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Let $F_2=\langle X,Y\rangle$ be the free group of rank $2$ and consider $A,B,C\in Aut(F_2)$ given by: $$A(X,Y)\mapsto(YX^{-1}Y^{-1},Y^{-1})$$ $$B(X,Y)\mapsto(X^{-1},Y^{-1})$$ $$C(X,Y)\mapsto(X^{-1},XY^{-1}X^{-1})$$ The abelianisation of $F_2$ is isomorphic to $\mathbb{Z}^2$ whose automorphism group is $GL(2,\mathbb{Z})$. Thus we get an induced surjective map $m\colon Aut(F_2)\to GL(2,\mathbb{Z})$ which maps an element in $Aut(F_2)$ to the matrix associated to the corresponding element in $Aut(\mathbb{Z}^2)$. I would like to show that the subgroup $\langle A,B,C\rangle$ is normal in $Aut(F_2)$ by proving that $m(\langle A,B,C\rangle)$ is the centre of $GL(2,Z)$. So far I wasn't able to have any good idea... could you help me? Thanks, bye.

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Just calculate the matrices: m(A) = m(B) = m(C) = -1 –  Jack Schmidt Nov 22 '11 at 18:22
You may want to use \langle and \rangle instead of < and >; the spacing is correct for the former. –  Arturo Magidin Nov 22 '11 at 18:27
how did you calculate explicitly the matrices? $A$,$B$ and $C$ do not belong to $Aut(\mathbb{Z}^2)$ so I should compute their image in the quotient, first. Do I have to write how, say $A$, acts on the cosets $F_2'X$ and $F_2'Y$ in $F_2/F_2'$? P.S. I did use \langle and \rangle! :) –  fatoddsun Nov 22 '11 at 19:22
@fatoddsun: Since $[F_2,F_2]$ is fully invariant, any homomorphism $f\colon F_2\to F_2$ induces a homomorphism $f_{\rm ab}\colon \mathbb{Z}^2\to\mathbb{Z}^2$ by $f_{\rm ab}(\overline{X}) = \overline{f(X)}$ and $f_{\rm ab}(\overline{Y}) =\overline{f(Y)}$. So just "abelianize" the image. –  Arturo Magidin Nov 22 '11 at 20:24
@fatoddsun: Note, however, that having the image be normal is not sufficient; that will only show that the preimage under the abelianization map $\mathrm{Aut}(F_2)\to\mathrm{GL}(2,\mathbb{Z})$ is normal, and it's not immediately clear to me that this preimage is just $\langle A,B,C\rangle$. –  Arturo Magidin Nov 22 '11 at 20:57

2 Answers 2

up vote 3 down vote accepted

Arturo Magidin's excellent answer explains how to show that $m(\langle A,B,C\rangle)=\{\pm 1\}$, and also why this does not in general imply that your subgroup is normal. However, I want to explain why in this case it is true that $\langle A,B,C\rangle$ is the full preimage of $\{\pm 1\}$, and thus is normal.

The key is a theorem of Nielsen (1) on what is now called $IA_n$: this is by definition the kernel of the natural surjection $Aut(F_n)\to GL_n\mathbb{Z}$, so it fits into a short exact sequence

$$1\to IA_n\to Aut(F_n)\to GL_n\mathbb{Z}\to 1$$

This group of automorphisms which are the Identity on the Abelianization has been well-studied, starting with Magnus (2), who proved that $IA_n$ is finitely generated by giving an explicit set of generators: type (A) which for given $i$ and $j$ conjugates $x_i$ by $x_j$ and fixes all other generators, and type(B) which for given $i$, $j$, and $k$ multiplies $x_i$ by $x_jx_kx_j^{-1}x_k^{-1}$ and fixes all other generators.

But anyway, we are interested here in $IA_2$, and Nielsen proves in (1) that $IA_2$ is just $Inn(F_2)$, the group of automorphisms obtained by conjugation. This is isomorphic to $F_2$ and generated by $conj_X\colon X\mapsto X, Y\mapsto XYX^{-1}$ and by $conj_Y\colon X\mapsto YXY^{-1}, Y\mapsto Y$.

We know that your group $\langle A,B,C\rangle$ subjects to $\{\pm 1\}$, so to show it is the full preimage of $\{\pm 1\}$ it suffices to show that $\langle A,B,C\rangle$ contains the full preimage of $\{1\}$, which is $IA_2$. But we can compute by hand that $AB=conj_Y$ and $CB=conj_X$. So by Nielsen's theorem your group contains $IA_2$, and thus fits into a short exact sequence

$$1\to IA_2\to \langle A,B,C\rangle\to \{\pm 1\}\to 1$$

In particular, your subgroup is normal in $Aut(F_2)$.

Finally, here is another method of proving that $\langle A,B,C\rangle$ is normal which does not require quoting any papers written in German. We know that the elementary Nielsen transformations provide a generating set for $Aut(F_n)$. This generating set becomes particularly simple in the case when $n=2$: we have three generators $R\colon X\mapsto X^{-1}, Y\mapsto Y$, $S\colon X\mapsto Y, Y\mapsto X$, and $T\colon X\mapsto XY,Y\mapsto Y$. To show that $\langle A,B,C\rangle$ is normal in $Aut(F_2)$ means that it is invariant under conjugation under any element of $Aut(F_2)$. But since $R,S,T$ generate $Aut(F_2)$, it is enough just to show that $\langle A,B,C\rangle$ is invariant under conjugation by $R$, by $S$, and by $T$.

To do this for $R$, for example, you just need to compute the conjugates $RAR^{-1}$, $RBR^{-1}$, and $RCR^{-1}$ and show that each can be written as a combination of $A$, $B$, and $C$. In this case this is quite easy: we have $RAR^{-1}=A$, $RBR^{-1}=B$, and $RCR^{-1}=C^{-1}$. Now all you have to do is do the same for $SAS^{-1}$, $SBS^{-1}$, $SCS^{-1}$, $TAT^{-1}$, $TBT^{-1}$, and $TCT^{-1}$, and you've proved that $\langle A,B,C\rangle$ is normal!

(1) Nielsen, Die Isomorphismen der allgemeinen unendlichen Gruppe mit zwei Erzeugenden, Math. Ann. 78 (1964), 385-397

(2) Magnus, Über n-dimensinale Gittertransformationen, Acta Math. 64 (1935), 353-367.

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Nice; thanks for filling the gap! –  Arturo Magidin Nov 22 '11 at 21:39

If $G$ is a group, $N$ is a normal subgroup of $G$, and $f\colon G\to G$ is a homomorphism with $f(N)\subseteq N$, then $f$ induces a homomorphism $\overline{f}\colon G/N\to G/N$ by $\overline{f}(gN) = f(g)N$. To see this, note that $\widehat{f}\colon G\to G/N$ given by $\widehat{f}(g) = gN$ is a homomorphism, and since $f(N)\subseteq N$, then $N$ is contained in the kernel of $\widehat{f}$, so $\widehat{f}$ factors through $G/N$.

Since $[F_2,F_2]$ is fully invariant, this argument will work for $A$, $B$, and $C$.

For $A$, $x$ is mapped to $yx^{-1}y^{-1}$, which gets mapped to $\overline{x}^{-1}$ in $F_2^{\rm ab}$; $y$ is mapped to $\overline{y}^{-1}$. So the induced map $A_{\rm ab}\colon F_2^{\rm ab}\to F_{2}^{\rm ab}$ sends $(a,b)$ to $(-a,-b)$.

This is the same as what $B_{\rm ab}$ does; again the same thing as $C_{\rm ab}$ does. So the image of the subgroup generated by $A$, $B$, and $C$ in $\mathrm{GL}(2,\mathbb{Z})$ is the subgroup generated by their images, which is just the subgroup generated by $-I$.

Note, however, that the fact that $m(\langle A,B,C\rangle) = \{I,-I\}$ does not show, by itself, that $\langle A,B,C\rangle$ is normal in $\mathrm{Aut}(F_2)$; it shows that the pre-image is normal. The preimage is the subgroup of $\mathrm{Aut}(F_2)$ generated by $A$, $B$, $C$, and all automorphisms that induce the identity on $F_2^{\rm ab}$. It is was not clear to me whether this is just $\langle A,B,C\rangle$. For this, see Tom Church's great answer.

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Thank you very much guys! –  fatoddsun Nov 23 '11 at 18:23

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