Let $G$ be a $p$-group for an odd prime $p$. The Frattini subgorup $\Phi(G)$ is defined as the intersection of all maximal subgroups or, equivalently, as the subgroup of non-generating elements, i.e. the elements $x \in G$ such that for every $R \subseteq G$ with $\langle x,R\rangle=G$ we must have that $G=\langle R\rangle$. The Agemo subgroups $\mho_k(G)$ is the subgroup generated by all $p$-th powers, namely $\mho_k(G)=\langle g^{p^k}|g\in G\rangle$. Finally I call $G'$ the derived subgroup of $G$. My question is the following:

When is true that $\Phi(G)=G'\mho_1(G)$?

In general is true that $G'\mho_1(G) \le \Phi(G)$.

I found that equation reading the proof of Hall's Theorem for $p$-groups on Huppert's book "Endliche Gruppen I", at page 358. There $G$ is non cyclic with every abelian characteristic subgroup being cyclic. As it shown in the proof, this hypothesis implies that

  1. $\Phi(G)$ is cyclic
  2. $G/\Omega_1(G)$ is cyclic
  3. $G$ is an extraspecial (and then regular?) $p$-group

I don't find out why that equation seems to be kind of general.

  • 2
    $\begingroup$ $\Phi(G) = G'G^p$ is true for all finite $p$-groups. $\endgroup$ – Derek Holt Nov 11 '18 at 15:48
  • $\begingroup$ I considered $V=G/G'\mho_p(G)$, that is a vector space. Each iperplane of $V$ correspond to a maximal subgroup, and their intersection, namely $G'\mho_p(G)$, contains the intersection of ALL maximal subgroup, that is $\Phi(G)$. Since the other inclusion is trivial we have the equality. Is this correct? $\endgroup$ – Lorban Nov 11 '18 at 18:40
  • $\begingroup$ Yes that sounds right! $\endgroup$ – Derek Holt Nov 11 '18 at 19:21
  • $\begingroup$ I don't know why some times I tale so long to undestand this not-so-hard things. Thank you!!! $\endgroup$ – Lorban Nov 11 '18 at 19:49
  • 1
    $\begingroup$ Your definition of "non-generators" is incorrect. $\Phi(G)$ is the group of all $x$ such that if $\langle R,x\rangle = G$, then $\langle R\rangle=G$; but what you wrote would make the Frattini subgroup equal to just $\{e\}$, for letting $R=\{e\}$ your condition $\langle x,R\rangle=\langle R\rangle$ would require $x=e$. $\endgroup$ – Arturo Magidin Nov 12 '18 at 1:37

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