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If the Frattini subgroup is trivial,for a group P, then P is elementary abelian. Where, Frattini subgroup is the intersection of all maximal subgroups. Please prove the above statement.

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  • $\begingroup$ It's not true. Take a look at the symmetric group $S_3$, of order $6$. Perhaps you meant for $P$ to have prime power order? $\endgroup$ – James Mar 21 '17 at 20:35
  • $\begingroup$ Yes, please answer if you can. If the group is Prime power order $\endgroup$ – student Mar 21 '17 at 20:37
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    $\begingroup$ Isn't this supposed to be a problem for you to solve yourself? As a hint, when $P$ is a group of prime power order, all maximal subgroups $M$ are normal and have index $p$. So, for all $g,h \in P$ we have $g^p \in M$ and $[g,h] \in M$. This is true for all such $M$, so, for all $g,h \in G$, we have $g^p \in \Phi(G)$ and $[g,h] \in \Phi(G)$ and hence $G/\Phi(G)$ is an elementary abelian $p$-group. $\endgroup$ – Derek Holt Mar 21 '17 at 20:49
  • $\begingroup$ You should provide some indication of the things you've tried and where you ran into trouble so that someone can formulate an answer to guide you. $\endgroup$ – James Mar 21 '17 at 20:49
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Let $G$ be a $p$-group. Then every maximal subgroup is normal.

Notice that $G/\Phi(G)=G/\bigcap_{i=1}^{n} M_i$ can be embedded into $G/M_1\times G/M_2... G/M_n$. (Show this !)

Since the target group is elementary abelian (Why?), $G/\Phi(G)$ is elementary abelian.

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