# 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 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 ProofThatAllSubgroupsOfACyclicGroupAreCyclic 2013-03-22 13:30:47 2013-03-22 13:30:47 Wkbj79 (1863) Wkbj79 (1863) 10 Wkbj79 (1863) Proof msc 20A05 ProofThatEverySubringOfACyclicRingIsACyclicRing