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Let $p$ be an odd prime and $G$ be a finite non-abelian $p$-group of order $p^4$ with the following presentation: $$\langle a, b, c, d\mid a^p=b^p=c^p=d^p=1, c^d=cb, b^d=ba, [a,d]=[b,c]=[a,c]=[a,b]=1\rangle.$$

According to this presentation $b=[c,d]$ and $a=[b,d]$. Is it true that $a, b \in \Phi(G)$? ($\Phi(G)$ denotes the Frattini subgroup of $G$).

Thanks for any comment or answer!

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    $\begingroup$ The Frattini subgroup of a finite $p$-group contains the derived group, so yes. $\endgroup$ – Derek Holt Jan 3 '15 at 19:19
  • $\begingroup$ Thanks for your nice point. $\endgroup$ – sebastian Jan 3 '15 at 19:29

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