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Magma
Magma
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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 (without proofproof 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 $\mathbb F_2 \oplus \mathbb Z$.

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 (without proof) 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 $\mathbb F_2 \oplus \mathbb Z$.

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 $\mathbb F_2 \oplus \mathbb Z$.

Source Link
Magma
Magma
  • 1k
  • 8
  • 11

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 (without proof) 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 $\mathbb F_2 \oplus \mathbb Z$.