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I know the group of outer automorphisms of $F_2$ (the free group on two generators) is isomorphic to $\text{GL}_2(\mathbb{Z})$. Let $\text{Aut}^+(F_2), \text{Out}^+(F_2)$ denote the subgroups of $\text{Aut}(F_2), \text{Out}(F_2)$ with image in $\text{SL}_2(\mathbb{Z})$. Then we have the exact sequences

$$1\longrightarrow \text{Inn}(F_2)\longrightarrow \text{Aut}(F_2)\longrightarrow\text{Out}(F_2)\longrightarrow 1$$ and $$1\longrightarrow \text{Inn}(F_2)\longrightarrow \text{Aut}^+(F_2)\longrightarrow\text{Out}^+(F_2)\longrightarrow 1$$

I think I have a topological proof that at least the second sequence is split. Ie, there exists a section $\text{Out}^+(F_2)\rightarrow \text{Aut}^+(F_2)$.

Can someone with more background in group theory than me provide a group theoretic answer? (ie, whether or not the two sequences are split).

thanks

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  • $\begingroup$ What is $F_2$? It's not clear. $\endgroup$ – Patrick Da Silva Mar 11 '14 at 21:31
  • $\begingroup$ Is it a free (abelian) group with two generators ? $\endgroup$ – mercio Mar 11 '14 at 21:32
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    $\begingroup$ $F_2$ is the free group on two generators $\endgroup$ – oxeimon Mar 11 '14 at 21:42
  • $\begingroup$ It is not unusual for group theoretic statements to have hard group theoretical proofs but very easy topological proofs. (E.g. a sub group of a free group is again free) Maybe you want to post your proof and we can check whether it is correct. $\endgroup$ – Peter Patzt Mar 11 '14 at 21:45
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    $\begingroup$ In principal you could prove it by writing down appropriate automorphisms $\alpha,\beta$ of $F_2$ such that $\alpha^4=\beta^3=1$ and $\alpha^2$ commutes with $\beta$. The obvious choice for $\alpha$ (with $F_2 = \langle x,y \rangle$) is $(x,y) \mapsto (y,x^{-1})$ so $\alpha^2$ is $(x,y) \mapsto (x^{-1},y^{-1})$. I tried $\beta:(x,y) \mapsto(x^{-1}y,x^{-1})$, which has order $3$, but that does not commute with $\alpha^2$. $\endgroup$ – Derek Holt Mar 12 '14 at 10:00
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Neither sequence splits. This is because $F_2$ has no automorphisms of order six but $\operatorname{SL}_2(\mathbb{Z})\cong C_4\ast_{C_2}C_6$ clearly does. For a reference that $\operatorname{Aut}(GF_2)$ has no element of order six see the paper Classification of automorphisms of free group of rank $2$ by ranks of fixed point subgroups by Oleg Bogopolski, J. Group Theory (2000).

(Note: Bogopolski's proof uses train track maps, if my memory serves me correctly. So his proofs are geometric. For an algebraic proof, I think you could use an outer automorphism $\widehat{\phi}$ of order six with representative $\phi$. If there is an automorphism of order $6$ then there exists a word $W(x, y)$ such that $(x, y)\mapsto (W(\phi(x))W^{-1}, W(\phi(y))W^{-1})$ has order six. This means that there exists some word $U$ such that $W=U\phi(U)\phi^2(U)\phi^3(U)\phi^4(U)\phi^5(U)$, and so I would induct on the length of a word $V$ to see how $\phi(V)$ can begin and end (if $V$ begins with $x$ then $\phi(V)$ begins with...etc). Then plug this together and work out if $W$ cannot be formed. This is a very messy approach though...I think, perhaps, the best way would be to think about fixed points (which is what Bogopolski is doing). Some automorphism of order six must fix the word $U\phi(U)\phi^2(U)\phi^3(U)\phi^4(U)\phi^5(U)$, so you have reduced the proof to thinking about fixed points. So look at this old MO question...)

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