Identify some Coxeter group As we all know, the weyl group of lie algebra of $B_{2}$ type is $\left\{s_{1},s_{2}|s_{1}^{2}=1, s_{2}^{2}=1, (s_{1}s_{2})^{4}=1\right\}$. How can we identify this with $Z^{2}_{2}\rtimes S_{2}$?
If I choose fundamental roots of $B_{2}$ as basis, then I can identify $s_{1}$ with matrix $\begin{bmatrix}  
    -1&1\\0&1 \end{bmatrix}$ and identify $s_{2}$ with matrix $\begin{bmatrix}  
    1&0\\2&-1 \end{bmatrix}$. Then our group can be thought as matrix group generated by these two matrices. But I don't know how to find semidirect product structure out there.
 A: I will describe this for type $B_n$ in general, since that actually makes it a bit more clear what is going on.
$$W(B_n) = \langle t_0, s_1,\dots, s_{n-1}\mid t_0^2, s_i^2, (t_0s_1)^4, (t_0s_i)^2\mbox{ for }i\geq 2, (s_is_{i+1})^3, (s_is_j)^2\mbox{ for }|i-j|>1\rangle$$
So it has generators $s_i$ for $i\in \{1,\dots, n-1\}$ which generate a copy of $S_n$ (just note that the $s_i$ precisely satisfy the relations of the type $A_{n-1}$ Coxeter group), together with an extra generator that I have here called $t_0$. The reason for this comes now:
For each $i\in \{1,\dots, n-1\}$ define recursively $t_i = s_it_{i-1}s_i$.
Now I claim that the subgroup $\langle t_0, t_1,\dots, t_{n-1}\rangle$ is isomorphic to $(\mathbb{Z}/2\mathbb{Z})^n$, that it is normalized by the subgroup $\langle s_1,\dots, s_{n-1}\rangle$, and that it intersects the latter trivially.
All of this leads to $W(B_n)$ being isomorphic to the semidirect product $(\mathbb{Z}/2\mathbb{Z})^n\rtimes S_n$ (in fact, a very nice such semidirect product: It is the wreath product of $\mathbb{Z}/2\mathbb{Z}$ with $S_n$).
I leave the proof of my claim about the subgroup as a good exercise.
Hint for showing that the copy of $S_n$ normalizes the subgroup generated by the $t$'s: Show explicitly that $$s_it_js_i = \begin{cases}t_{j-1} & \mbox{if }i=j \\ t_{j+1} & \mbox{if }i = j+1 \\ t_j & \mbox{else}\end{cases}$$
