# How to prove that $T_L(u) = T_L(w)$ implies that $u=w$?

For a reduced decomposition $u=s_1 \cdots s_k$, $u \in W$, $W$ is a Coxeter group. $T_L(u)$ is defined as $$T_L(u)=\{s_1 s_2 \cdots s_i \cdots s_2 s_1, 1 \leq i \leq k\}. \quad (1)$$ I am trying to solve the exercis 11 on page 23 of the book Combinatrics of Coxeter groups.

How to prove that $T_L(u) = T_L(w)$ implies that $u=w$?

We can verify this for many examples. For example, if $u=s_1 s_2 s_1$, $w=s_2 s_1 s_2$, then we have $$T_L(u)=\{s_1, s_1 s_2 s_1, s_1 s_2 s_1 s_2 s_1\} \\ =\{s_1, s_1 s_2 s_1, s_2 \} \\ T_L(w)=\{s_2, s_2 s_1 s_2, s_2 s_1 s_2 s_1 s_2\} \\ =\{s_2, s_1 s_2 s_1, s_1 \}.$$ But how to prove (1) in general? Thank you very much.

• When you write $u=w$, you presumably mean $u$ and $w$ represent the same element of $W$ rather than $u$ and $w$ are the same word? – Derek Holt Mar 31 '17 at 9:12
• @DerekHolt, yes. – LJR Mar 31 '17 at 9:20
• I don't have this book at hand. But is it explained there why and if $T_L(u)$ is independent from the choice of the reduced expression of $u$? – user213008 Apr 25 '17 at 15:20