A question about Coxeter groups.

Question

Let $G$ be a group, and $x,y,z \in G - \{1\}$, where $1$ is the identity element of $G$. Assume that $x^2=y^2=z^2=1$, $xy =yz$ and $xz$ is of infinite order.

1. Can $G$ be a Coxeter group?
2. Can $G$ be a Coxeter group if $G$ is presented by $G= \langle x,y,z | x^2=y^2=z^2 =1, xy=yz \rangle$?

More generally, what characterize a group that can be embedded in a Coxeter group?

Thanks to everyone.

Answer 1

Derek answered question 1. For question 2, at least any finite group can be embedded in a Coxeter group because the symmetric group is a Coxeter group. With a little more effort we can see that a (countable) profinite group can be embedded in a countably generated Coxeter group (the symmetric group on countably many generators).

Answer 2

The relation $xy=yz$ is equivalent to $z = y^{-1}xy$, so we can eliminate $z$ and the presentation reduces to $\langle x,y \mid x^2=y^2=1 \rangle$, which is indeed a Coxeter group. In fact it is the infinite dihedral group.