nyuuzyou/wikis huggingface dataset:
Generating Function of Multiple of Parameter

== Theorem ==
Let $G \left({z}\right)$ be the generating function for the sequence $\left\langle{a_n}\right\rangle$.
Let $c$ be a constant.
Then $G \left({c z}\right)$ be the generating function for the sequence $\left\langle{b_n}\right\rangle$ where:
:$\forall n \in \Z_{\ge 0}: b_n = c^n a_n$
== Proof ==
Hence the result by definition of generating function.
== Sources ==
* : $\S 1.2.9$: Generating Functions
Category:Generating Functions
Generating Function of Sequence by Index

== Theorem ==
Let $\map G z$ be the generating function for the sequence $\sequence {a_n}$.
Then:
:$z \map {G'} z$ is the generating function for the sequence $\sequence {n a_n}$
where $\map {G'} z$ is the derivative of $\map G z$ $z$.
== Proof ==
The result follows by definition of generating function.
== Sources ==
* : $\S 1.2.9$: Generating Functions
Category:Generating Functions
Generation of Linear Code from Standard Generator Matrix/Method 1

== Theorem ==
Let $G$ be a (standard) $k \times n$ generator matrix over $\Z_p$ for a linear code.
The following method can be used to generate from $G$ a linear $\tuple {n, k}$ code over $\Z_p$:
A linear code $C$ can be obtained from $G$ by:
:considering the rows of $G$ as codewords
:forming all possible linear combinations of those codewords, considering them as vectors of a vector space.
== Proof ==
== Sources ==
* : Chapter $6$: Error-correcting codes: Definition $6.11$
Category:Linear Codes
Generation of Linear Code from Standard Generator Matrix/Method 2

== Theorem ==
Let $G$ be a (standard) $k \times n$ generator matrix over $\Z_p$ for a linear code.
The following method can be used to generate from $G$ a linear $\tuple {n, k}$ code over $\Z_p$:
A linear code $C$ can be obtained from $G$ by:
:taking the set $U$ of all sequences of length $k$ over $\Z_p$ and expressing them as $1 \times k$ matrices
:forming all possible matrix products $u G$ for all $u \in U$.
== Proof ==
== Sources ==
* : Chapter $6$: Error-correcting codes: Definition $6.11$
Category:Linear Codes
Generator for Almost Isosceles Pythagorean Triangle

== Theorem ==
Let $P$ be a Pythagorean triangle whose sides correspond to the Pythagorean triple $T = \tuple {a, b, c}$.
Let the generator for $T$ be $\tuple {m, n}$.
Then:
:$P$ is almost isosceles
:$\tuple {2 m + n, m}$ is the generator for the Pythagorean triple $T'$ of another almost isosceles Pythagorean triangle $P'$.
== Proof ==
By definition of almost isosceles:
:$\size {a - b} = 1$
First note that, from Consecutive Integers are Coprime, an almost isosceles Pythagorean triangle is a primitive Pythagorean triangle.
Hence $T$ and $T'$ are primitive Pythagorean triples.
Thus it is established that by Solutions of Pythagorean Equation, both $P$ and $P'$ are of the form:
:$\tuple {2 m n, m^2 + n^2, m^2 - n^2}$
for some $m, n \in \Z_{>0}$ where:
:$m > n$
:$m \perp n$
:$m$ and $n$ are of opposite parity.
=== Necessary Condition ===
Let $\tuple {2 m + n, m}$ be the generator for the Pythagorean triple $T'$ of the almost isosceles Pythagorean triangle $P'$.
Let $p$ and $q$ be the legs of $P'$.
By Solutions of Pythagorean Equation:
:$p = 2 \paren {2 m + n} m$
:$q = \paren {2 m + n}^2 - m^2$
Thus:
That is, $P$ is almost isosceles.
=== Sufficient Condition ===
Let $P$ be almost isosceles.
, by Solutions of Pythagorean Equation:
:$a = 2 m n$
:$b = m^2 - n^2$
Thus:
By Solutions of Pythagorean Equation, $2 \paren {2 m + n} m$ and $\paren {2 m + n}^2 - m^2$ are the legs of a Pythagorean triangle $P'$ whose generator is $\paren {2 m + n, m}$.
But from the above, these legs differ by $1$.
Hence, by definition, $P'$ is an almost isosceles Pythagorean triangle $P'$.
== Sequence ==
== Sources ==
* : $13$
* : $13$
Category:Pythagorean Triangles
Generator for Almost Isosceles Pythagorean Triangle/Sequence

== Sequence ==
The sequence of almost isosceles Pythagorean triangles can be tabulated as follows:
:$\begin{array} {r r | r r | r r r}
m & n & m^2 & n^2 & 2 m n & m^2 - n^2 & m^2 + n^2 \\
\hline
2 & 1 & 4 & 1 & 4 & 3 & 5 \\
5 & 2 & 25 & 4 & 20 & 21 & 29 \\
12 & 5 & 144 & 25 & 120 & 119 & 169 \\
29 & 12 & 841 & 144 & 696 & 697 & 985 \\
70 & 29 & 4900 & 841 & 4060 & 4059 & 5741 \\
169 & 70 & 28 \, 561 & 4900 & 23 \, 660 & 23 \, 661 & 33 \, 461 \\
\hline
\end{array}$
The sequence of elements of the generators are the Pell numbers:
:$1, 2, 5, 12, 29, 70, 169, 408, \ldots$
== Sources ==
* 
* : $13$
* : $13$
* : $20$
Category:Pythagorean Triples
Category:Pythagorean Triangles
Generator for Product Sigma-Algebra

== Theorem ==
Let $\struct {X, \Sigma_1}$ and $\struct {Y, \Sigma_2}$ be measurable spaces.
Let $\GG_1$ and $\GG_2$ be generators for $\Sigma_1$ and $\Sigma_2$, respectively.
Then $\GG_1 \times \GG_2$ is a generator for the product $\sigma$-algebra $\Sigma_1 \otimes \Sigma_2$.
== Proof ==
We begin by stating that this result is in fact incorrect.
For a simple counter-example, consider $X = Y = \set {1, 2}$ both equipped with the power set sigma algebra, that is, $\Sigma_1 = \Sigma_2 = \set {\O, \set 1, \set 2, \set {1, 2} }$.
Then clearly, we can take $\GG_1 = \GG_2 = \set 1$.
However, note that $\GG_1 \times \GG_2 = \set {\tuple {1, 1} }$ and hence the sigma algebra generated by this set will have exactly four elements.
Clearly, $\Sigma_1 \times \Sigma_2$ contains more than four elements and hence the product sigma algebra, i.e., the sigma algebra generated by the product $\Sigma_1\times\Sigma_2$ will contain more than four elements.
We also state and prove a minor modification of the above stated result in this light.
Theorem
Let $X, Y, \Sigma_1, \Sigma_2, \GG_1, \GG_2$ be as before and let $X \in \GG_1$ and $Y \in \GG_2$.
Then $\Sigma_1 \otimes \Sigma_2$ is generated by $\GG_1 \times \GG_2$.
Proof:
We begin by defining $\Sigma$ as the sigma algebra generated by $G_1 \times G_2$.
It is clear that $\Sigma \subset \Sigma_1 \otimes \Sigma_2$.
For the reverse inclusion, set:
:$M_1 := \set {A \subset X: A \times Y \in \Sigma}$
Clearly, $\GG_1 \subset M_1$.
Therefore, $M_1$ is non-empty.
We claim that $M_1$ is a sigma algebra.
Note that $X \in M_1$ obviously.
Also, $A \in M_1$ suffices $\paren {X \setminus A} \times Y = \paren {X \times Y} \setminus \paren {A \times Y} \in \Sigma$, proving closure of $M_1$ under complements.
Also, if $A_n \in M_1$ for $n \in \N$, then:
:$\ds \paren {\bigcup_{n \mathop = 1}^\infty A_n} \times Y = \bigcup_{n \mathop = 1}^\infty \paren {A_n \times Y} \in M_1$
proving closure under countable unions and establishing our claim.
Now since $M_1$ is a sigma algebra containing $\GG_1$, $M_1 \supset \Sigma_1$.
Similarly, setting $M_2 := \set {B \subset Y: X \times B \in \Sigma}$ gives us that $M_2 \supset \Sigma_2$.
Now, if $C \in \Sigma_1$ and $D \in \Sigma_2$, then clearly $C \in M_1$ and $D \in M_2$ giving $C \times Y, X \times D \in \Sigma$
Hence:
:$C \times D = \paren {C \times Y} \cap \paren {X \times D} \in \Sigma$
proving:
:$\Sigma_1 \times \Sigma_2 \subset \Sigma$
and hence giving:
:$\Sigma_1 \otimes \Sigma_2 \subset \Sigma$
establishing the result.
== Sources ==
* : $13.3$
Category:Product Sigma-Algebras
Generator for Quaternion Group

== Theorem ==
The Quaternion Group can be generated by the matrices:
:$\mathbf a = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}
\qquad
\mathbf b = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix}$
where $i$ is the imaginary unit:
:$i^2 = -1$
== Proof ==
Note that:
:$\mathbf I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$
is the identity for (conventional) matrix multiplication of order $2$.
We have:
and so:
Next we have:
and so:
:$\mathbf b^2 = \mathbf a^2$
Then we have:
Thus $\gen {\mathbf a = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}, \mathbf b = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} }$ fulfils the conditions for the group presentation of $\Dic 2$:
Hence the result.
Category:Quaternion Group
Category:Examples of Generators of Groups
Generator of Additive Group Modulo m iff Unit of Ring

== Theorem ==
Let $m \in \Z: m > 1$.
Let $\struct {\Z_m, +_m}$ denote the additive group of integers modulo $m$.
Let $\struct {\Z_m, +_m, \times_m}$‎ be the ring of integers modulo $m$.
Let $a \in \Z_m$.
Then:
:$a$ is a generator of $\struct {\Z_m, +_m}$ 
:$a$ is a unit of $\struct {\Z_m, +_m, \times_m}$
== Proof ==
From Integers under Addition form Infinite Cyclic Group, the identity element $1_\Z$ of the ring of integers $\struct {\Z, +, \times}$ is a generator of the group $\struct {\Z, +}$.
Thus from Quotient Group of Cyclic Group, the identity element $1_{\Z_m}$ of the ring $\struct {\Z_m, +_m, \times_m}$ is a generator of the group $\struct {\Z_m, +_m}$.
Let $a \in \Z_m$.
Suppose $1_{\Z_m} \in \gen a$, where $\gen a$ signifies the group generated by $a$.
Then the smallest subgroup of $\struct {\Z_m, +_m}$ containing $1_{\Z_m}$, that is $\struct {\Z_m, +_m}$ itself, is contained in $\gen a$.
Thus $\gen a = \struct {\Z_m, +_m}$ iff $1_{\Z_m} \in \gen a$.
However, from Subgroup of Additive Group Modulo m is Ideal of Ring, $\gen a$ is an ideal of $\struct {\Z_m, +_m, \times_m}$, and hence is the principal ideal $\ideal a$ generated by $a$.
But from Principal Ideal from Element in Center of Ring, $1_{\Z_m} \in \gen a$ $a$ is a unit of the ring $\struct {\Z_m, +_m, \times_m}$.
Hence the result.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem: Theorem $25.9$
Category:Additive Groups of Integers Modulo m
Category:Ring Theory
Generator of Cyclic Group/Examples/C8

== Examples of Generators of Cyclic Groups ==
Let $C_8$ be generated by $x$:
:$C_8 = \gen x$
The set of generators of the cyclic group $C_8$ is:
:$\set {x, x^3, x^5, x^7}$
== Proof ==
From Power of Generator of Cyclic Group is Generator iff Power is Coprime with Order, $C_8$ is generated by any $x^n$ where $n \perp 8$.
Hence the result.
== Sources ==
* : Chapter $5$: Subgroups: Exercise $14$
Category:Examples of Generators of Cyclic Groups
Generator of Cyclic Group/Examples/Subgroup of Multiplicative Group of Real Numbers Generated by 2

== Example of Generators of Cyclic Group ==
Consider the multiplicative group of real numbers $\struct {\R_{\ne 0}, \times}$.
Consider the subgroup $\gen 2$ of $\struct {\R_{\ne 0}, \times}$ generated by $2$.
Then:
:$\dfrac 1 2$ is also a generator of $\gen 2$
but:
:$4$ is not a generator of $\gen 2$.
== Proof ==
We have that:
:$\dfrac 1 2 = 2^{-1}$
and so:
:$\dfrac 1 2 \in \gen 2$
Now let $x \in \gen 2$.
Then $x = 2^m$ for some $m \in \Z$.
It follows that:
:$x = \paren {\dfrac 1 2}^{-m}$
for $-m \in \Z$.
Similarly, consider $\gen {\dfrac 1 2}$.
Let $y \in \gen {\dfrac 1 2}$.
Then $y = \paren {\dfrac 1 2}^{-r}$ for some $r \in \Z$.
It follows that:
:$y = 2^{-r}$
for $-r \in \Z$.
Thus it is seen that:
:$\gen 2 = \gen {\dfrac 1 2}$
and so $\dfrac 1 2$ is a generator of $\gen 2$.
From the argument in Element of Cyclic Group is not necessarily Generator it is seen that $4$ is not a generator of $\gen 2$.
== Sources ==
* : $\S 39$. Cyclic Groups
Category:Examples of Generators of Cyclic Groups
Generator of Normal Subgroup

== Theorem ==
Let $G$ be a group.
Let $S \subseteq G$.
Let $\hat S = S \cup S^{-1}$, where $S^{-1}$ is the set of all the inverses of all the elements of $S$.
Let $\tilde S = \set {a s a^{-1}: s \in \hat S, a \in G}$.
Let $\map W {\tilde S}$ be the set of words of $\tilde S$.
Let $N$ be the smallest normal subgroup of $G$ that contains $S$.
Then $N = \gen S = \map W {\tilde S}$.
== Proof ==
Let $N$ be the smallest normal subgroup of $G$ that contains $S$, where $S \subseteq G$.
$N$ must certainly include $\hat S$, because any group containing $s \in S$ must also contain $s^{-1}$. 
Therefore, $N$ must be the smallest normal subgroup containing $\hat S$.
Since $N \lhd G$, it follows that $\forall a \in G: \forall s \in \hat S: a s a^{-1} \in N$.
Thus, $\forall x \in \tilde S: x \in N$.
Thus $\tilde S \subseteq N$.
By the closure axiom, $N$ must also contain all products of any finite number of elements of $\tilde S$.
Thus $\map W {\tilde S} \subseteq N$.
Now we prove that $\map W {\tilde S} \lhd G$.
$\tilde S \ne \O$, as $e \paren {s s^{-1} } e^{-1} = e \in \tilde S$.
By Conjugate of Set with Inverse Closed for Inverses, $\tilde S$ is closed under taking inverses.
So from Set of Words Generates Group: Corollary:
:$\map W {\tilde S} \le G$
From Conjugate of Set with Inverse is Closed:
:$\forall w \in\map W {\tilde S}: \forall a \in G: a w a^{-1} \in \map W {\tilde S}$
From Subgroup is Normal iff Contains Conjugate Elements:
:$\map W {\tilde S} \lhd G$.
The result follows by the definition of the minimality of $N$.
Category:Normal Subgroups
Generator of Quotient Groups

== Theorem ==
Let $N \lhd G$ be a normal subgroup of $G$.
Let:
:$N \le A \le G$
:$N \le B \le G$
For a subgroup $H$ of $G$, let $\alpha$ be the bijection defined as:
:$\map \alpha H = \set {h N: h \in H}$
Then:
:$\map \alpha {\gen {A, B} } = \gen {\map \alpha A, \map \alpha B}$
where $\gen {A, B}$ denotes the subgroup of $G$ generated by $\set {A, B}$.
== Proof ==
From the proof of the Correspondence Theorem:
:$\map \alpha H \subseteq G / N$
Then:
== Sources ==
* : Chapter $7$: Normal subgroups and quotient groups: Proposition $7.16 \ \text{(ii)}$
Category:Quotient Groups
Category:Generated Subgroups
Generator of Sigma-Algebra Separates Points

== Theorem ==
Let $X$ be a set.
Let $\GG \subseteq \powerset X$ be a collection of subsets of $X$.
Let $\map \sigma \GG$ be the $\sigma$-algebra generated by $\GG$.
Let $x, y \in X$.
Suppose that $\map \sigma \GG$ separates $x$ and $y$ in the sense:
:$\exists A \in \map \sigma \GG : \quad x \in A \quad \wedge \quad y \in X \setminus A$
Then $\GG$ also separates $x$ and $y$ in the sense:
:$\exists A_0 \in \GG : \quad x \in A_0 \quad \wedge \quad y \in X \setminus A_0$
== Proof ==
:
:$\forall A_0 \in \GG : \quad x, y \in A_0 \quad \vee \quad x, y \in X \setminus A_0$
Let:
:$\CC := \set { A \in \map \sigma {\GG} : \quad x, y \in A \quad \vee \quad x, y \in X \setminus A }$
Then $\CC$ is a $\sigma$-algebra.
Since:
:$\GG \subseteq \CC$
we have:
:$\map \sigma \GG \subseteq \CC$
This means:
:$\forall A \in \map \sigma \GG : \quad x, y \in A \quad \vee \quad x, y \in X \setminus A$
This is a contradiction.
Hence the result by Proof by Contradiction.
Category:Definitions/Sigma-Algebras
Generator of Subgroup/Examples/Positive Odd Numbers

== Example of Generator of Subgroup ==
Let $A$ be the set of positive odd integers.
Let $\struct {\Z, +}$ be the additive group of integers.
The subgroup of $\struct {\Z, +}$ generated by $A$ is $\struct {\Z, +}$ itself.
== Proof ==
From Generator of Subsemigroup: Positive Odd Numbers, the subsemigroup of $\struct {\Z, +}$ generated by $A$ is the semigroup of strictly positive integers under addition.
Then the set of integers $\Z$ is generated by $\struct {\Z, +}$ by various routes.
Hence the result.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings
Category:Examples of Generators of Subgroups
Generator of Subsemigroup/Examples/Positive Odd Numbers

== Example of Generator of Subsemigroup ==
Let $\struct {\Z, +}$ be the additive group of integers.
Let $A$ be the set of positive odd integers.
The subsemigroup of $\struct {\Z, +}$ generated by $A$ is the semigroup of strictly positive integers under addition.
== Proof ==
Let $\struct {S, +}$ be the subsemigroup of $\struct {\Z, +}$ generated by $A$.
First we note that $\struct {A, +}$ is not itself closed, as:
Then we note that:
:$\forall x \in \Z_{>0}: \begin {cases} x \in A & : \text {$x$ odd} \\ \exists y \in A: x = y + 1 & : \text {$x$ even} \end {cases}$
Hence $\Z_{>0}$ is the smallest subset of $\Z$ on which $+$ is a closed operation.
So:
:$A + A = \Z_{>0}$
and from Non-Zero Natural Numbers under Addition form Semigroup, it follows that $\struct {\Z_{>0}, +}$ is a semigroup.
Hence the result by definition of subsemigroup of $\struct {\Z, +}$ generated by $A$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings
Category:Examples of Subsemigroups
Generator of Vector Space Contains Basis

== Theorem ==
Let $E$ be a vector space of $n$ dimensions.
Let $G$ be a generator for $E$.
Then:
:$G$ contains a basis for $E$.
=== Generator of Vector Space Contains Basis/Infinite Dimensional Case ===
== Proof ==
From:
:Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set
:Bases of Finitely Generated Vector Space have Equal Cardinality
and
:Sufficient Conditions for Basis of Finite Dimensional Vector Space
all we need to do is show that every infinite generator $S$ for $E$ contains a finite generator.
Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ be an ordered basis of $E$.
For each $k \in \closedint 1 n$ there is a finite subset $S_k$ of $S$ such that $a_k$ is a linear combination of $S_k$.
Hence $\ds \bigcup_{k \mathop = 1}^n S_k$ is a finite subset of $S$ generating $E$, for the subspace it generates contains $\set {a_1, \ldots, a_n}$ and hence is $E$.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Theorem $27.14$
Category:Bases of Vector Spaces
Category:Generators of Vector Spaces
Category:Generator of Vector Space Contains Basis
Generator of Vector Space Contains Basis/Infinite Dimensional Case

== Theorem ==
Let $K$ be a field.
Let $X$ be a vector space over $K$. 
Let $G$ be a generator of $X$.
Then: 
:$G$ contains a basis for $X$.
== Proof ==
If $X = \set { {\mathbf 0}_X}$, then we must have $G \subseteq \set { {\mathbf 0}_X}$, while the only basis for $X$ is $\O$.
So in this case, we have the claim immediately. 
Now take $X \ne \set { {\mathbf 0}_X}$. 
Let: 
:$\SS = \set {L \subseteq G : L \text { is linearly independent} }$
Since $G$ generates $X$, it contains some $x \ne {\mathbf 0}_X$.
Then we have $\set x \in \SS$, so $\SS \ne \O$.
With view to apply Zorn's Lemma, we show that every non-empty $\subseteq$-chain in $\SS$ has an upper bound.
Let $\CC$ be a non-empty $\subseteq$-chain in $\SS$.
Let: 
:$\ds C = \bigcup \CC$
Let $x_1, x_2, \ldots, x_n \in C$ and $\alpha_1, \alpha_2, \ldots, \alpha_n$ such that: 
:$\ds \sum_{k \mathop = 1}^n \alpha_k x_k = 0$ 
For each $1 \le i \le n$, let $C_i \in \CC$ be such that $x_i \in C_i$. 
Since $\CC$ is a $\subseteq$-chain, there exists $1 \le j \le n$ such that: 
:$C_i \subseteq C_j$ for all $1 \le i \le n$. 
Then $x_i \in C_j$ for each $1 \le i \le n$. 
Since $C_j$ is linearly independent, we have: 
:$\alpha_1 = \alpha_2 = \ldots = \alpha_n = 0$
So $C$ is linearly independent with $C \subseteq G$. 
So $C \in \SS$, while $F \subseteq C$ for all $F \in \CC$. 
So $C$ is an upper bound for $\CC$.
So every non-empty $\subseteq$-chain in $\SS$ has an upper bound.
So by Zorn's Lemma, $\SS$ has a $\subseteq$-maximal element $\BB$. 
We show that $\BB$ is a basis for $E$.
Let: 
:$V = \map \span \BB$ 
Suppose that $V \ne X$.
We then have that $V$ cannot contain $G$, otherwise we would have $V = X$, since $G$ generates $X$. 
So there exists $y \in G$ such that $y \not \in V$.
From Vector not contained in Linear Span of Linearly Independent Set is Linearly Independent of Set, $\BB \cup \set y$ is linearly independent, contradicting the maximality of $\BB$.
So there exists no such $y$ and we have $V = X$.
Since $\BB$ is a linearly independent generator for $X$, it is a basis.
Since $\BB \in \SS$, we have $\BB \subseteq G$ as desired.
Category:Generator of Vector Space Contains Basis
Generator of Vector Space is Basis iff Cardinality equals Dimension

== Theorem ==
Let $E$ be a vector space of $n$ dimensions.
Let $G$ be a generator for $E$:
:$G$ is a basis for $E$ $\card G = n$.
== Proof ==
=== Necessary Condition ===
Let $G$ be a basis for $E$.
From Cardinality of Basis of Vector Space, $\card G = n$.
=== Sufficient Condition ===
Let $\card G = n$.
From Sufficient Conditions for Basis of Finite Dimensional Vector Space, $G$ is a basis for $E$.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Theorem $27.14$
* : Chapter $7$: Vector Spaces: $\S 34$. Dimension: Theorem $67 \ \text{(iv)}$
Category:Generators of Vector Spaces
Category:Bases of Vector Spaces
Category:Dimension of Vector Space
Generators of Additive Group of Integers

== Theorem ==
The only generators of the additive group of integers $\struct {\Z, +}$ are $1$ and $-1$.
== Proof ==
From Integers under Addition form Infinite Cyclic Group, $\struct {\Z, +}$ is an infinite cyclic group generated by $1$.
From Generators of Infinite Cyclic Group, there is only one other generator of such a group, and that is the inverse of that generator.
The result follows.
== Sources ==
* : Chapter $5$: Subgroups: Exercise $14$
* : Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem
Category:Integer Addition
Category:Infinite Cyclic Group
Category:Additive Group of Integers
Generators of Infinite Cyclic Group

== Theorem ==
Let $\gen g = G$ be an infinite cyclic group.
Then the only other generator of $G$ is $g^{-1}$.
Thus an infinite cyclic group has exactly $2$ generators.
== Proof ==
By definition, the infinite cyclic group with generator $g$ is:
:$\gen g = \set {\ldots, g^{-2}, g^{-1}, e, g, g^2, \ldots}$
where $e$ denotes the identity $e = g^0$.
The fact that $g^{-1}$ generates $G$ is shown by Inverse of Generator of Cyclic Group is Generator.
Futhermore:
:$\gen e = \set e \ne G$
By definition of infinite cyclic group:
:$g^i \ne g^j$ for all $i \ne j$
Let $n \in Z \setminus \set {-1, 0, 1}$.
Then:
:$\gen {g^n} = \set {\ldots, g^{-2 n}, g^{-n}, e, g^n, g^{2 n}, \ldots}$
But since $\order n > 1$, none of these elements is equal to $g$, because $1 \notin n \Z$.
So:
:$g \notin \gen {g^n} \implies \gen {g^n} \ne \gen g$
== Note ==
While for $n \in Z \setminus \set {-1, 0, 1}$ we have shown that $\gen {g^n}$ and $\gen g$ are different as sets, the two are isomorphic as abstract groups via:
:$\gen g \owns h \mapsto h^n \in \gen {g^n}$
== Sources ==
* : Chapter $2$: Subgroups and Cosets: $\S 44 \alpha$
* : Chapter $6$: An Introduction to Groups: Exercise $19$
Category:Infinite Cyclic Group
Category:Examples of Generators of Groups
Generators of Special Linear Group of Order 2 over Integers

== Theorem ==
Let:
:$ S = \begin{pmatrix}
0 & - 1 \\
1 & 0
\end{pmatrix}$
and:
:$T = \begin{pmatrix}
1 & 1 \\
0 & 1
\end{pmatrix}$
Then $S$ and $T$ are generators for the special linear group of order $2$ over $\Z$. 
== Proof ==
Let:
:$ g = \begin{pmatrix}
a & b \\
c & d
\end{pmatrix}$
be an element of $\SL {2, \Z}$.
Observe that:
:$T^n = \begin{pmatrix} 
1 & n \\ 
0 & 1
\end{pmatrix}$
so that:
:$T^n g = \begin{pmatrix}
a + nc & b + nd \\
c & d
\end{pmatrix}$
Also note that:
:$S^2 = -I$
and:
 
:$S g = \begin{pmatrix}
-c & -d \\
a & b
\end{pmatrix}$
We now describe an algorithm to reduce $g$ to the identity matrix only using multiplication by $S$ and $T$. 
This will show that $g^{-1}$ can be written as a word in $S$ and $T$. 
We can then take the inverse of this expression in $S$ and $T$ and it too will be a word in $S$ and $T$. 
Recall $g = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$. 
If $c = 0$ then, since $\map \det g = 1$, we must have $a d = 1$. 
Thus $a = d = \pm 1$. In this case:
:$g = \pm\begin{pmatrix}
1 & b' \\
0 & 1
\end{pmatrix} = \pm T^{b'}$
where $b' = \pm b$. 
Since $S^2 = -I$ we either have $g = T^{b'}$ or $g = S^2 T^{b'}$. 
If $c \ne 0$ we proceed as follows. 
, we may suppose $\size a \ge \size c$, for if $\size a < \size c$ we could apply $S$ to swap $a$ and $c$. 
By the Division Theorem, we can write $a = c q + r$ where $q, r \in \Z$ and $0 \le r < \size c$. 
Then $T^{-q} g$ has upper left entry $a - q c = r$ which is smaller than $\size c$.
Applying $S$ switches the rows (with a sign change), which then makes our lower left entry $-r$. 
We have applied a combination of $S$ and $T$ to turn $g$ into a matrix where the absolute value of the lower left entry is strictly less than the one we started with. 
If $r = 0$ then we're in our original case, and we know we can multiply by a multiple of $T$ and $S$ to get to the identity matrix. 
Note that this process can have at most $r$ steps, since there are only $r$ natural numbers between $r$ and $0$. 
Thus the process terminates in a finite number of steps.
Category: Special Linear Group