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# 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 remainder of the proof, it will be assumed that both $G$ and $H$ are nontrivial.

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. ∎

## Mathematics Subject Classification

20A05*no label found*

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## Comments

## A puzzle

About a year ago I had given a sketch proof of the infinitude of primes having the form x^2 +1 (using failure functions

^{}).Today when I searched for” sketch proof ” the search did not reveal this topic

Wonder whether this has been deleted, Could anyone pl enlighten me?

A.K. Devaraj