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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?

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 $\leq \mathrm{rank}(G)$.

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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.

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  • $\begingroup$ Nice example, thank you. $\endgroup$
    – AGenevois
    May 27 '20 at 18:55

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