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My question is the following: Let $G:=F_n$ If we look at the commutator subgroup $[G,G]$ of $G$, we get the canonical epimorphism

$\varphi: G\to G/[G,G]$

Since $[G,G]$ is characteristic in $G$, we know that $Aut(G)$ acts in a natural way on the factor group $G/[G,G]$ and we get a map:

$\Phi:\mathrm{Aut}(G)\to \mathrm{Aut} (G/[G,G]);\alpha(g) \mapsto \bar{\alpha}(g*[G,G]):=\alpha(g)*[G,G]$

But how can I show that $\Phi$ is an epimorphism?

Added.

When I was asking the question we were in the general case, where $G$ is an arbitrary group. Because of some answers, I edited the question into the case, where $G=F_n$, the free group of rank $n$.

Thanks to the last comment, I now know that there is a solution in the book "Combinatorial Group Theory" from Magnus. I don't have the book beside me. So does someone knows a proof for the existence of the epimorphism $\Phi$. I think, if we assume that $Aut(F_n)$ is generated by the right nielsen transformations, we only have to show that $Aut(F_n/[F_n,F_n])$ is generated by these trasnformations, since we know that every $\alpha\in Aut(F_n)$ induces an $\bar{\alpha}\in Aut(F_n/[F_n,F_n])$. Is this true? And how can we get this?

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    $F_n/[F_n,F_n]\cong \mathbb{Z}^n$, which has automorphism group $GL(n,\mathbb{Z})$. You need to show the image of the Nielsen transformations generates $GL(n,\mathbb{Z})$, which I think is not that difficult. In particular, transvections (along with an inversion to "get out" of $SL$) should be enough.2012-01-10

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It (being $\Phi$) isn't always surjective, and so isn't always an epimorphism. I cannot think of a non-complicated example...so I'll go for a vaguely complicated one! (I cite a paper, so it counts as complicated.)

If you take a Baumslag-Solitar group $BS(n, n)=\langle a, b;ab^na^{-1}=b^n\rangle$ with $n>1$ then it has abelianisation $\mathbb{Z}^2$, which has automorphism group $GL(2, \mathbb{Z})$. However, $BS(n, n)$ has outer automorphism group $D_{\infty}\times C_2$ (this is a result of Gilbert, Howie, Metaftsis and Raptis, and can be found in their paper "Tree actions of automorphism groups"), which is virtually cyclic. It is an easy exercise to prove that homomorphic images of virtually-cyclic groups are virtually cyclic, and it is well-known that $GL(2, \mathbb{Z})$ is not virtually cyclic.

So, noting that $\Phi$ must factor through the inner automorphism group if it is induced by the abelianisation map, this provides a counter-example.

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    Thanks. I edited the question.2012-01-10