# Relationship between Frattini subgroup and the first Agemo subgroup of a prime-power order group

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.

• $\Phi(G) = G'G^p$ is true for all finite $p$-groups. – Derek Holt Nov 11 '18 at 15:48
• 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? – Lorban Nov 11 '18 at 18:40
• Yes that sounds right! – Derek Holt Nov 11 '18 at 19:21
• I don't know why some times I tale so long to undestand this not-so-hard things. Thank you!!! – Lorban Nov 11 '18 at 19:49
• 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 $x=e$. – Arturo Magidin Nov 12 '18 at 1:37