Let $G=A_1*A_2*...*A_n*F_k$ be a free product, where each $A_i$ is free abelian of different rank (rank at least 2), and $F_k$ is a free group. Consider an arbitrary automorphism $\phi$ in $Aut(G)$, can we prove that $\phi(A_i)$ is conjugate to $A_i$ for all $i$? How?
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2$\begingroup$ This is an application of the Kurosh subgroup theorem. en.wikipedia.org/wiki/Kurosh_subgroup_theorem $\endgroup$– Lee MosherCommented May 15, 2017 at 4:42
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$\begingroup$ Can it be more specific? From Kurosh subgroup theorem, one can write $\phi(A_i)$ in the form of free products, but how to prove that this free product is exactly conjugate to $A_i$? $\endgroup$– PumazsqfCommented May 15, 2017 at 6:45
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$\begingroup$ Since $\phi(A_i)$ is of rank at least 2, it doesn't split as a free product. Therefore it is conjugate into some $A_j$. Now apply the same observation to $\phi^{-1}$ to conclude. Note that the conclusion is not that $\phi(A_i)$ is conjugate to $A_i$, but that there is a permutation $\sigma$ of the indices so that $\phi(A_i)$ is conjugate to $A_{\sigma(i)}$. $\endgroup$– HJRWCommented May 15, 2017 at 8:55
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$\begingroup$ @HJRW The requirement that "each $A_i$ is free abelian of different rank" keeps $\phi(A_i)$ from being conjugate to $A_{\sigma(i)}$ unless $\sigma(i)=i$. $\endgroup$– Lee MosherCommented May 15, 2017 at 12:28
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$\begingroup$ @LeeMosher -- so it does, I missed that hypothesis. Hmm... this looks increasingly like a homework problem. $\endgroup$– HJRWCommented May 15, 2017 at 14:25
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