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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?

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Well, just do what the theorem says: 1. find (any) torus, 2. find its normalizer, 3. make the quotient, 4. compare it with $S_n$. Which step are you having trouble with? –  Marek Jan 16 '13 at 19:13
@Marek: Number 4: compare with $S_{n}$. –  user58536 Jan 16 '13 at 23:20
I'd like to know 2,3 and 4 step. –  ArthurStuart Jan 17 '13 at 0:15
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up vote 1 down vote accepted

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 $S_n$.

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