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Let $G$ be a group with presentation $\langle S \ \vert \ R \rangle$. Then any $g \in G$ has (in case $R$ not trivial in general non unique) the form $g=s_1^{a_1}s_2^{a_2}...s_n^{a_n}$ with $s_i \in S$ and $a_i \in \{+1,-1\}$.
Let length of $g$ be given as $l(g)$ be $\min\{n \in \Bbb N_0 \ \vert \ g=s_1^{a_1}s_2^{a_2}...s_n^{a_n}\}$.

Obviously $l(gh) \le l(g)+l(h)$, so for any $s \in S$ we have $l(g)-1 \le l(gs) \le l(g)+1$.

In this question I wrongly conjectured that it cannot happen that $l(gs)=l(g)$ for $s \in S$.
Now I have follwing two questions:

(1): Is there something special about class of presentable groups for which for every $g \in G, s \in S$ holds precisely $l(gs)=l(g) \pm 1$? (Maybe, do they have a name in literature & do these share some common interesting properties?

(2): I believe that the Coxeter group (...in which I was primary interested when asking the linked question) has this property. Recall its a group with presentation $\langle s_1,s_2,\ldots,s_n \mid (s_is_j)^{m_{ij}}=1\rangle$ where $m_{ii}=1$ and $m_{ij} = m_{ji} \ge 2$.
How to see in detail that for this group ee have always $l(gs)=l(g) \pm 1$?

The problem is the following: Assume $g=s_1^{a_1}s_2^{a_2}...s_{l(g)}^{a_{l(g)}}$ (so with minimal length) and $s \in S$ under relations from usual Coxeter presentation. If $s=s_{l(g)}$ then $l(gs)=l(g)-1$ since according to Coxeter relations all $s \in S$ are idempotent, ie $s^2=e_G$ and we win.
But how to handle the case $s \neq s_{l(g)}$ for Coxeter group?

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    $\begingroup$ The fact that your Property (1) holds for Coxeter groups generalizes immediately to all groups defined by presentations in which all of the defining relators have even length. $\endgroup$ Commented Oct 23 at 19:09
  • $\begingroup$ @Derek Holt: Do such groups share some interesting structural properties? $\endgroup$ Commented Oct 25 at 13:47
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    $\begingroup$ Yes, as has been pointed out, they have a normal subgroup of index $2$. Conversely, any group with a normal subgroup of index $2$ has a generating set for which the group has that property. In fact any generating set with all generators outside of the normal subgroup will work. $\endgroup$ Commented Oct 25 at 14:36

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For the case of Coxeter groups $G$, each generator $s_i$ has order $2$ and the other relations are such that the map $s \to -1$ for $s \in S$ extends to a homomorphism $G \to \{ \pm 1 \}$ given by $g \mapsto (-1)^{\ell(g)}$. This means $\ell(gs)$ necessarily has different parity from $\ell(g)$ so they can't be equal.

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  • $\begingroup$ But is it clear that the extension of this map $s_i \mapsto -1$ to whole $G$ is really well defined? Obviously it gives a $1$ on any "standard relation", ie those of type $(s_is_j)^{m_{ij}}$, but does this already suffice to deduce formally that this map extends to $G$? In other words how to formally justify that it cannot happen that say I come with some word $s_1^{a_1}s_2^{a_2}...s_n^{a_{n}}$ which is accidentally mapped to $-1$, but which turns out to reduce to $e_G$ after clevery playing with relations? Ie, can it be turned in formal argument? $\endgroup$ Commented Oct 23 at 16:39
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    $\begingroup$ @user267839: To answer your questions in the comment: yes, this is a clear consequence of the definition of a presentation. If that is not clear to you, then it sounds like your real question might be about the behavior and properties of group presentations. $\endgroup$ Commented Oct 23 at 16:44
  • $\begingroup$ @Lee Mosher: More generally your point suggest following general principle about whose correctness I'm not sure: Assume $H$ is a group and $G= \langle s_1,s_2,\ldots,s_n \in S \mid r_1,r_2,..., r_m=1\rangle$ a presented group. Then does any set theoretic $S \to H$ which maps by natural word extension every $r_j$ to $e_H$ suffice to induce already a well defined homomorphism $G \to H$ extending the set map? $\endgroup$ Commented Oct 23 at 16:48
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    $\begingroup$ That is exactly correct, see this answer. $\endgroup$ Commented Oct 23 at 17:03
  • $\begingroup$ @Lee Mosher: Or, yes sorry that should be clear if I'm not missing something: Let $R:=\{r_1,...,r_m\}$ and let $N_R$ normal hull of $R$, ie normal closure of subgroup $\langle R \rangle (\subset F_S$) inside free group $F_S$ gen by $S$. Then $G=F_S/N_R$ and $S \to H$ extends to whole $G$ iff for extension $F_S \to H$ the whole subgroup $N_R$ is mapped to $1_H$. But that's equivalent to that all $r_j$ mapped to $e_H$ since conjugation leaves kernel invariant, so we can extend once we known all $r_j$ are mapped to $e_H$. $\endgroup$ Commented Oct 23 at 17:08

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