avewright/proofwiki-math huggingface dataset:
Theorem: Order of Subgroup Product

Let $G$ be a group.
Let $H$ and $K$ be subgroups of $G$.
Then:
:$\order {H K} = \dfrac {\order H \order K} {\order {H \cap K} }$
where:
:$H K$ denotes subset product
:$\order H$ denotes the order of $H$.
{{questionable|The {{RHS}} does not make sense if $\order {H \cap K} {{=}} +\infty$, because it is then $\dfrac {+\infty}{+\infty}$. Can someone check the source?}}

Proof:
The number of product elements $h k$ that can be formed where $h \in H$ and $k \in K$ is $\order H \order K$, although this may include duplication:
:$h_1 k_1 = h_2 k_2$ may be possible for $h_1, h_2 \in H, k_1, k_2 \in K$.
So, consider the Cartesian product $H \times K$.
From Cardinality of Cartesian Product of Finite Sets:
:$\size {H \times K} = \order H \times \order K$
Let us define a relation $\sim$ on $H \times K$ as:
:$\tuple {h_1, k_1} \sim \tuple {h_2, k_2} \iff h_1 k_1 = h_2 k_2$
As $\sim$ is based on the equality relation it is seen that $\sim$ is an equivalence relation:
:Reflexivity: $h_1 k_1 = h_1 k_1$
:Symmetry: $h_1 k_1 = h_2 k_2 \implies h_2 k_2 = h_1 k_1$
:Transitivity: $h_1 k_1 = h_2 k_2,  h_2 k_2 = h_3 k_3 \implies h_1 k_1 = h_3 k_3$
Each equivalence class of $\sim$ corresponds to a particular element of $H K$.
Hence $\size {H K}$ is the number of equivalence classes of $\sim$.
It remains to be shown that each of these equivalence classes contains exactly $\order {H \cap K}$ elements.
Let $E$ be the equivalence class of $\tuple {h k}$.
We aim to prove that:
:$(1): \quad E = \set {\tuple {h x^{-1}, x k}: x \in H \cap K}$
Let $x \in H \cap K$.
Then:
:$a x^{-1} \in H$
and:
:$x k \in K$
so:
:$\tuple {h x^{-1}, x k} \in H \times K$
Conversely:
:$\tuple {h, k} \sim \tuple {h_1, k_1} \implies h k = h_1 k_1$
and so:
:$x = h_1^{-1} h = k_1 k^{-1} \in A \cap B$
Thus:
:$h_1 = h x^{-1}$
and:
:$k_1 = x k$
and $(1)$ is seen to hold.
Finally:
:$\tuple {h x^{-1}, x k} = \tuple {h y^{-1}, y k} \implies x = y$
Thus $E$ has exactly $\order {H \cap K}$ elements and the proof is complete.
{{qed}}
Theorem: Index of Intersection of Subgroups

Let $G$ be a group.
Let $H, K$ be subgroups of finite index of $G$.
Then:
:$\index G {H \cap K} \le \index G H \index G K$
where $\index G H$ denotes the index of $H$ in $G$.
Note that here the symbol $\le$ is being used with its meaning '''less than or equal to'''.
Equality holds {{iff}} $H K = \set {h k: h \in H, k \in K} = G$.

Proof:
Note that $H \cap K$ is a subgroup of $H$.
From Tower Law for Subgroups, we have:
:$\index G {H \cap K} = \index G H \index H {H \cap K}$
From Index in Subgroup, also:
:$\index H {H \cap K} \le \index G K$
Combining these results yields the desired inequality.
Again from Index in Subgroup, it follows that:
:$\index H {H \cap K} = \index G K$
{{iff}} $H K = G$.
{{qed}}
Theorem: Intersection of Subgroups of Prime Order

Let $G$ be a group whose identity is $e$.
Let $H$ and $K$ be subsets of $G$ such that:
:$\order H = \order K = p$
:$H \ne K$
:$p$ is prime.
Then:
: $H \cap K = \set e$
That is, the intersection of two unequal subgroups of a group, both of whose order is the same prime, consists solely of the identity.

Proof:
From Intersection of Subgroups is Subgroup:
:$H \cap K \le G$
and:
:$H \cap K \le H$
where $\le$ denotes subgrouphood.
So:
{{begin-eqn}}
{{eqn | l = H \cap K
      | o = \le
      | r = H
      | c = Intersection of Subgroups is Subgroup
}}
{{eqn | ll= \leadsto
      | l = \order {H \cap K}
      | o = \divides
      | r = \order H
      | c = Lagrange's Theorem
}}
{{eqn | ll= \leadsto
      | l = \order {H \cap K}
      | o = \divides
      | r = p
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \order {H \cap K}
      | r = 1 \text{ or } p
      | c = $p$ is prime
}}
{{end-eqn}}
Because $H \ne K$ and $\order H = \order k$, it follows that $H \nsubseteq K$.
So:
{{begin-eqn}}
{{eqn | l = H \cap K
      | o = \ne
      | r = H
      | c = Intersection with Subset is Subset
}}
{{eqn | ll= \leadsto
      | l = H \cap K
      | o = \subset
      | r = H
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \order {H \cap K}
      | o = <
      | r = \order H = p
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \order {H \cap K}
      | r = 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = H \cap K
      | r = \set e
      | c = {{Defof|Trivial Group}}
}}
{{end-eqn}}
{{Qed}}
Theorem: Tower Law for Subgroups

Let $\struct {G, \circ}$ be a group.
Let $H$ be a subgroup of $G$ with finite index.
Let $K$ be a subgroup of $H$.
Then:
:$\index G K = \index G H \index H K$
where $\index G H$ denotes the index of $H$ in $G$.

Proof:
Let $p = \index G H$, $q = \index H K$.
By hypothesis these numbers are finite.
Therefore, there exist $g_1, \ldots, g_p \in G$ such that $G$ is a disjoint union:
$\ds G = \bigsqcup_{i \mathop = 1}^p g_i H$
Similarly, there exist $h_1,\ldots,h_q \in H$ such that $H$ is a disjoint union:
$\ds H = \bigsqcup_{j \mathop = 1}^q h_j K$
Thus:
{{begin-eqn}}
{{eqn | l = G
      | r = \bigsqcup_{i \mathop = 1}^p g_i H
}}
{{eqn | r = \bigsqcup_{i \mathop = 1}^p g_i \bigsqcup_{j \mathop = 1}^q h_j K
}}
{{eqn | r = \bigsqcup_{i \mathop = 1}^p \bigsqcup_{j \mathop = 1}^q g_i \paren {h_j K}
      | c = Product of Subset with Union
}}
{{eqn | r = \bigsqcup_{i \mathop = 1}^p \bigsqcup_{j \mathop = 1}^q \paren {g_i h_j} K
      | c = Subset Product within Semigroup is Associative: Corollary
}}
{{end-eqn}}
This expression for $G$ is the disjoint union of $p q$ cosets.
Therefore the number of elements of the coset space is:
:$\index G K = p q = \index G H \index H K$
{{qed}}
Theorem: Tower Law for Subgroups

Let $\struct {G, \circ}$ be a group.
Let $H$ be a subgroup of $G$ with finite index.
Let $K$ be a subgroup of $H$.
Then:
:$\index G K = \index G H \index H K$
where $\index G H$ denotes the index of $H$ in $G$.

Proof:
Assume $G$ is finite.
Then:
{{begin-eqn}}
{{eqn | l = \index G H
      | r = \frac {\order G} {\order H}
      | c = Lagrange's Theorem
}}
{{eqn | l = \index G K
      | r = \frac {\order G} {\order K}
      | c = Lagrange's Theorem
}}
{{eqn | ll= \leadsto
      | l = \index G K
      | r = \frac {\order H} {\order K} \times \index G H
      | c = 
}}
{{end-eqn}}
Since $K \le H$, from Lagrange's Theorem we have that $\dfrac {\order H} {\order K} = \index H K$.
Hence the result.
{{qed}}
Theorem: Morphism from Integers to Group

Let $G$ be a group whose identity is $e$.
Let $g \in G$.
Let $\phi: \Z \to G$ be the mapping defined as:
:$\forall n \in \Z: \map \phi n = g^n$.
Then:
:If $g$ has infinite order, then $\phi$ is a group isomorphism from $\struct {\Z, +}$ to $\gen g$.
:If $g$ has finite order such that $\order g = m$, then $\phi$ is a group epimorphism from $\struct {\Z, +}$ to $\gen g$ whose kernel is the principal ideal $\paren m$.
:Thus $\gen g$ is isomorphic to $\struct {\Z, +}$, and $m$ is the smallest (strictly) positive integer such that $g^m = e$.

Proof:
By Epimorphism from Integers to Cyclic Group, $\phi$ is an epimorphism from $\struct {\Z, +}$ onto $\gen g$.
By Kernel of Group Homomorphism is Subgroup, the kernel $K$ of $G$ is a subgroup of $\struct {\Z, +}$.
Therefore by Subgroup of Integers is Ideal and Principal Ideals of Integers, $\exists m \in \N_{>0}: K = \paren m$.
Thus $\gen g \cong \struct {\Z, +}$.
By Quotient Epimorphism from Integers by Principal Ideal:
:$\forall m \in \N_{>0}: \order {\Z_m} = m$
So, if $\gen g$ is finite, and if $\gen g \cong \struct {\Z, +}$, then $m = \order g$.
Furthermore, $m$ is the smallest (strictly) positive integer such that $g^m = e$, since $m$ is the smallest (strictly) positive integer in $\paren m$ from Principal Ideals of Integers.
If $\gen g$ is infinite, then $m = 0$ and so $\phi$ is a (group) isomorphism from $\struct {\Z, +}$ onto $\gen g$.
{{qed}}
Theorem: Identity is Only Group Element of Order 1

In every group, the identity, and only the identity, has order $1$.

Proof:
Let $G$ be a group with identity $e$.
Then:
:$e^1 = e$
and:
:$\forall a \in G: a \ne e: a^1 = a \ne e$.
Hence the result.
{{qed}}
Theorem: Group Element is Self-Inverse iff Order 2

Let $\struct {S, \circ}$ be a group whose identity is $e$.
An element $x \in \struct {S, \circ}$ is self-inverse {{iff}}:
:$\order x = 2$

Proof:
Let $x \in G: x \ne e$.
{{begin-eqn}}
{{eqn | l = \order x
      | r = 2
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | l = x \circ x
      | r = e
      | c = {{Defof|Order of Group Element}}
}}
{{eqn | ll= \leadstoandfrom
      | l = x
      | r = x^{-1}
      | c = Equivalence of Definitions of Self-Inverse
}}
{{end-eqn}}
{{qed}}
{{expand|Free generalisation to monoids}}
Theorem: Powers of Infinite Order Element

Let $G$ be a group whose identity is $e$.
Let $a \in G$ have infinite order in $G$.
Then:
:$\forall m, n \in \Z: m \ne n \implies a^m \ne a^n$

Proof:
Let $m, n \in \Z$.
Then:
{{begin-eqn}}
{{eqn | l = a^m
      | r = a^n
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a^m \paren {a^n}^{-1}
      | r = e
      | c = 
}}
{{eqn | lo= \land
      | l = a^n \paren {a^m}^{-1}
      | r = e
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a^{m - n}
      | r = a^{n - m} = e
      | c = 
}}
{{eqn | ll= \leadsto
      | l = a^{\size {m - n} }
      | r = e
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \size {m - n}
      | r = 0
      | c = {{Defof|Infinite Order Element}}: there exists no $x \in \Z_{>0}$ such that $a^x = e$
}}
{{eqn | ll= \leadsto
      | l = m
      | r = n
      | c = 
}}
{{end-eqn}}
The result follows from transposition.
{{qed}}
Theorem: Element of Finite Group is of Finite Order

In any finite group, each element has finite order.

Proof:
Let $G$ be a group whose identity is $e$.
From Element has Idempotent Power in Finite Semigroup, for every element in a ''finite'' semigroup, there is a power of that element which is idempotent.
As $G$, being a group, is also a semigroup, the same applies to $G$.
That is:
:$\forall x \in G: \exists n \in \N_{>0}: x^n \circ x^n = x^n$
From Identity is only Idempotent Element in Group, it follows that:
:$x^n \circ x^n = x^n \implies x^n = e$
So $x$ has finite order.
{{Qed}}
Theorem: Element of Finite Group is of Finite Order

In any finite group, each element has finite order.

Proof:
Follows as a direct corollary to the result Powers of Infinite Order Element.
{{qed}}
Theorem: Inverse Element is Power of Order Less 1

Let $G$ be a group whose identity is $e$.
Let $g \in G$ be of finite order.
Then:
:$\order g = n \implies g^{n - 1} = g^{-1}$

Proof:
{{begin-eqn}}
{{eqn | l = \order g
      | r = n
      | c = 
}}
{{eqn | ll= \leadsto
      | l = g^n
      | r = e
      | c = 
}}
{{eqn | ll= \leadsto
      | l = g^n g^{-1}
      | r = e g^{-1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = g^{n - 1}
      | r = g^{-1}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Equal Powers of Finite Order Element

:$g^r = g^s \iff k \divides \paren {r - s}$

Proof:
=== Necessary Condition ===
Suppose that $k \divides \paren {r - s}$.
From the definition of divisor:
:$k \divides \left({r - s}\right) \implies \exists t \in \Z: r - s = k t$
So:
:$g^{r - s} = g^{k t}$
Thus:
{{begin-eqn}}
{{eqn | l = g^r
      | r = g^{s + k t}
      | c = 
}}
{{eqn | r = g^s g^{k t}
      | c = 
}}
{{eqn | r = g^s \paren {g^k}^t
      | c = 
}}
{{eqn | r = g^s \paren e^t
      | c = 
}}
{{eqn | r = g^s
      | c = 
}}
{{end-eqn}}
{{qed|lemma}}
Theorem: Order of Element Divides Order of Finite Group

In a finite group, the order of a group element divides the order of its group:
:$\forall x \in G: \order x \divides \order G$

Proof:
Let $G$ be a group.
Let $x \in G$.
By definition, the order of $x$ is the order of the subgroup generated by $x$.
Therefore, by Lagrange's Theorem, $\order x$ is a divisor of $\order G$.
{{qed}}
Theorem: Recurrence Relation for Number of Derangements on Finite Set

The number of derangements $D_n$ on a finite set $S$ of cardinality $n$ is:
:$D_n = \paren {n - 1} \paren {D_{n - 1} + D_{n - 2} }$
where $D_1 = 0$, $D_2 = 1$.

Proof:
Let $\card S = 1$ such that $S = \set s$, say.
Then $\map f s = s$ is the only permutation on $S$.
This by definition is not a derangement.
Thus:
:$D_1 = 0$
Let $\card S = 2$.
Then $S = \set {s, t}$, say.
There are two permutations on $S$:
:$f = \set {\tuple {s, s}, \tuple {t, t} }$
and:
:$g = \set {\tuple {s, t}, \tuple {t, s} }$
and only the latter is a derangement.
So:
:$D_2 = 1$
Let $\card S > 2$.
Let $f: S \to S$ be a derangement.
We aim to count the total number of such $f$. 
{{WLOG}}, we can take:
:$S = \set {1, 2, \ldots, n + 1}$
Now, consider an arbitrary $s \in S$ such that $s \ne 1$.
Let $\map f s = 1$.
By the Sum Rule for Counting, the total number of $f$ will be: 
:$\paren {\text {the number of $f$ where } \map f 1 \ne s} + \paren {\text {the number of $f$ where} \map f 1 = s}$
Theorem: Element to Power of Group Order is Identity

Let $G$ be a group whose identity is $e$ and whose order is $n$.
Then:
:$\forall g \in G: g^n = e$

Proof:
Let $G$ be a group such that $\order G = n$.
Let $g \in G$ and let $\order g = k$.
From Order of Element Divides Order of Finite Group:
:$k \divides n$
So:
:$\exists m \in \Z_{>0}: k m = n$
Thus:
{{begin-eqn}}
{{eqn | l = g^n
      | r = \paren {g^k}^m
      | c = Powers of Group Elements: Product of Indices
}}
{{eqn | r = e^m
      | c = {{Defof|Order of Group Element}}: $g^k = e$
}}
{{eqn | r = e
      | c = Power of Identity is Identity
}}
{{end-eqn}}
{{qed}}
Theorem: Boolean Group is Abelian

Let $G$ be a Boolean group.
Then $G$ is abelian.

Proof:
By definition of Boolean group, all elements of $G$, other than the identity, have order $2$.
By Group Element is Self-Inverse iff Order 2 and Identity is Self-Inverse, all elements of $G$ are self-inverse.
The result follows directly from All Elements Self-Inverse then Abelian.
{{qed}}
Theorem: Boolean Group is Abelian

Let $G$ be a Boolean group.
Then $G$ is abelian.

Proof:
Let $ a, b \in G$.
By definition of Boolean group:
:$\forall x \in G: x^2 = e$
where $e$ is the identity of $G$.
Then:
{{begin-eqn}}
{{eqn | l = a b
      | r = a e b 
      | c = {{Group-axiom|2}}
}}
{{eqn | r = a \paren {a b}^2 b
      | c = as $\forall x \in G: x^2 = e$
}}
{{eqn | r = a \paren {a b} \paren {a b} b
      | c = 
}}
{{eqn | r = \paren {a a} \paren {b a} \paren {b b}
      | c = {{Group-axiom|1}}
}}
{{eqn | r = a^2 \paren {b a} b^2
      | c = 
}}
{{eqn | r = e \paren {b a} e
      | c = as $\forall x \in G: x^2 = e$
}}
{{eqn | r = b a
      | c = 
}}
{{end-eqn}}
Thus $a b = b a$ and therefore $G$ is abelian. 
{{qed}}
Theorem: Order of Group Element equals Order of Inverse

Let $G$ be a group whose identity is $e$.
Then:
: $\forall x \in G: \order x = \order {x^{-1} }$
where $\order x$ denotes the order of $x$.

Proof:
By Powers of Group Elements: Negative Index:
: $\paren {x^k}^{-1} = x^{-k} = \paren {x^{-1} }^k$
Hence:
{{begin-eqn}}
{{eqn | l = x^k
      | r = e
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {x^{-1} }^k
      | r = e^{-1}
      | c = 
}}
{{eqn | r = e
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \order {x^{-1} }
      | o = \le
      | r = \order x
      | c = {{Defof|Order of Group Element}}
}}
{{end-eqn}}
Similarly:
{{begin-eqn}}
{{eqn | l = \paren {x^{-1} }^k
      | r = e
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {\paren {x^{-1} }^{-1} }^k
      | r = e^{-1}
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x^k
      | r = e^{-1}
      | c = 
}}
{{eqn | r = e
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \order x
      | o = \le
      | r = \order {x^{-1} }
      | c = {{Defof|Order of Group Element}}
}}
{{end-eqn}}
A similar argument shows that if $x$ is of infinite order, then so must $x^{-1}$ be.
Hence the result.
{{Qed}}
Theorem: Even Order Group has Order 2 Element

Let $G$ be a group whose identity is $e$.
Let $G$ be of even order.
Then:
:$\exists x \in G: \order x = 2$
That is:
:$\exists x \in G: x \ne e: x^2 = e$

Proof:
In any group $G$, the identity element $e$ is self-inverse with Identity is Only Group Element of Order 1, and is the only such.
That leaves an odd number of elements.
Each element in $x \in G: \order x > 2$ can be paired off with its inverse, as $\order {x^{-1} } = \order x > 2$ from Order of Group Element equals Order of Inverse.
Hence there must be at least one element which has not been paired off with any of the others which is therefore self-inverse.
The result follows from Group Element is Self-Inverse iff Order 2.
{{qed}}
Theorem: Order of Conjugate Element equals Order of Element

Let $\struct {G, \circ}$ be a group whose identity is $e$.
Then
:$\forall a, x \in \struct {G, \circ}: \order {x \circ a \circ x^{-1} } = \order a$
where $\order a$ denotes the order of $a$ in $G$.

Proof:
Let $\order a = k$.
Then $a^k = e$, and:
:$\forall n \in \N_{>0}: n < k \implies a^n \ne e$
by definition of the order of $a$ in $G$
We have:
{{begin-eqn}}
{{eqn | l = \paren {x \circ a \circ x^{-1} }^k
      | r = x \circ a^k \circ x^{-1}
      | c = Power of Conjugate equals Conjugate of Power
}}
{{eqn | r = x \circ e \circ x^{-1}
      | c = 
}}
{{eqn | r = x \circ x^{-1}
      | c = 
}}
{{eqn | r = e
      | c = 
}}
{{end-eqn}}
Thus $\order {x \circ a \circ x^{-1} } \le \order a$.
Now suppose $a^n = y, y \ne e$.
Then:
:$x \circ a^n \circ x^{-1} = x \circ y \circ x^{-1}$
If $x \circ y = e$, then:
:$x \circ a^n \circ x^{-1} = x^{-1}$
If $y \circ x^{-1} = e$, then:
:$x \circ a^n \circ x^{-1} = x$
So:
:$a^n \ne e \implies x \circ a^n \circ x^{-1} = \paren {x \circ a \circ x^{-1} }^n \ne e$
Thus:
:$\order {x \circ a \circ x^{-1} } \ge \order a$
and the result follows.
{{qed}}
Theorem: Order of Homomorphic Image of Group Element

Let $G$ and $H$ be groups whose identities are $e_G$ and $e_H$ respectively.
Let $\phi: G \to H$ be a homomorphism.
Let $g \in G$ be of finite order.
Then:
:$\forall g \in G: \order {\map \phi g} \divides \order g$
where $\divides$ denotes divisibility.

Proof:
Let $\phi: G \to H$ be a homomorphism.
Let $\order g = n, \order {\map \phi g} = m$.
{{begin-eqn}}
{{eqn | l = \paren {\map \phi g}^n
      | r = \map \phi {g^n}
      | c = Homomorphism of Power of Group Element
}}
{{eqn | r = \map \phi {e_G}
      | c = 
}}
{{eqn | r = e_H
      | c = Homomorphism to Group Preserves Identity
}}
{{end-eqn}}
It follows from Element to Power of Multiple of Order is Identity that $m \divides n$.
{{Qed}}
Theorem: No Group has Two Order 2 Elements

A group can not contain exactly two elements of order $2$.

Proof:
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Suppose:
: $s, t \in \struct {G, \circ}\: s \ne t, \order s = \order t = 2$
That is. they are self-inverse:
:$s^2 = e = t^2$
The identity is of order $1$.
Hence $s$ nor $t$ is the identity
Hence, as $s \ne t$, then $s \circ t \in G$ is distinct from both $s$ and $t$.
Also $s \circ t \ne e$ because $s \ne t^{-1}$.
Suppose $s$ and $t$ commute.
Then $\paren {s \circ t}^2 = e$ from Self-Inverse Elements Commute iff Product is Self-Inverse.
Thus there is a third element (at least) in $G$ which is of order $2$.
Now suppose $s$ and $t$ do ''not'' commute.
Then from Commutation Property in Group, $s \circ t \circ s^{-1}$ is another element of $G$ distinct from both $s$ and $t$.
But from Order of Conjugate Element equals Order of Element:
:$\order {s \circ t \circ s^{-1} } = \order t$
and thus $s \circ t \circ s^{-1}$ is another element of order $2$.
Thus there is a third element (at least) in $G$ which is of order $2$.
{{Qed}}
Theorem: Odd Order Group Element is Square

Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $x \in G$.
Let the order $\order x$ be odd.
Then:
:$\exists y \in G: y^2 = x$

Proof:
Let $\order x$ be odd.
Then:
:$\exists n \in \Z_{> 0}: x^{2 n - 1} = e$
from the definition of the order of an element.
Conversely, suppose that 
:$\exists n \in \Z_{> 0}: x^{2 n - 1} = e$
Then $\order x$ is a divisor of $2 n - 1$ from Element to Power of Multiple of Order is Identity.
Hence $\order x$ is odd.
So $\order x$ is odd {{iff}}:
:$\exists n \in \Z_{> 0}: x^{2 n - 1} = e$
Then:
{{begin-eqn}}
{{eqn | l = \order x
      | r = 2 n - 1
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x^{2 n - 1}
      | r = e
      | c = 
}}
{{eqn | ll= \leadsto
      | l = x^{2 n}
      | r = x
      | c = 
}}
{{eqn | ll= \leadsto
      | l = \paren {x^n}^2
      | r = x
      | c = 
}}
{{eqn | ll= \leadsto
      | q = \exists y \in G
      | l = y^2
      | r = x
      | c = where $x^n$ is precisely that $y$
}}
{{end-eqn}}
Hence the result.
{{qed}}
Theorem: Group Isomorphism Preserves Order of Group Element

Let $G$ and $H$ be groups whose identities are $e_G$ and $e_H$.
Let $\phi: G \to H$ be a group isomorphism.
Then:
:$a \in G \implies \order {\map \phi a} = \order a$

Proof:
First, suppose $a$ is of finite order.
By definition, $\phi$ is bijective, therefore injective.
The result then follows from Order of Homomorphic Image of Group Element.
{{qed|lemma}}
Now suppose $a$ is of infinite order.
Suppose $\map \phi a$ is of finite order.
Consider the mapping $\phi^{-1}: H \to G$.
Let $b = \map \phi a$.
Let $\order b = m$.
Then:
:$\order a = \order {\map {\phi^{-1} } b} = m$
and that would mean $a$ was of finite order.
The result follows by transposition.
{{qed}}
Theorem: Non-Trivial Group has Non-Trivial Cyclic Subgroup

Let $G$ be a group whose identity element is $e$.
Let $g \in G$.
If $g$ has infinite order, then $\gen g$ is an infinite cyclic group.
If $\order g = n$, then $\gen g$ is a cyclic group with $n$ elements.
Thus, every group which is non-trivial has at least one cyclic subgroup which is also non-trivial.
In the case that $G$ is itself cyclic, that cyclic subgroup may of course be itself.

Proof:
=== Infinite Order ===
Suppose that $g$ has infinite order.
We have that $\gen g$ consists of all possible powers of $g$.
So $\gen g$ can contain a finite number of elements only if some of these were equal.
Then we would have:
:$\exists i, j \in \Z, i < j: g^i = g^j$
and so:
:$g^{j - i} = g^{i - i} = g^0 = e$
which would mean that $g$ was of finite order.
This contradiction leads to the conclusion that $\gen g$ must be an infinite cyclic group.
{{qed|lemma}}
Theorem: Epimorphism from Integers to Cyclic Group

Let $\gen a = \struct {G, \circ}$ be a cyclic group.
Let $f: \Z \to G$ be a mapping defined as:
$\forall n \in \Z: \map f n = a^n$.
Then $f$ is a (group) epimorphism from $\struct {\Z, +}$ onto $\gen a$.

Proof:
By Powers of Element form Subgroup:
:$\forall n \in \N: a^n \in \gen a$
Hence by the Index Law for Monoids: Negative Index:
:$\forall n \in \Z: a^n \in \gen a$
Also, by Index Law for Monoids: Sum of Indices, $f$ is a homomorphism from $\struct {\Z, +}$ into $\struct {G, \circ}$.
By Homomorphism Preserves Subsemigroups, its codomain $f \sqbrk \Z$ is therefore a subgroup of $\gen a$ containing $a$.
By Existence of Unique Subgroup Generated by Subset: Singleton Generator, $\gen a$ is the smallest subgroup of $G$ containing $a$.
So:
:$f \sqbrk \Z = \gen a$
Therefore $f$ is an epimorphism from $\struct {\Z, +}$ onto $\struct {G, \circ}$.
{{Qed}}
Theorem: Cyclic Group is Abelian

Let $G$ be a cyclic group.
Then $G$ is abelian.

Proof:
Let $G$ be a cyclic group.
All elements of $G$ are of the form $a^n$, where $n \in \Z$.
Let $x, y \in G: x = a^p, y = a^q$.
From Powers of Group Elements: Sum of Indices:
: $x y = a^p a^q = a^{p + q} = a^{q + p} = a^q a^p = y x$
Thus:
: $\forall x, y \in G: x y = y x$
and $G$ is by definition abelian.
{{qed}}
Theorem: Cyclic Group is Abelian

Let $G$ be a cyclic group.
Then $G$ is abelian.

Proof:
We have that Integers under Addition form Abelian Group.
The result then follows from combining:
:Epimorphism from Integers to Cyclic Group
:Epimorphism Preserves Commutativity.
{{qed}}