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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 :)

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    I'm sorry, to be honest, I still don't know what to do with that. Maybe I also get wrong how the elements look like; The Elements in $[G,G]\cap Z(G)$ are like $aba^{-1}b^{-1}(=e)$ right? I really don't know where to go with your info. I mean I know now that $\forall g\in G:$ $g[a,b]=[a,b]g$ but that does not mean we don't need it for the group generation, right? Sorry, I am a little blocked :-\ But thanks for the answer so far :)2012-11-05

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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.

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    If a maximal subgroup of a group is normal then the quotient has no non-trivial subgroups and thus it *must* be a (finite, of course) group of prime order. Of course it follows at once that the quotient is abelian, too.2012-11-05