# Weyl group of this root system is $S_n$?

Let $E=\{(x_1,\dots,x_n)\in \mathbb{R}^n:\sum x_i=0\}$ is of dimension $n-1$ take root system $R=\{e_1-e_j:1\ge i,j\le n\}$, I know that it is of rank $n-1$ root system, but why its weyl group is $S_n$? I have calculated $s_{e_i-e_j}((\dots,x_i,\dots,x_j\dots)=(\dots,x_j,\dots,x_i\dots)$ I mean refelcetion is just transposition, so weyl group should be $S_{n-1}$ why $S_n$?

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there are $n$ $x_i$'s, so the group is $S_n$ (notice that if you take an element of $E$ with all $x_i$'s different, then its orbit has $n!$ elements) –  user8268 Oct 1 '12 at 11:28

The root system is a subset of $R^n$. The simple reflections $s_{\alpha_i}$ where $\alpha_i=e_i- e_{i+1}$ generate the Weyl group. They correspond to the transpositions $(i, i+1)$ $i=1,2, \cdots, n-1$ in $S_n$ which generate $S_n$.