I have some difficulties in understanding why the Weyl group of algebraic group $SL_n(K)$ is isomorphic to symmetric group $S_n$.
Let $G=SL_n(K)$ be the simply-connected algebraic group over the algebraic closed filed $K$ with $charK>2$. Also let $T$ be the maximal tori of $G$ consists of diagonal matrices with determinant 1. In this case, we have $W=N_G(T)/T$ is the Weyl group of $G$ in which $N_G(T)$ is the group of generalized permutation matrices with determinant 1.
On the other hand, it is well-known that Weyl group of $G$ is isomorphic to symmetric group on $n$ letters. Hence $N_G(T)/T\cong S_n$.
Addendum and Edition:
Now let $S\in N_G(T)$ be corresponding to the cycle $(1~~ 2)\in S_n$. Considering the action of $S$ on $diag[a_1, a_2,..., a_n]\in T$, we get $diag[a_1, a_2,..., a_n]^S=diag[a_2, a_1, a_3,..., a_n]$. Since the determinant of a permutation matrix is just the Sign of the corresponded permutation, we have $det(S)=Sign(1~~2)=-1$ , which is impossible according to $S\in N_G(T)$.
This implies that $N_G(T)$ can not contain any element which is corresponded to $(1~~2)$. So how could I explain the isomorphism $N_G(T)/T\cong S_n$?
I would be grateful for any help.
Your statement about the determinant of the diagonal matrices is wrong! $\operatorname{det} \operatorname{diag}(a_1,\ldots, a_n)=\operatorname{det}\operatorname{diag}(a_2,a_1,a_3,\ldots,a_n)=\prod_{i=1}^n a_i$.
Re the addendum: it's not correct that a matrix $S$ conjugating those two matrices must be a permutation matrix. Indeed, in the case $n=2$ the matrix $$ \left( \begin{array}{ll}0 & -1 \\ 1 & 0 \end{array}\right)$$ works, as Jyrki points out in the comments, and in general you can find elements of $N_G(T)$ implementing the permutation you want similarly.