Subgroup of finite abelian group of order m

Question

I am trying to prove the below statement:

Let G be an abelian group of order m. If n divides m, show that G has a subgroup of order n.

I think the classification theorem for finite abelian groups would be helpful, but I'm not sure how. Not looking for an answer in particular, but would appreciate a hint.

Thanks.

Answer 1

Proceed in three steps.

1. Show the result for cyclic groups.

2. Show that if $m \mid n_1 \dots n_r$ then $m=m_1 \dots m_r$ with $m_i \mid n_i$ for each $i$.

3. Use the structure theorem for finite abelian groups to tie things together.

Answer 2

Without (directly) using the structure theorem for f.g. abelian groups, you can proceed as follows. Suppose the result holds for all $m'| m$, and choose some prime $p$ dividing $n > 1$. By Cauchy's theorem, there exists an element $\gamma\in G$ of order $p$. Put $N = \langle{\gamma}\rangle\subset G$ and $Q = G/N$. Then $Q$ has order $m/p$ and thus contains a subgroup $H\subset Q$ of order $n/p$. The subgroup $HN\subset G$ is then well-defined and has order $n$. The result follows by induction.