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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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The thing to do is likely to show elements of $[G,G]\cap Z(G)$ are non-generators of $G$, i.e. can be removed from any generating set. They're the commutators fixed by every conjugation, $aba^{-1}b^{-1}$ such that $caba^{-1}b^{-1}c^{-1}=aba^{-1}b^{-1}$ for every $c\in G$, for what it's worth. – Kevin Carlson Nov 5 '12 at 2:04
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 :) – Nadine Nov 5 '12 at 2:22
up vote 7 down vote accepted

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


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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Does that also work for infinite $G$? Must there exist a maximal $M\leq G$? Thanks for the answer! :) – Nadine Nov 5 '12 at 3:14
@Nadine When there is not a maximal $M\leq G$, $\Phi(G)$ is conventially selected to be $G$. The problem statement is trivial true in this case. – peoplepower Nov 5 '12 at 3:15
True -.- thanks a lot! :) – Nadine Nov 5 '12 at 3:16
Another question. How do we know, that $|G/M|$ has prime order? Is that known or shell I prove this? Can I also show that $G/M$ is abelian with the isomorphism theorem? That $G/M \cong A$ and with A abelian then also $G/M$? – Nadine Nov 5 '12 at 4:07
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. – DonAntonio Nov 5 '12 at 4:16

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