Fundamental theorem of cyclic groups

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In abstract algebra, the fundamental theorem of cyclic groups states that if $G\,$ is a cyclic group of order $n\,$ then every subgroup of $G\,$ is cyclic. Moreover, the order of any subgroup of $G\,$ is a divisor of $n\,$ and for each positive divisor $k\,$ of $n\,$ the group $G\,$ has at most one subgroup of order $k\,$ .

1 Proof
2 Alternate proof
3 Converse
4 See also

[edit] Proof

Let $G = \langle a\rangle\,$ be a cyclic group for some $a \in G$ and with identity $e\,$ and order $n\,$ , and let $H\,$ be a subgroup of $G\,$ . We will now show that $H\,$ is cyclic. If $H = \{ e \}\,$ then $H = \langle e \rangle\,$ . If $H \neq \{ e \}\,$ then since $G\,$ is cyclic every element in $H\,$ is of the form $a^t\,$ , where $t\,$ is a positive integer. Let $m\,$ be the least positive integer such that $a^m \in H$ .

We will now show that $H = \langle{a^m}\rangle\,$ . It follows immediately from the closure property that $\langle {a^m}\rangle \subseteq H$ .

To show that $H \subseteq \langle {a^m}\rangle$ we let $b \in H$ . Since $b \in G$ we have that $b = a^k\,$ for some positive integer $k\,$ . By the division algorithm, $k = mq + r\,$ with $0 \le r < m\,$ , and so $a^k = a^{mq + r} = a^{mq}a^r\,$ , which yields $a^r = a^{-mq}a^k\,$ . Now since $a^k \in H$ and $a^{-mq} = (a^{m})^{-q} \in H$ , it follows from closure that $a^r \in H$ . But $m\,$ is the least integer such that $a^m \in H$ and $0 \le r < m\,$ , which means that $r = 0\,$ and so $b = a^k = a^{mq} = (a^m)^q \in \langle a^m \rangle$ . Thus $H \subseteq \langle {a^m}\rangle$ .

Since $H \subseteq \langle a^m\rangle$ and $\langle {a^m}\rangle \subseteq H$ it follows that $H = \langle a^m\rangle \,$ and so $H\,$ is cyclic.

We will now show that the order of any subgroup of $G\,$ is a divisor of $n\,$ . Let $H\,$ be any subgroup of $G\,$ . We have already shown that $H = \langle a^m \rangle\,$ , where m is the least positive integer such that $a^m \in H$ . Since $e = a^n = a^m\,$ it follows that $n = mq\,$ for some integer $q\,$ . Thus $m | n\,$ .

We will now prove the last part of the theorem. Let $k\,$ be any positive divisor of $n\,$ . We will show that $\langle a^{n/k} \rangle\,$ is the one and only subgroup of $\langle a\rangle\,$ of order $k\,$ . Note that $\langle a^{n/k} \rangle\,$ has order ${n\over{gcd(n, {n \over {k}})}} = {n \over {n \over k}} = k\,$ . Let $H\,$ be any subgroup of $\langle a \rangle\,$ with order $k\,$ . We know that $H = \langle a^m \rangle\,$ , where $m\,$ is a divisor of $n\,$ . So $m = \operatorname{gcd}(n, m)\,$ and $k = | \langle a ^m \rangle | = | a ^m | = |a^{\operatorname{gcd}(n, m)}| = {n \over {\operatorname{gcd}(n, m)}} = {n \over m}\,$ . Consequently $m = {n \over k}\,$ and so $H = \langle a^{n \over k} \rangle\,$ , and thus the theorem is proved.

[edit] Alternate proof

Let $G = \langle a \rangle$ be a cyclic group, and let $H$ be a subgroup of $G$ . Define a morphism $\varphi: \mathbb{Z} \rightarrow G$ by $\varphi(n) = a^n$ . Since $G$ is cyclic generated by $a$ , $\varphi$ is surjective. Let $K = \varphi^{-1}(H) \subseteq \mathbb{Z}$ . $K$ is a subgroup of $\mathbb{Z}$ . Since $\varphi$ is surjective, the restriction of $\varphi$ to $K$ defines a surjective morphism from $K$ onto $H$ , and therefore $H$ is isomorphic to a quotient of $K$ . Since $K$ is a subgroup of $\mathbb{Z}$ , $K$ is $n\mathbb{Z}$ for some integer $n$ . If $n = 0$ , then $K = 0$ , hence $H = 0$ , which is cyclic. Otherwise, $K$ is isomorphic to $\mathbb{Z}$ . Therefore $H$ is isomorphic to a quotient of $\mathbb{Z}$ , and is necessarily cyclic.

[edit] Converse

The following statements are equivalent.

A group G of order $n\,$ is cyclic.
For every divisor $d\,$ of $n\,$ a group G has exactly one subgroup of order $d\,$ .
For every divisor $d\,$ of $n\,$ a group G has at most one subgroup of order $d\,$ .