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Suppose $G=\mathbb{Z}\ast C_n=\langle a, b; b^n\rangle$ is the free product of the infinite cyclic group with a finite cyclic group. Then $G$ has finite outer automorphism group. However, my proofs (and the proofs I have found in the literature*) of this fact use lots of Nielsen-esq automorphisms-of-free-groups theory (with a touch of Whitehead, if I recall correctly).

I was wondering if anyone knows of any short (two-liner), lower-level proofs?

*(Although I will admit I skipped the classic reference, which is a paper of Fouxe-Rabinovitch written in Russian with a MathSciNet entry in German. I speak neither language.)

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Are you willing to use the fact that all subgroups of order $n$ in $G$ are conjugate? If so, then we can assume that our automorphism fixes $b$. Then it is enough to prove that it must map $a$ to $b^iab^j$ for some $i,j$, and there are only finitely many such automorphisms. So it is enough to prove that, if $G = \langle g,b \rangle$, then $g = b^iab^j$. If not, then we can assume that the normal form of $g$ has the form $awa$ or $awa^{-1}$ for some nontrivial word $w$ containing $b^{\pm 1}$, and it is not hard to see you cannot generate $a$ from such a $g$ and $b$.

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  • $\begingroup$ Thanks, yes, I hadn't thought to use the normal form theorem (actually, you could translate this proof into Bass-Serre theory, because, as your observation about only having one subgroup points out, all automorphisms fix the graph of groups structure (if you view your group as an HNN-extension with trivial associated subgroups - is that allowed?)). Very nice. Although someone should say that this is only an index $\phi(n)$ subgroup of $\operatorname{Out}(G)$, where $\varphi$ is the Euler totient function. (I am sure you know this, but I am also sure that someone reading this will not.) $\endgroup$ – user1729 Sep 13 '13 at 17:31
  • $\begingroup$ Since $b^iab^j$ is conjugate to $b^ia$ by a power of $b$, representatives of the outer automorphisms seem to be $a \mapsto b^ia^{\pm 1}$, $b \mapsto b^j$ with $(j,n)=1$, which would make $|{\rm Out}(G)|= 2n\varphi(n)$ if I've got that right. $\endgroup$ – Derek Holt Sep 14 '13 at 8:46
  • $\begingroup$ Yeah, it is actually isomorphic to $D_n\rtimes\operatorname{Aut}(C_n)$. $\endgroup$ – user1729 Sep 14 '13 at 10:13
  • $\begingroup$ @DerekHolt Excuse me for asking so late. Could you provide some reference for the fact that all subgroups of order $n$ in $G$ are conjugate. Thanks in advance! $\endgroup$ – 1123581321 Oct 9 '20 at 12:49
  • $\begingroup$ @ΓιάννηςΠαπαβασιλείου Well, every finite subgroup of a free product is conjugate into one of the free factors (which is standard). As $n<\infty$ is the order of one of the free factors, the claim follows. (In fact, something stronger holds: if $\mathbb{Z\ast Z}_n$ contains a subgroup of order $m$ then all subgroups of order $m$ are conjugate. The proof is identical, except at the last step you observe that cyclic groups contain precisely one subgroup of each order.) $\endgroup$ – user1729 Oct 9 '20 at 13:12

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