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I am reading a proof over 4.2 in Suzuki group theory I and can't make sense of some parts.

I will just type the proof and then say what my query is.

Statement

Let $(W,S)$ be a Coxeter system. Let $T$ be the set of all elements of $W$ that are conjugate to some generator $s \in S$. Let $X=\{\pm 1 \} \times T$.

For a sequence $\mathfrak{s}=(s_1,\dotsc,s_q)$ define $\mathfrak{t}$ to be the sequence $(t_1,\dotsc,t_q)$ where $$t_i = (s_1\dotsm s_{i-1}) s_i (s_1\dotsm s_{i-1})^{-1} \in T. $$

For any $t \in T$ let $n(\mathfrak{s}, t)$ be the number of indices $j$ such that $t_j=t$.

Define $\eta(\mathfrak{s}, t)=(-1)^{n(\mathfrak{s},t)}$.

Define $U:S \rightarrow \operatorname{sym}(X) $, $s \mapsto U_s $ where $$U_s(\epsilon, t)=(\pm \epsilon, sts) $$ where the signature of the right side of the equation is negative only when $s=t$.

Then the mapping $U $ can be extended to a homomorphism from all of $W$ to $\operatorname{sym}(X)$.

Proof

We have $U_s^2=1$ and so $U_s$ is a permutation of $X$. We just need to show that this map $U$ preserves the relations of the group.

Let $s, s' \in S$ and let $m$ be the order of $ss'$. Then for any $k<m $ we get $$(U_{s'}U_s)^k(\epsilon, t) = (\pm \epsilon, (ss')^{-k}t(ss')^k).$$

Now the sign of $\epsilon $ changes at $k+1$ if and only if $(ss')^{-k}t(ss')^k=s$ or $(ss')^{-k}t(ss')^k = ss's$. (*) In these cases we have either $t=(ss')^{2k}s $ or $t=(ss')^{2k+1}s$. (**) If $t=(ss')^l s $ for some $l < m $ then $t=(ss')^{l+m}s$. So a change of sign occurs twice. Thus $(U_{s'}U_s)^m=1 $ and the mapping preserves the defining relations.

Queries

(*) Firstly, how has he gone from $(ss')^{-k}t(ss')^k = s$ to $t=(ss')^{2k}s $, when surely it should be $t=(ss')^k s (ss')^{-k} $?

(**) Secondly, what does he mean by “the change of sign occurs twice” and how does this imply that $(U_{s'}U_s)^m=1$?

Any help would be greatly appreciated.

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  • $\begingroup$ In the first sentence of the fourth paragraph, "For a sequence $\mathfrak s = (s_1, \dotsc, s_q)$ For $s \in S$ …", I think that it is only the first quantification that belongs. Is that correct? $\endgroup$
    – LSpice
    Commented Aug 12, 2023 at 19:08
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    $\begingroup$ Also, I think you meant not $(U_{s'}U_s)^k (\pm \epsilon, (ss')^{-k}t)(ss')^k)$ (which has an odd number of parentheses!), but $(U_{s'}U_s)^k(\epsilon, t) = (\pm\epsilon, (s s')^{-k}t(s s')^k)$. $\endgroup$
    – LSpice
    Commented Aug 12, 2023 at 19:22
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    $\begingroup$ Yes you’re right for both $\endgroup$ Commented Aug 12, 2023 at 20:25
  • $\begingroup$ Re, OK, I have edited accordingly. $\endgroup$
    – LSpice
    Commented Aug 12, 2023 at 22:18

1 Answer 1

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For (*), certainly $t$ equals $(s s')^k s(s s')^{-k}$, as you expect; but $s(s s')^{-1}s$ equals $s s'$, so that $(s s')^k s(s s')^{-k}$ equals $(s s')^k(s(s s')^{-1}s)^k s = (s s')^k(s s')^k s$.

For (**), Suzuki is making the point that something happens for every index $k$ such that $t$ equals $(s s')^{2k}s$ or $(s s')^{2k + 1}s$. If $t$ equals $(s s')^l s$ with $l < m$, then we get one such index $k = \lfloor l/2\rfloor$; but we also get another such index, $k = \lfloor(l + m)/2\rfloor$. These can never be equal when $l$ and $m$ are positive integers, and $l$ is less than $m$. Then the observation is that a sign change occurring an even number of times is the same as no sign change occurring at all.

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  • $\begingroup$ I’m still confused how your first paragraph gives the result. With the second paragraph I see that if a sign changes at $l$ then a sign change also occurs at $l+m$ like you say. But how is this showing that $(U_{s’}U_s)^m =1 $? $\endgroup$ Commented Aug 12, 2023 at 20:28
  • $\begingroup$ @Anonmath101, re, as in the answer, because it means that the number of sign changes is even, and making an even number of sign changes is the same as making none. (The only other thing that the $m$-fold composition is doing is conjugating by $(s s')^m$ on the second factor, but $(s s')^m$ equals $1$.) $\endgroup$
    – LSpice
    Commented Aug 12, 2023 at 22:20
  • $\begingroup$ So it shows that $(U_{s’}U_s)^k \neq 1 $ for $k<m$ as it flips the sign at some $t$ and then because there’s an even number of sign flip from $k$ ranging from $1$ to $m-1$ it shows that $(U_{s’}U_s)^m =1 $. Is this correct? $\endgroup$ Commented Aug 13, 2023 at 10:50
  • $\begingroup$ Re, it is irrelevant whether or not $(U_{s'}U_s)^k$ equals $1$ for $k < m$; we only have to show that $(U_{s'}U_s)^m$ equals $1$, and that is true for the reason that you said (together with the fact that $(s s')^m$ equals $1$). $\endgroup$
    – LSpice
    Commented Aug 13, 2023 at 22:47

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