Isomorphic Coxeter Groups Is it true that two Coxeter groups having the same Coxeter matrix (equivalently, the same Coxeter graph) isomorphic? Because otherwise, the definition of a Coxeter group from its Coxeter matrix does not make sense to me. If so, what is the proof? Actually, it will be of great help, if I understand why the Coxeter group A(n-1) is isomorphic to the symmetric group S(n).
 A: Indeed, Coxeter groups having the same Coxeter matrix are isomorphic. When we specify a Coxeter matrix, we are specifying relations on the generators of the group. The group is completely determined by these relations. Say the two Coxeter systems are $(W,S)$ and $(W',S')$ and the Coxeter matrix is
$$(m_{ij})$$
where $m_{ii}=1$, $m_{ij}=m_{ji}$, and $m_{ij}\in ([2,\infty)\cap\mathbb{N})\cup\{\infty\}$ if $i\neq j$. The generating set can be indexed by any set, say $I$, and since the groups have the same matrix this means they can be indexed by the same set, yielding the same Coxeter matrix via that indexing. Then we are saying that for each $i,j\in I$, where $s_i,s_j\in S$ and $s_i',s_j'\in S'$ we have
$$(s_is_j)^{m_{ij}}=1$$
and
$$(s_i's_j')^{m_{ij}}=1$$
provided $m_{ij}<\infty$ (and if $m_{ij}=\infty$ then $s_is_j$, $s_i's_j'$ are of infinite order). These are the only relations in the two groups, as $W$ is the quotient of the free group on $S$ by these relations and $W'$ is the quotient of the free group on $S'$ by these relations. Thus there is a homomorphism $W\to W'$ given by the unique extension of the map $S\to S'$ such that $s_i\mapsto s_i'$. Since the homomorphism maps a generating set onto a generating set, it is surjective, and it is injective because any product of generators in $W'$ that yields the identity also yielded the identity in $W$ due to the fact that the relations are identical.
For a slightly more precise proof of the latter statement, denote by $F(X)$ the free group on a set $X$. There is an isomorphism $F(S)\to F(S')$ given by extending the map $s_i\mapsto s_{i}'$ simply by the definition of a free group. The kernel of the quotient map $F(S)\to W$ is $K=\langle (s_is_j)^{m_{ij}}\rangle$ and the kernel of $F(S')\mapsto W'$ is $K'=\langle (s_i's_j')^{m_{ij}}\rangle$. The isomorphism $F(S)\to F(S')$ maps $K\to K'$ bijectively. If $s_{i_1}'s_{i_2}'\cdots s_{i_k}'=1\in W$, then $s_{i_1}'s_{i_2}'\cdots s_{i_k}'\in K'$, hence $s_{i_1}s_{i_2}\cdots s_{i_k}\in K$, so $s_{i_1}s_{i_2}\cdots s_{i_k}=1\in W$. Thus the homomorphism $W\to W'$ has trivial kernel.
