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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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    Please don't use math mode for emphasis or italics; use `*` or `_` in text. Also, try to avoid titles that are almost entirely made up of $\LaTeX$.2012-01-10
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    A silly counterexample is the direct product of the dihedral groups on 3, 5, 7, 11, 13, 17, and 19 points. Each of the dihedral groups is characteristic (having trivial center), so the automorphism group of the product is the product of the automorphism groups, each of order (p-1)*p. However, the automorphism of the abelianization is GL(7,2), which is over 20 times larger. In fact the image of Φ is trivial, which is pretty far from surjective.2012-01-10
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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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