hcju/proofwikiqa huggingface dataset:
=== Lemma === {{:Order of Subgroup Product/Lemma}}{{qed|lemma}} We have that $H K$ is the union of all left cosets $h K$ with $h \in H$: :$\displaystyle H K = \bigcup_{h \mathop \in H} h K$ From Left Coset Space forms Partition, unequal $h K$ are disjoint. From Cosets are Equivalent, each $h K$ contains $\order K$ elements. From the Lemma, the number of different such left cosets is: :$\index H {H \cap K}$ where $\index H {H \cap K}$ denotes the index of $H \cap K$ in $H$. From Lagrange's Theorem: $\index H {H \cap K} = \dfrac {\order H} {\order {H \cap K} }$ Hence: :$\order {H K} = \dfrac {\order H \order K} {\order {H \cap K} }$ {{qed}}
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: :$\size {H \times K} = \order H \times \order K$ 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}}
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}}
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}}
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: $\displaystyle G = \bigsqcup_{i \mathop = 1}^p g_i H$ Similarly, there exist $h_1,\ldots,h_q \in H$ such that $H$ is a disjoint union: $\displaystyle H = \bigsqcup_{j \mathop = 1}^q h_j K$ 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}}
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}}
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 $\gen g \rangle \cong \struct {\Z, +}$. By Canonical 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}}
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}}
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}}
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}}
Follows as a direct corollary to the result Powers of Infinite Order Element. {{qed}}
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}}
=== 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}} === Sufficient Condition === Let $g^r = g^s$. Then: : $g^{r - s} = g^r g^{-s} = g^s g^{-s} = e$   By the Division Theorem: : $r - s = q k + t$ for some $q \in \Z, 0 \le t < k$. Thus: :$e = g^{r - s} = g^{k q + t} = \paren {g^k}^q g^t = e^q g^t = g^t$ So by the definition of $k$: :$\paren {t < k} \land \paren {e = g^t} \implies t = 0$ So: :$r - s = q k + 0 = q k \implies k \divides \paren {r - s}$ {{qed}}
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}}
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}$ === Case 1  === :$\map f 1 \ne s$ Take: :$T_1 = \set {1, 2, \dotsc, s - 1, s + 1, \dotsc, n + 1} = S \setminus \set s$ Define the derangement $g_1: T_1 \to T_1$ by: :$\forall t \in T_1: \map {g_1} t = \map f t$ Then $g_1$ is a derangement on a set of cardinality $n$. Thus there are $D_n$ such $g_1$.  Note that: :$f = g_1 \cup \map f s$ We have that $s$ can be chosen in $n$ ways. By the Product Rule for Counting there are in total $n D_n$ such $f$ where $\map f s \ne 1$. === Case 2 === :$\map f 1 = s$ Take: :$T_2 = \set {2, 3, \dotsc, s - 1, s + 1, \dotsc, n + 1} = S \setminus \set {1, s}$ Define the derangement $g_2: T_2 \to T_2$ by: :$\forall t \in T_2: \map {g_2} t = \map f t$ Then $g_2$ is a derangement on a set of cardinality $n - 1$. So there are $D_{n - 1}$ such $g_2$.  Note that: :$f = g_2 \cup \set s \cup \map f s$ We have that $s$ can be chosen in $n$ ways By the Product Rule for Counting there are in total $n D_{n - 1}$ such $f$ where $\map f s = 1$. Summing the results from both cases, we get the total number of derangements $f$ on a set of cardinality $n + 1$ is: :$D_{n + 1} = n D_n + n D_{n - 1} = n \paren {D_n + D_{n - 1} }$ as was to be proved. {{qed}} Category:Counting Arguments Category:Number of Derangements on Finite Set r2mrcuj4x7ja1j53ganx2y98kpw4oic
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}}
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}}
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 $G2$: Properties of Identity }} {{eqn | r = a \left({a b}\right)^2 b       | c = as $\forall x \in G: x^2 = e$ }} {{eqn | r = a \left({a b}\right) \left({a b}\right) b       | c =  }} {{eqn | r = \left({a a}\right) \left({b a}\right) \left({b b}\right)       | c = Group Axiom $G1$: Associativity }} {{eqn | r = a^2 \left({b a}\right) b^2       | c =  }} {{eqn | r = e \left({b a}\right) 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}}
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}}
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}}
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}}
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}}
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}}
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= \leadstoandfrom       | l = x^{2 n - 1}       | r = e       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x^{2 n}       | r = x       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \paren {x^n}^2       | r = x       | c =  }} {{end-eqn}} Hence the result. {{qed}}
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}}
=== 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}} === Finite Order === If $\order g = n$, we have from Equal Powers of Finite Order Element that there are exactly $n$ different elements of $G$ of the form $g^i$. Hence $\gen g$ is a cyclic group with $n$ elements. {{qed}}
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}}
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}}
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}}
Let $G_1$ and $G_2$ be cyclic groups, both of finite order $k$. Let $G_1 = \gen a, G_2 = \gen b$. Then, by the definition of a cyclic group: :$\order a = \order b = k$ Also, by definition: :$G_1 = \set {a^0, a^1, \ldots, a^{k - 1} }$ and: :$G_2 = \set {b^0, b^1, \ldots, b^{k - 1} }$ Let us set up the obvious bijection: :$\phi: G_1 \to G_2: \map \phi {a^n} = b^n$ The next task is to show that $\phi$ is an isomorphism. Note that $\map \phi {a^n} = b^n$ holds for all $n \in \Z$, not just where $0 \le n < k$, as follows: Let $n \in \Z: n = q k + r, 0 \le r < k$, by the Division Theorem. Then, by Element to Power of Remainder: :$a^n = a^r, b^n = b^r$ Thus: :$\map \phi {a^n} = \map \phi {a^r} = b^r = b^n$ Now let $x, y \in G_1$. Since $G_1 = \gen a$, it follows that: :$\exists s, t \in \Z: x = a^s, y = a^t$ Thus: {{begin-eqn}} {{eqn | l = \map \phi {x y}       | r = \map \phi {a^s a^t}       | c =  }} {{eqn | r = \map \phi {a^{s + t} }       | c =  }} {{eqn | r = b^{s + t}       | c =  }} {{eqn | r = b^s b^t       | c =  }} {{eqn | r = \map \phi {a^s} \map \phi {a^t}       | c =  }} {{eqn | r = \map \phi x \map \phi y       | c =  }} {{end-eqn}} So $\phi$ is a homomorphism. As $\phi$ is bijective, $\phi$ is an isomorphism from $G_1$ to $G_2$. Thus $G_1 \cong G_2$ and the result is proved. {{Qed}}
The fact that $H$ is cyclic follows from Subgroup of Cyclic Group is Cyclic. We need to show that $H$ has $\dfrac n d$ elements. Let $\order H = k$. By Non-Trivial Group has Non-Trivial Cyclic Subgroup: :$k = \order a$ where $\order a$ denotes the order of $a$. That is: :$a^k = e$ We have that $a = g^i$. So: {{begin-eqn}} {{eqn | l = a^k       | r = e       | c =  }} {{eqn | ll= \leadstoandfrom       | l = \paren {g^i}^k       | r = e = g^n       | c =  }} {{eqn | ll= \leadstoandfrom       | l = n       | o = \divides       | r = i k       | c = {{Defof|Order of Group Element}} }} {{end-eqn}} We now need to calculate the smallest $k$ such that: : $n \divides i k$ where $\divides$ denotes divisibility. That is, the smallest $t \in \N$ such that $n t = i k$. Let $d = \gcd \set {n, i}$. Thus: :$t = \dfrac {k \paren {i / d} } {n / d}$ From Integers Divided by GCD are Coprime, $\dfrac n d$ and $\dfrac i d$ are coprime. Thus from Euclid's Lemma: :$\dfrac n d \divides k$ As $a \divides b \implies a \le b$, the smallest value of $k$ such that $\dfrac k {\paren {n / d} } \in \Z$ is $\dfrac n d$. Hence the result. {{Qed}}
Follows directly from Order of Subgroup of Cyclic Group: :$\order {\gen {g^{n/d} } } = \dfrac n {\gcd \set {n, n/d} } = d$ Thus from List of Elements in Finite Cyclic Group: :$\gen {g^{n/d} } = \set {e, g^{n/d}, \paren {g^{n/d} }^2, \ldots, \paren {g^{n/d} }^{d - 1} }$ and the result follows. {{qed}}
Let $G$ be generated by $g$, such that $\order g = n$. From Number of Powers of Cyclic Group Element, $g^{n/d}$ has $d$ distinct powers. Thus $\gen {g^{n / d} }$ has $d$ elements. Now suppose $H$ is another subgroup of $G$ of order $d$. Then by Subgroup of Cyclic Group is Cyclic, $H$ is cyclic. Let $H = \gen y$ where $y \in G$. Thus $\order y = d$. Thus $\exists r \in \Z: y = g^r$. Since $\order y = d$, it follows that $y^d = g^{r d} = e$. From Equal Powers of Finite Order Element: : $n \divides r d$ Thus: {{begin-eqn}} {{eqn | lo= \exists k \in \N:       | l = k n       | r = r d       | c =  }} {{eqn | r = k \paren {\dfrac n d} d       | c =  }} {{eqn | ll= \leadsto       | l = r       | r = k \paren {\dfrac n d}       | c =  }} {{eqn | ll= \leadsto       | l = \dfrac n d       | o = \divides       | r = r       | c =  }} {{end-eqn}} Thus $y$ is a power of $g^{n / d}$. Hence $H$ is a subgroup of $\gen {g^{n / d} }$. Since both $H$ and $\gen {g^{n / d} }$ have order $d$, they must be equal. {{qed}}
From the argument in Subgroup of Finite Cyclic Group is Determined by Order, it follows that $x$ satisfies the equation $x^d = e$ {{iff}} $x$ is a power of $g^{n/d}$. Thus there are $d$ solutions to this equation. {{qed}}
Let $a \in G: a \ne e$ where $e$ is the identity of $G$. From Group of Prime Order p has p-1 Elements of Order p, $a$ has order $p$. Hence by definition, $a$ generates $G$. Hence also by definition, $G$ is cyclic. {{qed}}
Let $G$ be a non-abelian group. From Non-Abelian Group has Order Greater than 4, the order of $G$ must be at least $5$. But $5$ is a prime number. By Prime Group is Cyclic it follows that a group of order $5$ is cyclic. By Cyclic Group is Abelian this group is abelian. Hence the result. {{qed}}