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).

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.