# Subgroups of Abelian Groups, Theorem of Finite Abelian Groups

If the order of a group G is divisible by $p^n$ for some prime p and natural number n, then prove G has a subgroup of order $p^n$.

My try has been by induction on the exponent of divisor $p^n$. I use Cauchy's Theorem for Abelian case which basically is a corollary of Theorem of Finite Abelian Groups. Note that I want to prove this without Sylow's theorems.

$n = 1$ it is clearly true, as Cauchy's Theorem for Abelian Groups says that if order of G is divisible by p, then it contains an element a of order p which clearly generates a cyclic group $<a>$ of order p. Inductive assumption will then of course be that if order of G is divisible by $p^{n-1}$, then G contains a subgroup of this order.

Assume order of G is divisible by $p^n$. Then it certainly also is divisible by $p^{n-1}$, so it contains a nontrivial subgroup of order $p^{n-1}$. Call this subgroup for S. Then, since G is abelian, quotient group $G/S$ certainly exists, and we have relation of orders of groups: $|G| = |G/S||S|$ Since order of G is divisible by $p^n$, then it follows from the relation that p divides $|G/S|$ and by Cauchy's Theorem, G/S contains an element of order p. Could I somehow 'construct' a group of order $p^n$ by direct product of the S and the group generated by element of order P in G/S?

I'm pretty much stuck. Is there a way to go on, or have I hit a wall? If somebody knows, could you point me to a place on the internet where this or similar proof has been written out?

• Are you allowed to use the classification of finite abelian groups? Aug 16, 2011 at 14:44
• Look at the correspondence theorem between groups and their quotient groups.
– Joe
Aug 16, 2011 at 14:51
• By the classification, $G=A\times B$ where $A$ is a $p$-group and the order of $B$ is prime to $p$, and $A=A_1\times\cdots\times A_n$ where each $A_i$ is a cyclic $p$-group. The order of $A$ is $p^k$ with $k\ge n$. Take $H=A_1\times\cdots\times A_i$ such that the order $p^ j$, say, of $H$ divides $p^n$, and $i$ is maximal for this property. Then $A_{i+1}$ has a subgroup $K$ of order $p^ {n-j}$, and $H\times K$ does the job. Aug 16, 2011 at 15:22
• Variation: $A_i$ is cyclic of order $p^ {f(i)}$. The sum of the $f(i)$ is $\ge n$. For all $i$ choose $0\le g(i)\le f(i)$ such that the sum of the $g(i)$ is $n$. Each $A_i$ has a subgroup of order $p^ {g(i)}$. Take the product of these subgroups. Aug 16, 2011 at 15:37
• If it looks complicated, it means that I didn’t express myself well. A way to think about the argument is to start with particular cases. Case 1: $G$ is a cyclic $p$-group. Case 2: $G$ is a product of cyclic $p$-groups. Case 3 (general case): $G$ is the product of a product of cyclic $p$-groups by a group of order prime to $p$ (this last factor won’t count). [About the notification system: sometimes it works even without the @, but not always. Also it suffices to write @x where x is the first part of the user's name (for instance, for me, you can write @Pierre-Yves, or even @Pierre).] Aug 16, 2011 at 15:51

You are almost there. So you have a surjective homomorphism of abelian groups $G\longrightarrow G/S$, which simply sends $g$ to the coset $gS$. The group $G/S$ has a subgroup or order $p$. What can you say about the pre-image of this subgroup under the above homomorphism?
• I've tried to follow up on your thought. Since $G/S$ contains a subgroup of order $p$, this subgroup is cyclic and generated by some element $a$. Then $(Sa)^p = Sa^p = S$, so we see that $a^p \in S$. Also, $a$'s order in G is a multiple of p. Actually this is true for any member $a^d$ of subgroup generated by $a$, as for example if $d = 2$, $(Sa^2)^p = Sa^{2p} = S$. Am I on the right track, is there some fact I'm missing? Aug 16, 2011 at 15:58
• @Barre If $\overline{a}=aS$ generates a cyclic subgroup of $G/S$ of order $p$, then the preimage of that subgroup in $G$ is simply the disjoint union $$\bigsqcup_{i=0}^{p-1} a^iS.$$ What is its order? Show that it's a subgroup of $G$ and you are done. Aug 16, 2011 at 16:34