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> 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?
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**Motivation.** In order to optimize some of the results from [my preprint](https://arxiv.org/abs/1807.00622), 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 $\leq \mathrm{rank}(G)$. 



Rank of a sum with free products
No. 


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.


