Maximal torus and Lie group Let $T$ be a maximal torus of a compact Lie group $G$. We define the Weyl group of $G$ as the quotient space $N(T)/T$ where $N(T)$ is the normalizer of $T$ in $G$. I have to prove that if $G$ is $U(n)$ we can identify $W$ with the permuation group $S_{n}$. Could you give me a reference or a proof of this claim?

First let's fix some torus, e.g. one consisting of the matrices of the form
$$T(\phi_1, \dots, \phi_n) = 
\pmatrix {
e^{i\phi_1} & 0 & \cdots & 0 \cr
0& e^{i\phi_2} &  \cdots & 0 \cr
\vdots & \vdots & \ddots & 0 \cr
0 & 0 & \cdots & e^{i\phi_n} \cr
}$$
where $\phi_i \in [0, 2\pi)$ for all $1 \leq i \leq n$.
Now, you should be familiar with the fact that for any $A, B \in GL(n, \mathbb C)$ we have that $B$ and $ABA^{-1}$ have the same eigenvalues. In particular, this holds when $B \in T$ and so every $C \in N_{U_n}(T)$ must therefore preserve the eigenvalues of $B$. Since eigenvalues of $CBC^{-1}$ and $B$ are the same, we must have $CBC^{-1} = T(\phi_{j_1}, \dots \phi_{j_n})$ for some permutation $j \in S_n$. Therefore $N_{U_n}(T) / T \leq S_n$. Moreover, it is easy to see that each permutation of entries of $T(\phi_1, \dots, \phi_n)$ gives us a different element of $N_{U_n}(T) / T$ and therefore the Weyl group is indeed equal to $S_n$.
