TITLE: Does the Frattini subgroup $\Phi(G)$ contain the intersection $Z(G)\cap [G,G]$.
QUESTION [6 upvotes]: I want to show for any group $G$ that $[G,G]\cap Z(G)\subseteq \Phi(G)$.
But I don't really know why that works.
I looked at the definition of the different groups:
$[G,G]=\langle[a,b] | a,b\in G\rangle$, $[a,b]=aba^{-1}b^{-1}$.
So when the elements in the intersection are the $a,b\in G$ s.t $[a,b]=e$.
The thing is that all the usefull Lemma's & co. only for finite $G$ are, and I don't know how to show that the intersection must lay in $\Phi(G)$.
I hope someone is willing to give some hints :)

REPLY [9 votes]: This is a theorem by Gaschütz: 
if $\,A:=[G:G]\cap Z(G)\rlap{\;\,/}\subset \Phi(G)\,$ then there exists a maximal $\,M\leq G\,\,s.t.\,\,A\rlap{\;\,/}\subset M\,$ , and from here
$$G=MA\Longrightarrow \forall\,g\in G\,\,\exists\,m\in M\,\,,\,a\in A\,\,s.t.\,\,g=ma$$
and since $\,a\in Z(G)\,$ we get
$$g^{-1}Mg=a^{-1}m^{-1}Mma=m^{-1}Mm=M$$
so that $\,M\triangleleft G\Longrightarrow G/M\,$ has prime order and is thus abelian, but this means $\,[G:G]\leq M\,$ , and since $A\leq [G:G]\Longrightarrow A\leq M\,$, contradiction.