Let $\left(W,S\right)$ be a non-affine, irreducible Coxeter system and assume that $W$ contains a copy of $\mathbb{Z}\oplus\mathbb{Z}$ (this is equivalent to $W$ being not word hyperbolic). Does this imply that $W$ contains a copy of $\mathbb{Z}\oplus\mathbb{F}_{2}$ as well? Here $\mathbb{F}_{2}$ denotes the free group in two generators. If this is not true: Is it at least true for $S$ large enough?
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$\begingroup$ If a group $G$ contains a copy of $\mathbb{Z}^2$ then it is not word-hyperbolic. Did you mean the negation of your first line? $\endgroup$– Carl-Fredrik Nyberg BroddaCommented Mar 6, 2020 at 10:27
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$\begingroup$ Yes, of course. Thank you! $\endgroup$– worldreporterCommented Mar 6, 2020 at 10:28
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$\begingroup$ What do you mean by that? Does this extra condition imply that $W$ contains $\mathbb{Z} \oplus \mathbb{F}_2$? $\endgroup$– worldreporterCommented Mar 6, 2020 at 19:47
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Consider the free product $W = \tilde A_2 * A_1 * A_1 * \ldots * A_1$. This Coxeter group is not affine, and it has a copy of $\mathbb Z^2$ within the $\tilde A_2$ part, so it satisfies your condition.
Now assume $W$ contains a copy of $\mathbb F_2 \oplus \mathbb Z$, such that $u,v$ generate $\mathbb F_2$ and $w$ generates $\mathbb Z$. This means that $A := \langle u,w\rangle$ and $B := \langle v,w\rangle$ are two disjoint copies of $\mathbb Z^2$ whose intersection is $\langle w\rangle = \mathbb Z$.
Now I claim (proof in comments) that all copies of $\mathbb Z^2$ in $W$ are finite-index subgroups of some conjugate of the $\tilde A_2$ part. So if $A$ and $B$ lie within the same conjugate, they intersect in another finite-index subgroup of that conjugate (which is $\mathbb Z^2$ again), but if $A$ and $B$ lie within different conjugates, they have trivial intersection. In neither case the intersection is $\mathbb Z$.
This is a contradiction to $A \cap B = \mathbb Z$, hence our assumption is false and $W$ does not contain a copy of $\mathbb F_2 \oplus \mathbb Z$.
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2$\begingroup$ One directly sees that if a free product $A\ast B$ contains a 1-ended group $C$, then $C$ is conjugate to a subgroup of $A$ or $B$. This applies to $C=F_2\times Z$, or more generally to any direct product of two infinite f.g. groups. $\endgroup$– YCorCommented Mar 6, 2020 at 13:39
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$\begingroup$ @YCor How does one see this directly? $\endgroup$– MagmaCommented Mar 6, 2020 at 13:46
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2$\begingroup$ If $C$ is a 1-ended f.g. group, every free inversion-free action on a tree fixes a vertex. Apply this to the action of $A\ast B$ on its Bass-Serre tree: $C$ fixes a vertex, and the vertex stabilizers are the conjugates of $A$ and $B$ precisely. $\endgroup$– YCorCommented Mar 6, 2020 at 13:48
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