Let $G$ be a finite $p$-group and let $n > 0$. Let $G^n$ be the direct product of $n$ copies of $G$.

Are all subgroups of $G^n$ isomorphic to $H_1 \times \dotsm \times H_n$ for some subgroups $H_1, \dots, H_n$ of $G$?

Comments. The question is being isomorphic to a direct product of subgroups, and not being equal to a direct product of subgroups. A negative answer to a similar question for $G = S_3$ and $n = 2$ was given here. This question and its answer, which relies on Goursat's lemma might also be relevant.

P.S. I am especially interested in the case $p = 2$.

  • $\begingroup$ Have you run any computer checks for small values of $|G|$ and $n$? $\endgroup$ – the_fox Nov 22 '18 at 12:25
  • $\begingroup$ I tried a few examples on GAP without finding any counterexample so far. $\endgroup$ – J.-E. Pin Nov 22 '18 at 12:34
  • $\begingroup$ I doubt this is true, but I could be wrong. Include the code you have used to check this, if you want, and we can split the workload. (I was just about to start thinking how to write a piece of GAP code that would work, so it would save me some time.) $\endgroup$ – the_fox Nov 22 '18 at 12:41
  • $\begingroup$ I have replaced "Is any" in your question by "Are all". The word "any" is very often ambiguous. $\endgroup$ – Derek Holt Nov 22 '18 at 12:49
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    $\begingroup$ I think $D_8 \times D_8$ has a subdirect product of index 2 that does not decompose as a nontrivial direct product. (Here $D_8$ means dihedral group of order $8$.) $\endgroup$ – Derek Holt Nov 22 '18 at 12:55

Let $G=D_8 = \langle a,b \mid a^4=b^2=(ab)^2=1 \rangle$, $G^2 = \langle a_1,b_1\rangle \times \langle a_2,b_2 \rangle$ and $H = \langle a_1a_2,b_1,b_2,a_1^2,a_2^2 \rangle \le G$ with $|H|=32$.

If $H \cong H_1 \times H_2$, with $H_1$ and $H_2$ isomorphic to subgroups of $D_8$, then we must have $|H_1|=8$, $|H_2|=4$ (or vice versa), so $H_2$ is abelian, and hence $|Z(H_1 \times H_2)| =8$. But you can check that $Z(H)=Z(G) = \langle a_1^2,a_2^2 \rangle$ has order $4$.

In fact $H$ is indecomposable.


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