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I showed using a somewhat lengthy but elementary computation that the automorphism group $\operatorname{Aut}(D_{\infty})$ of the infinite dihedral group $D_{\infty} = \mathbb Z_2 * \mathbb Z_2$ splits as a semidirect product as follows: $$ \operatorname{Aut}(D_{\infty}) \cong \operatorname{Inn}(D_{\infty}) \rtimes \mathbb Z_2 $$ Here $\operatorname{Inn}(G)$ denotes the group of inner automorphisms of a group $G$. The $\mathbb Z_2$ factor in the above decomposition comes from the group homomorphism that interchanges the factors of $\mathbb Z_2 * \mathbb Z_2$. Since $D_{\infty}$ has trivial center, it follows that $\operatorname{Inn}(D_{\infty}) \cong D_{\infty}$. In particular we see that the group of outer automorphisms is given by $\operatorname{Out}(D_{\infty}) \cong \mathbb Z_2$.

What would be an approach to derive these results using some more theory (which might also be applicable to compute automorphism groups of more difficult groups)?

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    $\begingroup$ The proof given in section $1.4$ here is perhaps more interesting for you, and enables to compute the whole automorphism tower for $D_{\infty}$. $\endgroup$ – Dietrich Burde Nov 6 '14 at 13:52
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    $\begingroup$ There is a paper by M. Pettet, called The automorphism group of a graph product of groups, and a similar paper by Gilbert, Howie, Metaftsis and Raptis, Tree actions of automorphism groups, which you might find interesting. The group $D_{\infty}$ acts on a tree (the $\mathbb{Z}$-tree), and these papers look at groups whose automorphism groups induce an action on the tree. Unfortunately, these methods do not work here - swapping the generators messes up the tree. But the ideas are interesting! $\endgroup$ – user1729 Nov 6 '14 at 14:07
  • $\begingroup$ (Hmm...actually, I think the ideas do apply, because $C_2$ is finite so actually swapping the two copies of $C_2$ doesn't mess the tree up. (In Pettet's paper he talks about Property FA being sufficient for this to work, and finite $\Rightarrow$ Property FA...but I have a niggling feeling that I miss-read this passage in Pettet before and that there is something more going on. I need to think about it. But provisionally, Pettet's paper is applicable. Also: look at the diagram in Gilbert, Howie, Metaftsis and Raptis' paper. It is pretty, and explains it nicer-ly than Pettet.)) $\endgroup$ – user1729 Nov 6 '14 at 14:22

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