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Let $G$ be a non-Abelian group and let $n$ be a (large) positive integer (eventually going to infinity). For simplicity, let's take $G=\mathbb{D}_6$ (Dihedral group of order $6$). Is there a way to find (all or a lot of) subgroups of $G^n$? One way is to take images of homomorphisms into $G^n$ or kernels of homomorphisms from $G^n$ but neither of these gives you all of the subgroups. Another way is to use Goursat's lemma sequentially for $i=2,3,\cdots,n$ but since $n$ is large this is not feasible. Any other ideas?

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This might be a difficult problem. The number of subgroups for the simplest nontrivial (abelian) example $G=C_2$ are tabulated at oeis.org/A006116 and begin $2,5,16,67,374,2825,29212,417199\dots$. – Gerry Myerson Dec 4 '12 at 23:24
...which brings up the question, are you interested in counting the subgroups or just knowing which isomorphism types occur? – anon Dec 4 '12 at 23:26
I'm interested in counting them. My motivation is coding theory in which a subgroup of $G^n$ is regarded as a code. The idea is to find a "big" collection of subgroups of $G^n$ and somehow analyze the average "coding performance" of the collection. – Jim Dec 4 '12 at 23:31
Thanks Gerry, that's a very interesting post. – Jim Dec 4 '12 at 23:31
The number is going to blow up very quickly. For every prime divisor $p$ of $|G|$ you have a subgroup of order $p$ in $G$, so $G^n$ contains at least one copy of the elementary abelian group $(C_p)^n$, which by @Gerry already gives rise to a ton of subgroups. – Alexander Gruber♦ Dec 5 '12 at 6:47

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