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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 $\LaTeX$. –  Arturo Magidin Jan 10 '12 at 15:20
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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. –  Jack Schmidt Jan 10 '12 at 15:39
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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. –  user641 Jan 10 '12 at 16:09

1 Answer 1

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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$D_8$ has 8 automorphisms, while its abelianization $C_2\times C_2$ has 6... –  user641 Jan 10 '12 at 12:12
    
That is...somewhat simpler... –  user1729 Jan 10 '12 at 12:14
    
Nice. Thanks. I'm interested in the Situation where $G=F_n$ where $F_n$ is the free group of rank n. In my paper it is only mentioned, that there is an epimorphism $\Phi:Aut(F_n )\to Aut(F_n/[F_n,F_n])$. So I thougt that this epimorphism between the Automorphismgroups could be a general result for characteristic subgroups. But like you showed, it isn't. So why does this hold in the case where $G=F_n$? –  Peter Jan 10 '12 at 13:25
    
Do you perhaps want to edit your question to ask this? You can find an answer to this question in Section 3.3 of Magnus, Karrass and Solitar, "Combinatorial Group Theory" (Corollary 3.5.1). –  user1729 Jan 10 '12 at 13:35
    
Thanks. I edited the question. –  Peter Jan 10 '12 at 15:44

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