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If a finitely generated Coxeter group $W$ is finite, its integer-valued Coxeter length has a maximum. Conversely, if some $w_0$ has globally maximal length $N$, every group element has a word of length at most $N$. There are only finitely many words of bounded length in the finite set of simple generators, so $W$ is finite.

Realize the finite group as the reflection group of a root system with fundamental system $\Delta$ and positive system $\Pi$. Maximality and the fact that multiplication by a simple generator changes Coxeter length by one giveThe positive-root criterion for Coxeter length therefore gives $w_0\Delta \subseteq -\Pi$. Since $w_0\Delta$ is itself fundamental, it must be the simple system $-\Delta$ of the positive system $-\Pi$.

If $v_0$ is another maximal-length element, the same argument gives $v_0\Delta =-\Delta =w_0\Delta$. Hence $v_0^{-1}{w}_0$ stabilizes $\Delta$, and part c gives $v_0^{-1}{w}_0=1$. The Longest element of a finite Coxeter group is therefore unique.

By the assumed transitivity on fundamental systems, some $w_0\in W$ sends $\Delta$ to $-\Delta$. It therefore sends the entire positive system of a root system $\Pi$ to $-\Pi$. Part d then givesFor every $w\in W$, its inversion set is contained in $\Pi$, so $\ell (w)\le |\Pi |$ and $w_0$ has maximal length.

If $u$ also has maximal length, then $N(u)=\Pi$, so $u(\Pi )=-\Pi$. Hence $w_0^{-1}u$ preserves $\Pi$ and has no inversions. Part d makes its Coxeter length zero, so it is the identity. Thus $u=w_0$, proving that the Longest element of a finite Coxeter group is unique and has length $|\Phi |/2$.

There is a missing hypothesis in the printed claim: it is false when every irreducible component of $W$ has type $A_1$. The intended statement holds as soon as $W$ has an irreducible component of rank at least two, which we now assume.

Since $q=p$, every Hecke parameter of a BN-pair vanishes in $k$, and $H$ is the 0-Hecke algebra withLet $w_0$ be the Longest element of a finite Coxeter group. Choose a simple generator $s$ in a component of rank at least two, putThe element $r$ is again a simple generator. The identities $rv=w_0$ and $\ell (w_0)=\ell (v)+1$ giveIf $t\ne r$ is simple, then $t$ is a left descent of both $v$ and $w_0$: using $\ell (w_{0}u)=\ell (w_0)-\ell (u)$, one gets $\ell (tv)=\ell (v)-1$. HenceThe one-dimensional subspace $kX$ is therefore a left ideal. It is nonzero because $T_v$ and $T_{w_0}$ are distinct basis elements.

Every reduced expression in a Coxeter group for $v=w_{0}s$ contains $r$: in an irreducible finite component of rank at least two, deleting one final generator from $w_0$ does not remove any vertex from its support. A reduced expression for $w_0$ contains $r$ as well. Since $T_{r}X=0$, associativity now givesIf $H$ were a semisimple algebra, the left ideal $kX$ would be a direct summand of the regular module. The corresponding projection would produce a nonzero idempotent in $kX$, impossible because $(kX{)}^2=0$. Thus $H$ is not semisimple.

For completeness, if $W\cong (A_1{)}^m$, thenwhich is semisimple. This is the counterexample showing why the omitted rank condition is necessary.