# Rank of a sum with free products

Let $$G$$ be a finitely generated group. Does there exist a constant $$\kappa$$ depending only on the rank of $$G$$ such that, if $$G \simeq F_1 \oplus \cdots \oplus F_n$$, then at most $$\kappa$$ factors are non-trivial free products?

Motivation. In order to optimize some of the results from my preprint, I would to like to say that a finitely generated group can be decomposed as a graph product over a finite graph with factors which do not split non-trivially as graph products. This cannot be done as stated because there exist finitely generated groups isomorphic to their own square, so I introduced the concept of graphically irreducible groups: a group $$G$$ is graphically irreducible if, for every graph $$\Gamma$$ and every collection of groups $$\mathcal{G}$$ indexed by $$V(\Gamma)$$ such that $$G$$ is isomorphic to the graph product $$\Gamma \mathcal{G}$$, the graph $$\Gamma$$ must be complete.

Expected result: A finitely generated group decomposes as a graph product over a finite graph with graphically irreducible factors.

The idea is to argue by induction over the rank. If $$G$$ is not graphically irreducible, then it is isomorphic to a graph product $$\Gamma \mathcal{G}$$ where $$\Gamma$$ is finite and not complete. The graph $$\Gamma$$ can be decomposed as a join $$\Gamma_0 \ast \Gamma_1 \ast \cdots \ast \Gamma_n$$ where $$\Gamma_0$$ is complete and where each graph among $$\Gamma_1, \ldots, \Gamma_n$$ contains at least two vertices and is not a join. It is not difficult to see that the factors indexed by the vertices in $$\Gamma_1 \cup \cdots \cup \Gamma_n$$ have smaller ranks (compared to $$G$$). But we have no information on the subgroup $$\langle \Gamma_0 \rangle$$ generated by the factors indexed by $$V(\Gamma_0)$$. However, a positive answer to the question mentioned above would imply that we can take $$n$$ maximum in the previous decomposition, and the expected result is proved. Indeed, for every $$1 \leq i \leq n$$, the graph $$\Gamma_i$$ is not complete, so the subgroup $$\langle \Gamma_i \rangle$$ surjects onto a non-trivial free product $$F_i$$. Consequently $$G \simeq \Gamma \mathcal{G} \simeq \langle \Gamma_0 \rangle \oplus \langle \Gamma_1 \rangle \oplus \cdots \oplus \langle \Gamma_n \rangle$$ surjects onto $$F_1 \oplus \cdots \oplus F_n$$, which implies that this sum has rank $$\leq \mathrm{rank}(G)$$.

Indeed, write $$C_p\ast C_p=\langle u_p,v_p\mid u_p^p=v_p^p=1\rangle$$.
Let $$J$$ be any finite set of primes, and $$G_J=\prod_{p\in J}C_p\ast C_p$$. Then $$G$$ has generating rank two, regardless of $$J$$: indeed, it is generated by $$u_J=\prod_{p\in J}u_p$$ and $$v_J=\prod_{p\in J}v_p$$. While $$G_J$$ is a direct product of $$|J|$$ groups, each of which is a nontrivial free product.