
This is an instance of Ring of Square Matrices over Field is Ring with Unity.
{{qed}}
Theorem: Product Formula for Norms on Non-zero Rationals

Let $\Q_{\ne 0}$ be the set of non-zero rational numbers.
Let  $\Bbb P$ denote the set of prime numbers.
Let $a \in \Q_{\ne 0}$.
Then the following infinite product converges:
:$\size a \times \ds \prod_{p \mathop \in \Bbb P}^{} \norm a_p = 1$
where:
:$\size {\,\cdot\,}$ is the absolute value on $\Q$
:$\norm {\,\cdot\,}_p$ is the $p$-adic norm on $\Q$ for prime number $p$

Proof:
=== Lemma ===
{{:Product Formula for Norms on Non-zero Rationals/Lemma}}{{qed|lemma}}
Let $a = \dfrac b c$, where $b, c \in \Z_{\ne 0}$.
From the Lemma, the following infinite products converge:
:$\size b \ds \times \prod_{p \mathop \in \Bbb P} \norm b_p = 1$
:$\size c \ds \times \prod_{p \mathop \in \Bbb P} \norm c_p = 1$
From Quotient Rule for Real Sequences, the following infinite product converges:
:$\ds \dfrac {\size b} {\size c} \times \prod_{p \mathop \in \Bbb P } \dfrac {\norm b_p} {\norm c_p} = \dfrac 1 1 = 1 $
We have:
{{begin-eqn}}
{{eqn | l = \size a \times \mathop \prod_{p \mathop \in \Bbb P} \norm a_p
      | r = \size {\dfrac b c} \times \prod_{p \mathop \in \Bbb P} \norm {\dfrac b c}_p
}}
{{eqn | r = \dfrac {\size b} {\size c} \times \prod_{p \mathop \in \Bbb P} \dfrac {\norm b_p} {\norm c_p}
      | c = Norm of Quotient
}}
{{eqn | r = 1
}}
{{end-eqn}}
{{qed}}
Theorem: Ring is Subring of Itself

Let $R$ be a ring.
Then $R$ is a subring of itself.

Proof:
$R$ is a ring and $R \subseteq R$.
It follows by definition that $R$ is a subring of $R$.
{{qed}}
Theorem: Null Ring is Subring of Ring

Let $R$ be a ring.
Then the null ring is a subring of $R$.

Proof:
From Null Ring is Ring, the null ring $\struct {\set {0_R}, +, \circ}$ is a ring.
We also have that $\set {0_R}$ is a subset of $R$
Hence the result by definition of subring.
{{qed}}
Theorem: Integers form Subring of Reals

The ring of integers $\struct {\Z, +, \times}$ forms a subring of the field of real numbers.

Proof:
We have that the set of integers $\Z$ are a subset of the real numbers $\R$.
The field of real numbers is, a fortiori, also a ring.
Hence the result, by definition of subring.
{{qed}}
Theorem: Unity of Subring is not necessarily Unity of Ring

Let $\struct {S, +, \circ}$ be a ring with unity whose unity is $1_S$.
Let $\struct {T, + \circ}$ be a subring of $\struct {S, + \circ}$ whose unity is $1_T$.
Then it is not necessarily the case that $1_T = 1_S$.

Proof:
Let $\struct {S, + \times}$ be the ring formed by the set of order $2$ square matrices over the real numbers $R$ under (conventional) matrix addition and (conventional) matrix multiplication.
Let $T$ be the subset of $S$ consisting of the matrices of the form $\begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}$ for $x \in \R$.
We have that:
:$\begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix} + \begin{bmatrix} -y & 0 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} x - y & 0 \\ 0 & 0 \end{bmatrix}$
and so $T$ is closed under subtraction.
From Matrices of the Form $\begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}$ we have that $\struct {T, \times}$ is a subsemigroup of $\struct {S, \times}$.
It follows from the Subring Test that $\struct {T, + \circ}$ be a subbring of $\struct {S, + \circ}$.
From (some result somewhere) $\struct {S, \times}$ has an identity $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ which is not in $\struct {T, \times}$.
However, note that:
{{begin-eqn}}
{{eqn | l = \begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}
      | r = \begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}
      | c = 
}}
{{eqn | r = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} x & 0 \\ 0 & 0 \end{bmatrix}
      | c = 
}}
{{end-eqn}}
demonstrating that $\begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$ is the identity of $\struct {T, \times}$.
The result follows.
{{qed}}
Theorem: Zero of Subfield is Zero of Field

The zero of $\struct {K, +, \times}$ is also $0$.

Proof:
By definition, $\struct {F, +, \times}$ and $\struct {K, +, \times}$ are both rings.
Thus $\struct {K, +, \times}$ is a subring of $\struct {F, +, \times}$
The result follows from Zero of Subring is Zero of Ring.
{{qed}}
Theorem: Zero of Subfield is Zero of Field

The zero of $\struct {K, +, \times}$ is also $0$.

Proof:
By definition, $\struct {K, +, \times}$ is a subset of $F$ which is a field.
By definition of field, $\struct {K, +}$ and $\struct {F, +}$ are groups such that $K \subseteq F$.
So, by definition, $\struct {K, +}$ is a subgroup of $\struct {F, +}$.
By Identity of Subgroup, the identity of $\struct {F, +}$, which is $0$, is also the identity of $\struct {K, +}$.
{{qed}}
Theorem: Unity of Subfield is Unity of Field

The unity of $\struct {K, +, \times}$ is also $1$.

Proof:
By definition, $\struct {K, +, \times}$ is a subset of $F$ which is a field.
By definition of field, $\struct {K^*, \times}$ and $\struct {F^*, \times}$ are groups such that $K \subseteq F$.
So $\struct {K^*, \times}$ is a subgroup of $\struct {F^*, \times}$.
By Identity of Subgroup, the identity of $\struct {F^*, \times}$, which is $1$, is also the identity of $\struct {K^*, \times}$.
{{qed}}
Theorem: Zero of Subring is Zero of Ring

The zero of $\struct {S, +, \times}$ is also $0$.

Proof:
By definition, $\struct {S, +, \times}$ is a subset of $R$ which is a ring.
By definition of ring, $\struct {S, +}$ and $\struct {R, +}$ are groups such that $S \subseteq R$.
So, by definition, $\struct {S, +}$ is a subgroup of $\struct {R, +}$.
By Identity of Subgroup, the identity of $\struct {S, +}$, which is $0$, is also the identity of $\struct {R, +}$.
{{qed}}
Category:Subrings
eu7v5l65ix0xuk1vhigy84hewdr458h
Theorem: Zero of Subfield is Zero of Field/Proof 1

Let $\struct {F, +, \times}$ be a field whose zero is $0$.
Let $\struct {K, +, \times}$ be a subfield of $\struct {F, +, \times}$.
{{:Zero of Subfield is Zero of Field}}

Proof:
By definition, $\struct {F, +, \times}$ and $\struct {K, +, \times}$ are both rings.
Thus $\struct {K, +, \times}$ is a subring of $\struct {F, +, \times}$
The result follows from Zero of Subring is Zero of Ring.
{{qed}}
Theorem: Zero of Subfield is Zero of Field/Proof 2

Let $\struct {F, +, \times}$ be a field whose zero is $0$.
Let $\struct {K, +, \times}$ be a subfield of $\struct {F, +, \times}$.
{{:Zero of Subfield is Zero of Field}}

Proof:
By definition, $\struct {K, +, \times}$ is a subset of $F$ which is a field.
By definition of field, $\struct {K, +}$ and $\struct {F, +}$ are groups such that $K \subseteq F$.
So, by definition, $\struct {K, +}$ is a subgroup of $\struct {F, +}$.
By Identity of Subgroup, the identity of $\struct {F, +}$, which is $0$, is also the identity of $\struct {K, +}$.
{{qed}}
Theorem: P-adic Norm not Complete on Rational Numbers/Proof 1/Case 2

Let $\norm {\,\cdot\,}_p$ be the $p$-adic norm on the rationals $\Q$ for $p = 2$ or $3$.
Then:
:$\struct {\Q, \norm {\,\cdot\,}_p}$ is not a complete normed division ring.
That is, there exists a Cauchy sequence in $\struct {\Q, \norm{\,\cdot\,}_p}$ which does not converge to a limit in $\Q$.

Proof:
{{WIP}}
Theorem: Element in Integral Domain is Divisor iff Principal Ideal is Superset

:$x \divides y \iff \ideal y \subseteq \ideal x$

Proof:
Let that $x \divides y$.
Then by definition of divisor: 
{{begin-eqn}}
{{eqn | l = x \divides y
      | o = \leadsto
      | r = \exists t \in D: y = t x
      | c = {{Defof|Divisor of Ring Element}}
}}
{{eqn | o = \leadsto
      | r = y \in \ideal x
      | c = {{Defof|Principal Ideal of Ring}}
}}
{{eqn | o = \leadsto
      | r = \ideal y \subseteq \ideal x
      | c = {{Defof|Principal Ideal of Ring}}: $\ideal y$ is the smallest ideal containing $y$
}}
{{end-eqn}}
Conversely:
{{begin-eqn}}
{{eqn | l = \ideal y \subseteq \ideal x
      | o = \leadsto
      | r = y \in \ideal x
      | c = as $y \in \ideal y$
}}
{{eqn | o = \leadsto
      | r = \exists t \in D: y = t x
      | c = {{Defof|Principal Ideal of Ring}}
}}
{{eqn | o = \leadsto
      | r = x \divides y
      | c = {{Defof|Divisor of Ring Element}}
}}
{{end-eqn}}
So:
:$x \divides y \iff \ideal y \subseteq \ideal x$
{{qed}}
Theorem: Element in Integral Domain is Unit iff Principal Ideal is Whole Domain

:$x \in U_D \iff \ideal x = D$

Proof:
{{begin-eqn}}
{{eqn | l = x
      | o = \in
      | r = U_D
      | c = 
}}
{{eqn | ll= \leadsto
      | q = \exists u \in U_D
      | l = u
      | o = \in
      | r = \ideal x
      | c = Ideal of Unit is Whole Ring
}}
{{eqn | ll= \leadsto
      | l = \ideal x
      | r = D
      | c = Ideal of Unit is Whole Ring
}}
{{end-eqn}}
Conversely:
{{begin-eqn}}
{{eqn | l = \ideal x
      | r = D
      | c = 
}}
{{eqn | ll= \leadsto
      | l = 1_D
      | o = \in
      | r = \ideal x
      | c = 
}}
{{eqn | ll= \leadsto
      | q = \exists t \in D
      | l = 1
      | r = t x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = u
      | o = \in
      | r = U_D
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Residue of Gamma Function

Let $\Gamma$ be the Definition:Gamma Function. 
Let $n$ be a non-negative integer.
Then: 
:$\Res \Gamma {-n} = \dfrac {\paren {-1}^n} {n!}$

Proof:
By Poles of Gamma Function, $\Gamma$ has simple poles at the non-positive integers, so $-n$ is a simple pole of $\Gamma$. 
Then:
{{begin-eqn}}
{{eqn	| l = \Res \Gamma {-n}
	| r = \lim_{z \mathop \to -n} \paren {z - \paren {-n} } \map \Gamma z
	| c = Residue at Simple Pole
}}
{{eqn	| r = \lim_{z \mathop \to -n} \paren {z + n} \paren {\frac {z \paren {z + 1} \ldots \paren {z + n} } {z \paren {z + 1} \ldots \paren {z + n} } } \map \Gamma z
        | c = multiplying by $1$
}}
{{eqn	| r = \lim_{z \mathop \to -n} \paren {z + n} \frac {\map \Gamma {z + n + 1} } {z \paren {z + 1} \ldots \paren {z + n} }
        | c = Gamma Difference Equation
}}
{{eqn	| r = \lim_{z \mathop \to -n} \frac {\map \Gamma {z + n + 1} } {z \paren {z + 1} \ldots \paren {z + n - 1} }
        | c = cancelling $z + n$
}}
{{eqn	| r = \frac {\map \Gamma 1} {-n \paren {-n + 1} \ldots \paren {-1} }
}}
{{eqn	| r = \frac 1 {\paren {-1}^n \paren {n \paren {n - 1} \ldots \paren 1} }
	| c = $\map \Gamma 1 = 1$
}}
{{eqn	| r = \frac 1 {\paren {-1}^n n!}
	| c = {{Defof|Factorial}}
}}
{{eqn	| r = \frac {\paren {-1}^n} {n!}
}}
{{end-eqn}}
{{qed}}
Category:Gamma Function
digmic2l0cw8ish4k9k2itzobimd9zi
Theorem: Periodic Element is Multiple of Period

Let $f: \R \to \R$ be a real periodic function with period $P$.
Let $L$ be a periodic element of $f$.
Then $P \divides L$.

Proof:
{{AimForCont}} that $P \nmid L$.
Then by the Division Theorem we have $L = q P + r$ where $q \in \Z$ and $0 < r < P$.
And so:
{{begin-eqn}}
{{eqn | l = \map f {x + L}
      | r = \map f {x + \paren {q P + r} }
}}
{{eqn | r = \map f {\paren {x + r} + q P}
}}
{{eqn | r = \map f {x + r}
      | c = General Periodicity Property
}}
{{eqn | r = \map f x
      | c = {{Defof|Periodic Element}}
}}
{{end-eqn}}
But then $r$ is a periodic element of $f$ that is less than $P$.
Therefore $P$ cannot be the period of $f$.
The result follows from Proof by Contradiction.
{{qed}}
Category:Periodic Functions
9cg1yl2s47aaxgforllnw3llud4ruej
Theorem: Equivalence of Definitions of Associate in Integral Domain

{{TFAE|def = Associate in Integral Domain|view = associate|context = Integral Domain|contextview = integral domains}}
Let $\struct {D, +, \circ}$ be an integral domain.
Let $x, y \in D$.

Proof:
=== $(1)$ is Equivalent to $(2)$ ===
{{:Equivalence of Definitions of Associate in Integral Domain/Definition 1 Equivalent to Definition 2}}
Theorem: Finite Set of Elements in Principal Ideal Domain has GCD

Let $\struct {D, +, \circ}$ be a principal ideal domain.
Let $a_1, a_2, \dotsc, a_n$ be non-zero elements of $D$.
Then $a_1, a_2, \dotsc, a_n$ all have a greatest common divisor.

Proof:
Let $0_D$ and $1_D$ be the zero and unity respectively of $D$.
Let $J$ be the set of all linear combinations in $D$ of $\set {a_1, a_2, \dotsc, a_n}$.
From Set of Linear Combinations of Finite Set of Elements of Principal Ideal Domain is Principal Ideal:
:$J = \ideal x$
for some $x \in D$, where $\ideal x$ denotes the principal ideal generated by $x$.
We have that each $a_i$ can be expressed as a linear combination of $\set {a_1, a_2, \dotsc, a_n}$:
:$a_i = 0_D a_1 + 0_D a_2 + \dotsb + 1_D a_i + \dotsb + 0_D a_n$
Thus:
:$\forall i \in \set {0, 1, \dotsc, n}: a_i \in J$
and so by definition of $J$:
:$\forall i \in \set {0, 1, \dotsc, n}: a_i = t_i x$
for some $t_i \in D$.
Thus $x$ is a common divisor of $a_1, a_2, \dotsc, a_n$.
As $x \in \ideal x = J$, we have:
:$x = c_1 a_1 + c_2 a_2 + \dotsb + c_n a_n$
for some $c_1, c_2, \dotsc, c_n \in D$.
Thus every common divisor of $a_1, a_2, \dotsc, a_n$ also is a divisor of $x$.
Thus $x$ is a greatest common divisor of $a_1, a_2, \dotsc, a_n$.
{{qed}}
Theorem: Greatest Common Divisors in Principal Ideal Domain are Associates

Let $\struct {D, +, \circ}$ be a principal ideal domain.
Let $S = \set {a_1, a_2, \dotsc, a_n}$ be a set of non-zero elements of $D$.
Let $y_1$ and $y_2$ be greatest common divisors of $S$.
Then $y_1$ and $y_2$ are associates.

Proof:
From Finite Set of Elements in Principal Ideal Domain has GCD we have that at least one such greatest common divisor exists.
So, let $y_1$ and $y_2$ be greatest common divisors of $S$.
Then:
{{begin-eqn}}
{{eqn | l = y_1
      | o = \divides
      | r = y_2
      | c = as $y_2$ is a greatest common divisor
}}
{{eqn | l = y_2
      | o = \divides
      | r = y_1
      | c = as $y_1$ is a greatest common divisor
}}
{{end-eqn}}
Thus we have:
:$y_1 \divides y_2$ and $y_2 \divides y_1$
where $\divides$ denotes divisibility.
Hence the result, by definition of associates.
{{qed}}
Theorem: Bézout's Identity/Principal Ideal Domain

Let $\struct {D, +, \circ}$ be a principal ideal domain.
Let $S = \set {a_1, a_2, \dotsc, a_n}$ be a set of non-zero elements of $D$.
Let $y$ be a greatest common divisor of $S$.
Then $y$ is expressible in the form:
:$y = d_1 a_1 + d_2 a_2 + \dotsb + d_n a_n$
where $d_1, d_2, \dotsc, d_n \in D$.

Proof:
From Finite Set of Elements in Principal Ideal Domain has GCD we have that at least one such greatest common divisor exists.
So, let $y$ be a greatest common divisor of $S$.
Let $J$ be the set of all linear combinations in $D$ of $\set {a_1, a_2, \dotsc, a_n}$.
From Set of Linear Combinations of Finite Set of Elements of Principal Ideal Domain is Principal Ideal:
:$J = \ideal x$
for some $x \in D$, where $\ideal x$ denotes the principal ideal generated by $x$.
From Finite Set of Elements in Principal Ideal Domain has GCD, $x$ is a greatest common divisor of $S$.
From Greatest Common Divisors in Principal Ideal Domain are Associates, $y$ is an associate of $x$.
By definition of associate:
:$\ideal y = \ideal x$
Therefore:
:$y \in J$
and so by definition, $y$ is expressible in the form:
:$y = d_1 a_1 + d_2 a_2 + \dotsb + d_n a_n$
where $d_1, d_2, \dotsc, d_n \in D$.
{{qed}}
Theorem: Complete Factorizations of Proper Element in Principal Ideal Domain are Equivalent

Let $\struct {D, +, \circ}$ be a principal ideal domain.
Let $x \in D$ be a proper element of $D$.
Let there be two complete factorizations of $x$:
:$x = u_y \circ y_1 \circ y_2 \circ \cdots \circ y_m = F_1$
:$x = u_z \circ z_1 \circ z_2 \circ \cdots \circ z_n = F_2$
Then $F_1$ and $F_2$ are equivalent.

Proof:
{{ProofWanted|Whitelaw leaves this unresolved at the end of $\S 62$ as an exercise for the student. I haven't read ahead that far, but it may be proved in the exercises. Will return to this later.}}#
Theorem: Moment Generating Function of Poisson Distribution

Let $X$ be a discrete random variable with a Poisson distribution with parameter $\lambda$ for some $\lambda \in \R_{> 0}$.
Then the moment generating function $M_X$ of $X$ is given by: 
:$\map {M_X} t = e^{\lambda \paren {e^t - 1} }$

Proof:
From the definition of the Poisson distribution, $X$ has probability mass function:
:$\map \Pr {X = n} = \dfrac {\lambda^n e^{-\lambda} } {n!}$
From the definition of a moment generating function:
:$\ds \map {M_X} t = \expect {e^{t X} } = \sum_{n \mathop = 0}^\infty \map \Pr {X = n} e^{t n}$
So: 
{{begin-eqn}}
{{eqn | l = \map {M_X} t
      | r = \sum_{n \mathop = 0}^\infty \frac {\lambda^n e^{-\lambda} } {n!} e^{t n}
}}
{{eqn | r = e^{-\lambda} \sum_{n \mathop = 0}^\infty \frac {\paren {\lambda e^t}^n} {n!}
}}
{{eqn | r = e^{-\lambda} e^{\lambda e^t}
      | c = Power Series Expansion for Exponential Function
}}
{{eqn | r = e^{\lambda \paren {e^t - 1} }
      | c = Exponential of Sum
}}
{{end-eqn}}
{{qed}}
Theorem: Moment Generating Function of Binomial Distribution

Let $X$ be a discrete random variable with a binomial distribution with parameters $n$ and $p$ for some $n \in \N$ and $0 \le p \le 1$:
:$X \sim \Binomial n p$
Then the moment generating function $M_X$ of $X$ is given by: 
:$\map {M_X} t = \paren {1 - p + p e^t}^n$

Proof:
From the definition of the Binomial distribution, $X$ has probability mass function:
:$\map \Pr {X = k} = \dbinom n k p^k \paren {1 - p}^{n - k}$
From the definition of a moment generating function:
:$\ds \map {M_X} t = \expect {e^{t X} } = \sum_{k \mathop = 0}^n \map \Pr {X = k} e^{t k}$
So: 
{{begin-eqn}}
{{eqn | l = \map {M_X} t 
      | r = \sum_{k \mathop = 0}^n \binom n k p^k \paren {1 - p}^{n - k} e^{t k}
}}
{{eqn | r = \sum_{k \mathop = 0}^n \binom n k \paren {p e^t}^k \paren {1 - p}^{n - k}
}}
{{eqn | r = \paren {1 - p + p e^t}^n
      | c = Binomial Theorem
}}
{{end-eqn}}
{{qed}}