0

If $G$ is a cyclic group that is generated by $g\in G$ so that $G= <g>$, how would I show that for any subgroup $H \subseteq G$ then $H$ must be cyclic?

I've already attempted to start this proof:

Let $G$ be a cyclic group and $H$ any subgroup of $G$. I claim that for some $m \in \mathbb{Z}$ we have $H=<g^m>$.

Case 1: If $H=\left \{ 1 \right \}$ then $H=<g^0>$ and so H is cyclic.

Case 2: If $H=G$ then $H=G=<g^1>$ and so H is cyclic,

Case 3: $H$ is a proper, non trivial subgroup. Then $\exists \ k\in H $ such that $k\neq 1 $ and since $k\in G \implies k=g^m$ for some $m \in Z$ and so $g^m \in H$. It then follows by closure under the binary operation on $H$ that $\forall \ n\in \mathbb{Z}, (g^m)^n \in H \implies <g^m> \subseteq H$

My question is how do I show that $H \subseteq <g^m>$ in order to show that they are equal?

BigWig
  • 309

1 Answers1

0

Let's show the opposite direction. Pick a minimal $m$ so that $g^m \in H$. Suppose $k \in H$, we want to show $k \in \langle g^m \rangle$. The idea is that $k=g^n$ for $n>m$. Write $n=mq+r$, where $r< m$. Then $k=g^n=g^{mq}g^r$. Can you see why $r$ must be zero now and finish the argument?

Sheel Stueber
  • 924
  • would it need to be thar r=0 to avoid a contradiction so that means k is an element of the cyclic group and since it was chosen arbitrarily it means that H is contained in the cyclic group and so they are equal? – BigWig May 25 '18 at 17:52
  • yes exactly, $r=0$ because it is less than $m$ and $m$ is minimal. – Sheel Stueber May 25 '18 at 17:54