*This is a crosspost from [this MSE question](https://math.stackexchange.com/q/2389376/39599) from a year ago.*




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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](http://www.math.harvard.edu/~ctm/home/text/others/hirshon/cancel/cancel.pdf) 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](http://www.ams.org/journals/proc/1956-007-03/S0002-9939-1956-0078370-X/S0002-9939-1956-0078370-X.pdf)).


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 $A \cong B$.



Is there a non-trivial group $C$ such that $A*C \cong B*C$ implies $A \cong B$?
For $C=\mathbb{Z}/2\mathbb{Z}$, it follows from the [Kurosh subgroup theorem](https://en.wikipedia.org/wiki/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](https://en.wikipedia.org/wiki/Kurosh_subgroup_theorem#Statement_of_the_theorem) (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 $A\cong B$. 


