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Let $F_2 = \langle a, b \rangle$, the free group of rank $2$.

Is there an automorphism $\phi$ of $F_2$ such that $\phi(b) = b$, $\phi^2 = 1$ and $\phi(a) = wa^{-1}bw^{-1}$ for some $w \in F_2$?

First I checked the induced map on the abelianizations but it did not help. Then I expressed $w$ as a product of powers of $a$ and $b$. There seems to be too many cases to consider.

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This cannot happen. Consider the abelinisation. One route is to realise that taking $w=1$ is an automorphism, and thus you would need $w=b^i$ by Nielsen's old result, which says that if two automorphisms of $F_2$ act identically on the abelinisation then they are equal in $\operatorname{Out}(F_2)$ (note: this doesn't work in $F_n$ for $n>2$ because of Torelli). You can find Nielsen's Theorem in the book Combinatorial Group Theory by Magnus, Karrass and Solitar. The result can then be worked out as follows:

Take $w=b^i$. Then because $a\phi=b^ia^{-1}b^{1-i}$, $b\phi=b$ we have the following working. $$\begin{align*} a\phi^{2} &=(b\phi)^i(a\phi)^{-1}(b\phi)^{1-i}\\ &=b^i(b^ia^{-1}b^{1-i})^{-1}b^{1-i}\\ &=b^ib^{i-1}ab^{-i}b^{1-i}\\ &=b^{2i-1}ab^{1-2i}\\ \end{align*}$$ Therefore we require $i=1/2$, which cannot happen.

It is interesting to note that $\phi$ has order two on $\operatorname{Out}(F_2)$ but not in $\operatorname{Aut}(F_2)$. Indeed, every element of $\operatorname{Out}(F_2)\cong GL_2(\mathbb{Z})$ is contained in a subgroup isomorphic to $D_6$ or $D_4$, because $GL_2(\mathbb{Z})\cong D_6\ast_{D_2}D_4$ (the reference is B. Zimmermann, Finite groups of outer automorphisms of free groups, Glasgow Mathematical Journal, although I think there is a paper of Culler which does the same thing). Therefore, find these (unique-up-to-conjugacy) subgroups and you find all possible finite order elements of $\operatorname{Aut}(F_2)$. You can then test these to see which actually do have finite order in $\operatorname{Aut}(F_2)$. I did this once, but I have no idea where I wrote it down... It is a nice exercise though, and I would recommend you try it.

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    $\begingroup$ Thanks a lot for the answer, especially for the reference. I'll accept your answer after I fully understand the details. $\endgroup$
    – user33321
    Aug 19, 2013 at 19:10
  • $\begingroup$ The notation $a\phi$ which I assume stands for $\phi(a)$ seems strange to me. $\endgroup$ Aug 19, 2013 at 19:12
  • $\begingroup$ @AlexBecker some authors use it. It does seem a strange choice as the OP used the other, and much more common notation. $\endgroup$ Aug 19, 2013 at 19:13
  • $\begingroup$ @AlexBecker I prefer writing my maps on the right. Some text books do it, but I will admit that it is far, far from standard. I recently got a paper returned (not rejected ;-)) for using "non-standard" notation. Think it might have been for writing my maps on the right, but the editor wasn't saying anything... I prefer writing them on the right because I work with automorphism groups and it puts an emphasis on the maps not on the objects you are mapping. Really, it is just personal preference. I just wish everyone else would do what I do... $\endgroup$
    – user1729
    Aug 19, 2013 at 19:16
  • $\begingroup$ A similar discussion on the way I write my maps occurred here. $\endgroup$
    – user1729
    Aug 19, 2013 at 19:22

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