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I'm studying Humphreys' Lie algebra, but I'm stuck in finding the Weyl group of type D. In the book, the contents are written by :

Type D$_l$ : Let E=$\mathbb{R}^l$ and let $\Phi:=\{\pm(\epsilon_i\pm\epsilon_j)\: : \: i\neq j\}$ (The $\epsilon_i$ are the standard basis of $E$). For a base take the $l$ independent vectors $\epsilon_1-\epsilon_2, \cdots, \epsilon_{l-1}-\epsilon_l,\epsilon_{l-1}+\epsilon_l$ (so $D_l$ results). The Weyl group is the group of permutations and sign changes involving only even numbers of signs of the set $\{\epsilon_1,\cdots,\epsilon_l\}$. So the Weyl group is isomorphic to the semidirect product of $(\mathbb{Z}/2\mathbb{Z})^{l-1}$ and the symmetric group of degree $l$.

How to act the Weyl group on $E$? Also, I don't understand why the Weyl group is isomorphic to the above semidirect product. Please give me a hint or solution. Thanks in advance.

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    Oh, it's a typo. Thanks. – 2017-02-14

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Given a root system $R$ for a simple Lie group $G$, with maximal torus $T$, then the Weyl group $W$ is always isomorphic to $Norm_G(T)/T$. Hence, with $R$ given as in Humphrey's book, for type $D_n$ we obtain that $W$ consists of all permutations and an even number sign changes in $n$ coordinates. Hence we have $W\cong (\mathbb{Z}/2)^{n-1}\ltimes S_n$. For more details see here.

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    Oh, I'm sorry but I don't know the argument for Lie group. Could you explain it without the argument? – 2017-02-14
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    Oh, sorry, we only need to see that "W" consists of all permutations and an even number sign changes in $n$ coordinates. This follows directly from the given root system and the definition of $W$. Then $W$ just acts this way on $R$ (the action of $W$ on $T$ by conjugation gives an action of $W$ on $R$, again by the natural permutation action). – 2017-02-14
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    Ah, my question is why W consists of all permutations and even number of sign changes. The permutations may be proven : set $\alpha_i=\epsilon_i-\epsilon_{i+1}$ for $1\le i\le l-1$. Then, the corresponding reflection of $\alpha_i$ can be identified with the transposition i and i+1 in the symmetric group of degree l. So, they generate the permutations, I think. – 2017-02-14
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    Yes, and in addition "sign changes involving only even numbers of signs", accounting for the factor $(\mathbb{Z}/2)^{n-1}$. – 2017-02-14
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    Yes, I guess but I don't know why ;( – 2017-02-14