Any subgroup of a cyclic subgroup is cyclic

Question

I proved that any subgroup of a cyclic group is cyclic. Did I prove this correctly? Are there any missteps /ambiguities?

Let $G = \{a,a^2, a^3,\dots,a^n\}$ be a cyclic group of order $n$. Let $H$ be a subgroup of $G$. Let $k>1$ be the smallest integer such that $a^k \in H$. We claim that $k\vert n$. If $k\nmid n$, $mk = n+l$ for some $l< k$. Then $a^{mk} = a^{n+l} = a^n.a^l = e.a^l = a^l$. $\because$ $a^{mk} \in H$, $a^l \in H$, which contradicts our claim that $k$ is the smallest integer such that $a^k \in H$.

Now we assume that $H$ is not cyclic and let $n = ks$. Then $H$ has at least one element of the form $a^{kq + r}$ such that $q < s$ and $r$ are integers and $r < k$. But $\because a^{-kq} = (a^{kq})^{-1}\in H$, $a^{(s-q)k} \in H$, $\implies a^{kq + r}.a^{(s-q)k} = a^{r + sk} = a^r.e = a^r$. Thus, $a^r \in H$, which contradicts our claim that $k$ is the smallest integer such that $a^k \in H$. Hence we conclude that $H$ is cyclic.

EDIT:

Let $k$ be the smallest integer such that $a^k\in H$. If $H$ is not cyclic, then we have at least one element of the form $a^{kq +r} \in H$ where $1 \leq r < k$. But $$a^k\in H \implies a^{kq} \in H \implies (a^{kq})^{-1} = a^{-kq} \in H \implies a^{kq + r - kq} \in H \implies a^r \in H $$

This contradicts our assumption that $k$ is the smallest integer such that $a^{k} \in H$.

More can be said about the division algorithm in the proof, $kq+r$, see here. — Dietrich Burde, Feb 21 '17 at 15:31

score 1 · Answer 1 · answered Feb 21 '17 at 15:22

1

It looks fine, although you only did it for finite cyclic groups.

answered Feb 21 '17 at 15:22

Asinomás

105,651

Does the part after edit work for infinite cyclic groups? – saad Feb 22 '17 at 15:41
yes, it looks good. Although instead of saying if $G$ is not cyclic I would say if $g^k$ does not generate $G$. – Asinomás Feb 22 '17 at 16:10
and you may want to consider the case when no such $k$ exists – Asinomás Feb 22 '17 at 16:23

score 0 · Answer 2 · answered Feb 22 '17 at 16:15

It's correct up to a couple of minor details.

When you introduce $l$, you should say that $0<l<k$.
When you say “Assume $H$ is not cyclic” you should say “Assume $H$ is not generated by $a^k$”, which is what you're actually using in the argument.

There is a conceptually better proof.

Let's prove that every subgroup of $\mathbb{Z}$ is cyclic. Indeed, the statement is obvious for $\{0\}$. Suppose $H\ne\{0\}$ is a subgroup. Then $H$ contains a positive element (because together with $z\ne0$ it contains $-z$ and one of them is positive). Let $k$ be the least positive element in $H$.

Then $k\mathbb{Z}\subseteq H$.

If $h\in H$, then $h=qk+r$, with $0\le r<k$. Then $r=h-qk\in H$; by minimality of $k$, we conclude $r=0$. Thus $h\in k\mathbb{Z}$.

Let $G$ be a cyclic group with generator $g$. Then the map $\varphi_g\colon n\mapsto g^n$ is a surjective homomorphism $\varphi_g\colon\mathbb{Z}\to G$.

The inverse image $K=\varphi^{-1}(H)$ of a subgroup $H$ of $G$ is a cyclic subgroup of $\mathbb{Z}$. Since $H=\varphi_g(K)$, we conclude that $H$ is cyclic as well.

Any subgroup of a cyclic subgroup is cyclic

2 Answers2