# Subgroup of finite abelian group of order m

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.

• Do you know the classification of subgroups of cyclic groups? – Joshua Mundinger Nov 12 '14 at 16:26
• See this. – user 170039 Nov 12 '14 at 16:28
• @Alqatrkapa what do you mean? – jstnchng Nov 12 '14 at 16:35

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.

• quid, I really appreciate the hints. For #2), aren't n and m supposed to be reversed? Assuming m = m1...mr with m = |G|, n|m1...mr implies n|m (as given). – jstnchng Nov 12 '14 at 16:58
• I've also shown the first two parts to be true. Could you shed some light on how I should use the structure theorem? Thanks. – jstnchng Nov 12 '14 at 17:03
• Unfortunately I switched the characters relative to your usage. Yes $n \mid m_1 \dots m_r = m$ implies $n=n_1 \dots n_r$ with $n_i \mid m_i$ for each $i$. If you have some $n \mid m$ write $m= m_1 \dots m_r$ where the $m_i$ are the order in the decomposition of the group into cyclic subgroups you get from structuretheorem. Use 2. Then by 1 you get a subgroup of order $n_i$ of the cyclic group of order $m_i$. Take these cyclic subgroups together to get the subgroup of order $n$. – quid Nov 12 '14 at 17:10
• Yes, essentially. (I would rather call it the direct sum, but this is a detail you might well ignore.) – quid Nov 12 '14 at 17:27
• This is the detail I meant. It depends how you have learned the structure theorem. I will assume it is in the form G is isomorphic to $Z/m_1Z \times \dots \times Z/m_rZ$. Now for each of the cyclic subgroups you have a cyclic subgroup of oder $n_i$. The group generated by their union is a direct product of cyclic group of order $n_i$ and the preimage under the isomorphism is a subgroup of $G$. – quid Nov 12 '14 at 17:39

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.

• This is more elegant proof than mine. – quid Nov 12 '14 at 17:11