Say $\rho(G)$ is the minimum number of relations required to present the group $G$. Is $\rho(A*B)= \rho(A)+\rho(B)$? What can be said about $\rho(A*B)$?
A while ago I was thinking about $C_3*C_4$, thinking it was obvious that this group should not be a one-relator group. I am pretty sure at one point I came up with an argument but it seemed awfully complicated, and using specific information about one-relator groups for something that "feels" obvious.
Using the comments below as a guide, one can prove that if the abelianization of $A,B$ is finite and $\rho(A)=r(A),\rho(B)=r(B)$ where $r(G)$ is the rank of the group (minimum number of generators), then we can get that $\rho(A*B)=\rho(A)+\rho(B)$.
- It is known that $r(A*B) =r(A)+r(B)$, this can be found in Ch IV Cor 1.9 in Lyndon and Schupp.
- We also have that $\rho(A*B) \leq \rho(A)+\rho(B)=r(A)+r(B)$.
- If $\rho(A*B) < r(A*B)$, then a presentation witnessing that inequality can be shown to have an infinite abelianization, since the number of generators would be strictly greater than the number of relations (see this answer). This contradicts that the abelianization of $A*B$ is finite.
So finite free products of finite cyclic groups works and other groups too.
I could see a careful analysis the abelianization of f.g. groups giving more more information. And it does seem that a somewhat related concept, group deficiency, is related to some cohomology groups (I don't know any cohomology, but I would welcome an answer, even if it used cohomology).