proof that all subgroups of a cyclic group are cyclic

The following is a proof that all subgroups of a cyclic group are cyclic.

Proof.

Let $G$ be a cyclic group and $H\leq G$ . If $G$ is trivial, then $H=G$ , and $H$ is cyclic. If $H$ is the trivial subgroup, then $H=\{e_{G}\}=\langle e_{G}\rangle$ , and $H$ is cyclic. Thus, for the of the proof, it will be assumed that both $G$ and $H$ are nontrivial.

Let $g$ be a generator of $G$ . Let $n$ be the smallest positive integer such that $g^{n}\in H$ .

Claim: $H=\langle g^{n}\rangle$

Let $a\in\langle g^{n}\rangle$ . Then there exists $z\in{\mathbb{Z}}$ with $a=(g^{n})^{z}$ . Since $g^{n}\in H$ , we have that $(g^{n})^{z}\in H$ . Thus, $a\in H$ . Hence, $\langle g^{n}\rangle\subseteq H$ .

Let $h\in H$ . Then $h\in G$ . Let $x\in{\mathbb{Z}}$ with $h=g^{x}$ . By the division algorithm, there exist $q,r\in{\mathbb{Z}}$ with $0\leq r<n$ such that $x=qn+r$ . Thus, $h=g^{x}=g^{qn+r}=g^{qn}g^{r}=(g^{n})^{q}g^{r}$ . Therefore, $g^{r}=h(g^{n})^{-q}$ . Recall that $h,g^{n}\in H$ . Hence, $g^{r}\in H$ . By choice of $n$ , $r$ cannot be positive. Thus, $r=0$ . Therefore, $h=(g^{n})^{q}g^{0}=(g^{n})^{q}e_{G}=(g^{n})^{q}\in\langle g^{n}\rangle$ . Hence, $H\subseteq\langle g^{n}\rangle$ .

This proves the claim. It follows that every subgroup of $G$ is cyclic. ∎

Title	proof that all subgroups of a cyclic group are cyclic
Canonical name	ProofThatAllSubgroupsOfACyclicGroupAreCyclic
Date of creation	2013-03-22 13:30:47
Last modified on	2013-03-22 13:30:47
Owner	Wkbj79 (1863)
Last modified by	Wkbj79 (1863)
Numerical id	10
Author	Wkbj79 (1863)
Entry type	Proof
Classification	msc 20A05
Related topic	ProofThatEverySubringOfACyclicRingIsACyclicRing