I asked this question a few days ago in MO but got no answer, so I try here. Any hint will be appreciated.

Let $$M(p^3)=\langle a, b\mid a^{p^2}=b^p=1, a^b=a^{p+1}\rangle$$ and $$G=\langle a, b\mid a^{p^n}=b^{p^m}=1, a^b=a^{p^{n-m}+1}\rangle.$$ Does there exist a normal subgroup $N$ of $G$ such that $G/N\cong M(p^3)$? ($p$ is an odd prime and $n>m\geq 1$).

  • $\begingroup$ A good reference. $\endgroup$ – mrs Sep 24 '16 at 6:12
  • $\begingroup$ I haven't worked it out in detail, but I think the answer is yes if and only if $n-m=1$. If $n-m > 1$ then all quotients of order $p^3$ are abelian. $\endgroup$ – Derek Holt Sep 24 '16 at 8:51

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