# Is the minimum number of relations in a free product, the sum of the minimum number of relations in the free factors?

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).

• By the way it is known that the minimum number of generators is the sum of the minimum number of generators in the free factors ( Ch IV Cor 1.9 in Lyndon and Schupp for example) – Paul Plummer Aug 7 '15 at 4:28
• I would guess that the answer is yes, but I don't know a general proof. It's not hard to prove that $C_3*C_4$ cannot be a $1$-relator group. It is nonabelian and hence not cyclic, so any presentation has at least $2$ generators. A group defined by a presentation with fewer relations than generators has infinite abelianization, but the abelianization of $C_3*C_4$ is cyclic of order $12$. – Derek Holt Aug 7 '15 at 8:13
• @DerekHolt My first guess was that it should be true too. Thanks for the proof! It is interesting how much specific info is still used though. It looks like that this proof extends to finite groups $G$ where $\rho(G)=r(G)$ (the rank of $G$). By that corollary 1.9 we have $r(A*B)=r(A)+r(B) \geq \rho(A*B)$, but if it is strictly greater, it would have infinite abelianization, contradicting that the group has finite abelianization. I am pretty sure that most finite groups have $r(A)<\rho(A)$, so that is not terribly general. – Paul Plummer Aug 7 '15 at 15:16
• It looks like $(C_2 \times C_2) * (C_3 \times C_3)$ could be a counter example, depending on some things I am not sure on right now the their minimum number of relations, and using some results from ON THE NUMBER OF RELATIONS IN FREE PRODUCTS OF ABELIAN GROUPS by V. G. Bardakov and M. V. Neshchadim. – Paul Plummer Aug 8 '15 at 4:02
• I mistakenly attributed the "results" in my previous comment to ON THE NUMBER OF RELATIONS IN FREE PRODUCTS OF ABELIAN GROUPS. That is where I found the result mentioned, but it was actually in Presentation classes, 3-manifolds and free products by Cynthia Hog, Martin Lustig and Wolfgang Metzler, which appears in my answer to the question. – Paul Plummer Aug 9 '15 at 1:33

$\DeclareMathOperator{\def}{def}$ It turns out that there are groups $A*B$ with $\rho(A*B) < \rho(A)+\rho(B)$!
• $(C_2 \times C_2) * (C_3 \times C_3)$ has a 5 relation presentation $$\langle x,y,z,t \mid x^2,z^3,xyx^{-1}y^{-3}, ztz^{-1}t^{-4}, y^2t^{-3} \rangle$$ by theorem 3 in Presentation classes, 3-manifolds and free products by Cynthia Hog, Martin Lustig and Wolfgang Metzler.
• The Schur multiplier of a finite group $G$ gives information about the deficiency of the group, where the deficiency is the maximum value of generators minus relators over all presentation of the group which will be called $\def G$. In particular, the Schur multiplier can be generated by $-\def G$ generators. Both $C_2 \times C_2$ and $C_3 \times C_3$ have nontrivial Schur multiplier, so neither can have deficiency equal to $0$. That implies that they do not have two relator, two generator presentations.
• The groups $C_i \times C_i$ is not cyclic and can not have more generators than relations, since they are finite. So the fewest number of relations comes from the standard presentation $$C_i \times C_i\cong\langle x,y \mid x^i, y^i, xyx^{-1}y^{-1} \rangle.$$
This means that $$\rho {\big{(}}(C_2 \times C_2) *(C_3 \times C_3) {\big )}\leq 5 <6=\rho(C_2 \times C_2) + \rho(C_3 \times C_3).$$
• I probably won't accept this answer for a while, as I think there could be some stuff said about "what can we say about $\rho(A*B)$?", which is not exactly attacked in this answer (so far), and I am not familiar with the theory of the Shur multiplier. So I welcome more answers. – Paul Plummer Aug 9 '15 at 1:38