Let $p$ be a prime number. For which finite $p$-groups $H$ is there a finite $p$-group $G$ such that $[G,G] \cong H$?
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$\begingroup$ Could you provide some motivation for the question? $\endgroup$– Giuliano BiancoOct 22, 2014 at 8:22
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$\begingroup$ @GiulianoBianco I read Lemma 4.4 in isibang.ac.in/~manish/Publication/fg08.pdf and thought whether a generalization is possible. I also feel that this can be relevant for pro-$p$ groups but nothing too concrete is on my mind. $\endgroup$– PabloOct 22, 2014 at 8:29
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1$\begingroup$ It seems optimistic to hope for a general answer to such a broad question! $\endgroup$– Derek HoltOct 22, 2014 at 8:31
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$\begingroup$ @DerekHolt: well maybe. I at least hope to see some $H$ which does not appear as a commutator if this is possible. On the other side, I would like to know of some big family of $p$-groups appearing as commutators. $\endgroup$– PabloOct 22, 2014 at 8:33
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2$\begingroup$ The analogous question exists for nilpotent Lie algebras, where we have some results, e.g., see here. In particular, a filiform nilpotent Lie algebra is not of the form $[L,L]$ if and only if it is a CNLA. $\endgroup$– Dietrich BurdeOct 22, 2014 at 8:37
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2 Answers
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There are some results for special cases. Burnside has proved in $1912$ that, if $G$ is a non-metabelian $p$-group, then the centre of the derived group of $G$ cannot be cyclic. In particular, a non-Abelian group of order $p^3$ cannot be the derived group of a $p$-group. Blackburn later described the 2-generator groups that arise as commutator subgroups of $2$-generator $p$-groups, see here.
answered Oct 22, 2014 at 9:27
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(Essentially, Burnside) If $H$ is a $p$-group containing a nonabelian characteristic subgroup with cyclic center, then there is no $p$-group $G$ such that $H$ is a $G$-invariant subgroup of $\Phi(G)$. In particular, $H\ne G'$, $H\ne \Phi(G)$. Next, if a two-generator $H$ is $G$-invariant subgroup of $\Phi(G)$, then $H$ is metacyclic.
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$\begingroup$ If H is a two-generator p-group and $H\cong\Phi(G)$ for some p-group G, then H is metacyclic (known result). $\endgroup$– YakovDec 26, 2015 at 12:30