Frattini Subgroup is trivial hen the group is elementary abelian

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.

• 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? Mar 21, 2017 at 20:35
• Yes, please answer if you can. If the group is Prime power order Mar 21, 2017 at 20:37
• 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. Mar 21, 2017 at 20:49
• 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. Mar 21, 2017 at 20:49
• @Derek Holt..why would that imply that G is elementary abelian Jun 9, 2021 at 4:19

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.