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The question is about an exercise I found in Robinson's book A Course in the Theory of Groups. It says that, given a presentation $\pi$ from a free group $F$ to a group $G$ and said $R=$ker$\pi$, one can find a canonical homomorphism between the subgroup of all the automorphisms of $F$ fixing $R$, say $A$, and $Aut(G)$. Once such a homomorphism is found, the requirement is to find an outer automorphism of the alternating group $A_4$.

Now, if I'm doing well, the canonical automorphism should be the following: $$\alpha\in A\longrightarrow\overline{\alpha}\in Aut(F/R)$$ where $$\overline{\alpha}:xR\in F/R\longrightarrow x^{\alpha}R\in F/R.$$

Then I consider $A_4=\langle x,y|x^2=y^3=(xy)^3=1\rangle$, so $R=\langle x^2,y^3,(xy)^3\rangle^{F_2}$, and $\alpha\in Aut(F_2)$ such that $x^\alpha=x^{-1}$ and $y^\alpha=xy$. It clearly fixes $R$ and so should induce an automorphism on $A_4$. But it happens that, after reifying $x$ with $(12)(34)$ and $y$ with $(123)$, we easily have that $\alpha$ fixes every element of order $2$ in $A_4$, which is impossible. There must be a shameful mistake somewhere, but I cannot find it!

(Moreover, since we know that there is an outer automorphism of $A_4$ which fixes $(12)(34)$ and inverts $(123)$, i.e. $(12)$, I would be tempted to choose $\alpha$ such that $x^\alpha=x$ and $y^\alpha=y^{-1}$, but 1) that would be cheating, avoiding solving my mistake, and 2) I don't know if $\alpha$ fixes $R$, though I am pretty confident it does).

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  • $\begingroup$ As far as I can see, you have made no mistake, but $\alpha$ induces an inner automorphism of $A_4$. Why do you think that you have made a mistake? If you are required to find an outer automorphism of $A_4$ then you must use a different automorphism of $F$. $\endgroup$ – Derek Holt Jul 23 '17 at 18:39
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    $\begingroup$ Since, for example, $(123)^\alpha=(12)(34)(123)=(134)=(123)^{(234)}$ and $(12)(34)^\alpha=(12)(34)$, I cannot guess which element of $A_4$ it corresponds to (provided, clearly, that $A_4\simeq Inn(A_4)$)! $\endgroup$ – Alex Doe Jul 23 '17 at 19:47
  • $\begingroup$ I take it you have functions and permutations act from the right? $\endgroup$ – arctic tern Jul 23 '17 at 20:27
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    $\begingroup$ The automorphism of $A_4$ induced by $\alpha$ is conjugation by $(1,3)(2,4)$. $\endgroup$ – Derek Holt Jul 23 '17 at 20:28
  • $\begingroup$ @arctictern: I use to do this way, you are right. $\endgroup$ – Alex Doe Jul 23 '17 at 20:45
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So, as stated in the comments, the solution was really at hand and mine was half a false problem.

If we take $\alpha\in Aut(F)$ such that $x^\alpha=x$ and $y^\alpha=xy$, it induces on $A_4$ an automorphism fixing every element of order $2$, which is possible if we consider any element of the Klein $4$-group in $A_4$ acting by conjugation. In this case $\alpha$ induces the inner automorphism induced by $(13)(24)$. That is to say that $\alpha$ is not a total mistake, but is not what we are looking for.

If we hence take $\beta\in Aut(F)$ such that $x^\beta=x^{-1}$ and $y^\beta=y^{-1}$, it fixes $R$ (in fact $(x^2)^\beta=x^2$, $(y^3)^\beta=y^{-3}$ and $((xy)^3)^\beta=x^{-1}y^{-1}x^{-1}y^{-1}x^{-1}y^{-1}=(yx)^{-3}=((xy)^{-3})^{x^{-1}}\in R$). In fact, $\beta$ induces on $A_4$ the conjugation by $(13)$ and so is what we were looking for. A little remark is that if we define a $\gamma$ such that $x^\beta=x$ and $y^\beta=y^{-1}$, it would induce exactly the same automorphism on $A_4$, provided that $x^2$ belongs to $R$, which makes $x$ and $x^{-1}$ essentially the same when talking about the presented group.

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