Reflection cardinalities of the three connected finite rank-three types

By [F4] each diagram is connected with three vertices, and its Coxeter group is finite. Thus the system is irreducible finite type with $n=3\ge 2$. Number the path vertices $1,2,3$ in order. The classes $J=\{s_1,s_3\}$ and $K=\{s_2\}$ form a bipartition. Since the endpoints are not joined, $m(s_1,s_3)=2$ and the Coxeter relations give $s_1{s}_3=s_3{s}_1$. Hence $c=(s_1{s}_3)s_2$ is a bipartite Coxeter element in the convention of the spectrum and root-enumeration suppliers. Reversing the two classes replaces $ab$ by $ba=a^{-1}(ab)a$, so it does not change the order.

The operators in [F3] therefore have exact orders $q=4,6,10$, respectively. For the classical cases, complexification has the same real matrix entries in the simple-root basis, so ${\rho }_{\mathbb{C}}(c{)}^k=I$ if and only if $\rho (c{)}^k=I$ for every positive integer $k$. For all three cases, $\mathrm{\mid }S\mathrm{\mid }=3$ permits [F5], and injectivity gives $\rho (c{)}^k=I$ if and only if $c^k=1$. Consequently the group element has exact order $h=\mathrm{ord}(c)=q$, namely $4$, $6$ or $10$.

By [F1] and [F2], $\mathrm{\mid }T\mathrm{\mid }=\mathrm{\mid }{\mathrm{\Phi }}_{+}\mathrm{\mid }=3h\mathrm{/}2$. Substitution gives $3\cdot 4\mathrm{/}2=6$, $3\cdot 6\mathrm{/}2=9$ and $3\cdot 10\mathrm{/}2=15$. The use of canonical roots only evaluates the abstract conjugate set $T$; it does not assume an unproved identification of the original arrangement with a canonical representation.