Why isn't a coxeter group a HNN-extension?

Question

A doctorate told me to think about why there is no mapping from coxeter groups to $\mathbb{Z}$. This makes sense since HNN-extensions are of the form $$A\star_{\{(\varphi_1 , C,\varphi_2 )\}}=\langle A,t\mid R_A,R_C\rangle$$ where $R_C=${$ t\varphi_1(c)t^{-1}=\varphi_2(c) \mid c\in C $} and $A=\langle a_1,a_2,...\mid R_A\rangle$, $R_A$ the set of relations in $A.$

Since $t$ doesn't have any relations in the HNN-extension there exists a map $\alpha: \langle t\rangle \to \mathbb{Z}, t\mapsto 1$, so $\mathbb{Z}$ is basically embedded in the HNN-extension (correct me if I'm wrong)?

So when thinking about the initial question I thought that I just have to proof that coxeter groups don't have any elements of infinite order but upon researching that fact I learned that it's simply not true. So shouldn't therefore exist an element $t$ in a coxeter group such that $\alpha$ exists?

Please don't put full solutions to this question on here, I want to solve it myself but just need some intuition about coxeter groups.

Answer 1

You are confused by embeddings vs homomorphic images. Not every homomorphic image is isomorphic to a subgroup, and not every subgroup is isomorphic to a homomorphic image. In particular, Coxeter groups in general contain subgroups isomorphic to $\mathbb{Z}$, but have no homomorphic image isomorphic to $\mathbb{Z}$.

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A Coxeter group has presentation of the form $$\mathcal{P}=\langle a_1, \ldots, a_n\mid \mathbf{r}, a_i^2 (i=1, \ldots, n)\rangle$$ for set of relators $\mathbf{r}$; the specific form of relators in $\mathbf{r}$ does not matter for this problem.

Any such presentation $\mathcal{P}$ has an abelianisation where every non-trivial element has order $2$; this is because if $\hat{a_i}$ is the image of $a_i$ in the abelianisation then any word $W(\hat{a_1}, \ldots, \hat{a_n})$ over these generators has square $$W(\hat{a_1}, \ldots, \hat{a_n})W(\hat{a_1}, \ldots, \hat{a_n})=W(\hat{a_1}^2, \ldots, \hat{a_n}^2)=W(1, \ldots, 1)=1.$$ In particular, the abelianisation is finite (as there are finitely many generators). As every abelian quotient of a group factors through its abelianisation, every abelian quotient of every Coxeter group is finite. Therefore, no Coxeter group surjects onto $\mathbb{Z}$.

On the other hand, every HNN-extension surjects onto $\mathbb{Z}$. This is because they have (relative) presentation of the form $$ H=\langle G, t\mid t^{-1}xt=y, x, y\in G\rangle $$ so we can define a map $H\to\mathbb{Z}$ by $t\mapsto 1$, $g\mapsto 0$ for all $g\in G$.

Therefore, no Coxeter group can be an HNN-extension.