This is a crosspost from this MSE question from a year ago.

Finite groups are cancellable from direct products, i.e. if $F$ is a finite group and $A\times F \cong B\times F$, then $A \cong B$. A proof can be found in this note by Hirshon. In the same note, it is shown that $\mathbb{Z}$ is not cancellable, but if we only allow $A$ and $B$ to be abelian, it is (see here).

I would like to know if there are any groups that can be cancelled from free products rather than direct products. That is:

Is there a non-trivial group $C$ such that $A*C \cong B*C$ implies $A \cong B$?

It is certainly not true that every group is cancellable in free products. For example, if $A$, $B$, $C$ are the free groups on one, two, and infinitely many generators respectively, then $A*C \cong C \cong B*C$ but $A\not\cong B$. Many non-examples can be constructed this way, but they are all infinitely generated.

As is discussed in the original MSE question, it follows from Grushko's decomposition theorem that if $A$, $B$, and $C$ are finitely generated, then $A*C \cong B*C$ implies $A \cong B$.

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    $\begingroup$ In the final sentence, did you mean to say "it follows from Grushko's decomposition theorem that the answer is yes", because the cancellativity even holds for every $C$? $\endgroup$ – Tobias Fritz Sep 5 '18 at 21:09
  • $\begingroup$ @TobiasFritz: Sorry, you're correct. I will edit. $\endgroup$ – Michael Albanese Sep 5 '18 at 21:12
  • $\begingroup$ (Say that $C$ is freely cancelable if it satisfies your property.) Actually, at the opposite I'm not aware of any finitely generated group, nor even any finite free product of freely indecomposable groups, that is not freely cancelable. Do you know any? $\endgroup$ – YCor Sep 5 '18 at 21:23
  • $\begingroup$ @YCor: I do not. All the non-freely decomposable groups I can think of are not of the type you mention, in particular, they are all infinitely generated. $\endgroup$ – Michael Albanese Sep 5 '18 at 21:28

For $C=\mathbb{Z}/2\mathbb{Z}$, it follows from the Kurosh subgroup theorem that $A\ast C \cong B\ast C$ implies that $A\cong B$. Denote $C_1 \cong C_2\cong \mathbb{Z}/2\mathbb{Z}$, and let $\varphi: A\ast C_1 \to B\ast C_2$ be an isomorphism.

Then $\varphi(C_1)\leq B\ast C_2$ is an injection. By Kurosh (and since $C$ is freely indecomposable), there is $C_3\leq B$ or $C_4\leq C_2$ and $g\in B\ast C_2$ such that $\varphi(C_1)=gC_3g^{-1}$ or $\varphi(C_1)=gC_4g^{-1}$. Replacing the isomorphism $\varphi$ with $g^{-1}\varphi g$, we may assume that $\varphi(C_1)=C_3$ or $C_4$. In the second case, we have $C_4\cong C_2$, and hence $\varphi(C_1)=C_2$. Taking the quotient by the normal closure of $C_1$ and $C_2$ on the left and right respectively, we see that $A\cong B$.

In the case $\varphi(C_1)=C_3< B$, $\varphi^{-1}(C_2) = hC_5h^{-1}$, $C_5\leq A, h\in A\ast C_1$ (since otherwise we would be in the other case again). Then we see that $A \cong A\ast C_1 / \langle\!\langle C_1\rangle\!\rangle \cong B \ast C_2 /\langle\!\langle C_3\rangle\!\rangle \cong (B/\langle\!\langle C_3\rangle\!\rangle)\ast C_2$ (we use $G/\langle\!\langle H\rangle\!\rangle$ to denote the quotient by the normal subgroup generated by $H$). Similarly, $B\cong (A/\langle\!\langle C_5\rangle\!\rangle)\ast C_1$. Let $A_1=B/\langle\!\langle C_3\rangle\!\rangle, B_1=A/\langle\!\langle C_5\rangle\!\rangle$. Then $A\cong A_1\ast C_2, B\cong B_1\ast C_1$, and $\varphi: A_1\ast C_2\ast C_1 \to B_1\ast C_1\ast C_2$, taking $C_i$ to a conjugate of $C_i$. Take the quotient by the normal closures of $C_1$ and $C_2$ on both sides, we see that $A_1\cong B_1$, so $A\cong B$.

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    $\begingroup$ This is really nice. It seems that you only use two properties of $\mathbb{Z}/2\mathbb{Z}$: it is freely indecomposable, and it is not isomorphic to a proper subgroup of itself. Is that correct? If so, then your argument applies to many groups, in particular, all finite groups. $\endgroup$ – Michael Albanese Sep 10 '18 at 16:42
  • $\begingroup$ @MichaelAlbanese: Yes, it should work for finite groups, and possibly all finitely generated groups (but it might take a bit more work to deal with the case with $\mathbb{Z}$ factors if one want to take advantage of Kurosh). $\endgroup$ – Ian Agol Sep 10 '18 at 17:39

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