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Let $H$ be a subgroup of $G$. Does that imply, that $\Phi(H)\le \Phi(G)$? If not, then what properties $G$ must have for it to be true.

$\Phi$ stands for Fattini subgroup .

Chain Markov
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user148528
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    It is not true in general. For instance, a finite simple group $G$ satisfies $\Phi(G)=1$ (why?); but for any Sylow $p$-subgroup $H$ of $G$ which is not elementary abelian, $\Phi(H)\neq 1$. If $H$ is normal, the answer is affirmative in general (at least when $G$ is finite). – Yassine Guerboussa Aug 18 '15 at 12:24
  • I think G has to be non-abelian finite p-group but i'm not sure it is enough or right – user148528 Aug 18 '15 at 12:29
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    I think yes, every subgroup is subnormal, and we can use induction together with the positive answer for normal subgroups. Or directly use the fact that $\Phi(G)=G'G^p$. Try to think about more general classes. – Yassine Guerboussa Aug 18 '15 at 12:37

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