How to write presentation of a group? Let $G=(\mathbb{Z}/2\mathbb{Z})^n \rtimes S_n$, where $(\mathbb{Z}/2\mathbb{Z})^n$ is the direct product of $n$ copies of $\mathbb{Z}/2\mathbb{Z}$, $S_n$ is the symmetric group of degree $n$, and $\rtimes$ is a semi-direct product. How to write a presentation for $G$? Thank you very much.
 A: If $H= \langle X | R \rangle$ and $K= \langle Y |S \rangle$, then $H \rtimes_{\phi} K = \langle X,Y | R,S, (khk^{-1}= \phi(k) \cdot h,h \in X, k \in Y) \rangle$. 
A: This is the Coxeter group of type $B_n$. It therefore admits a particularly nice presentation encoded by its Coxeter diagram, namely
$$\langle s_1 ... s_n | s_i^2, (s_1 s_2)^3, (s_2 s_3)^3, ... (s_{n-2} s_{n-1})^3, (s_{n-1} s_n)^4 \rangle.$$
The elements $s_1, ... s_{n-1}$ generate the symmetric group $S_n$ while $s_n$ flips the sign of one vector. 
A: To combine the other two answers: @Seirios gives the standard presentation of any semi-direct product, and @Qiaochu Yuan gives the standard presentation of this Coxeter group (also known as a Weyl group, or the hyperoctahderal group).
If we take a standard presentation for $(\mathbb{Z}/2\mathbb{Z})^n$, $$\left\langle e_1, e_2, \ldots, e_n : e_i^2 = 1, e_i e_j = e_j e_i \mid 1 \leq i,j \leq n \right\rangle$$
and a standard presentation for $S_n$ (but with funny names for the generators, $s_i = (i-1,i)$ for $2 \leq i \leq n$),
$$\left\langle s_2, s_3, \ldots, s_n : s_i^2 = (s_i s_{i+1})^3 = (s_i s_j)^2 = 1 \mid 2 \leq i \leq n, i+2 \leq j \leq n \right\rangle$$
Then the semi-direct product $(\mathbb{Z}/2\mathbb{Z})^n \rtimes S_n$ where $S_n$ acts by permuting the basis $e_i$ has presentation
$$\left\langle\begin{array}{ll}
& e_i^2 = 1, ~e_{i+1} e_i = e_i e_{i+1}, ~e_j e_i = e_i e_j  \\
e_1, \ldots, e_n, s_2, \ldots, s_n \quad: 
& s_i^2 = (s_i s_{i+1})^3 = (s_i s_j)^2 = 1 \\
& 
s_j e_i = e_i s_j, ~ s_i e_{i-1} = e_i s_i, s_i e_i = e_{i-1} s_i
\end{array} \right\rangle$$
where the index $i$ varies over $1$ to $n$ as long as $s_1$ is not involved, and the index $j$ varies over $1$ to $n$ as long as $j \notin \{i-1,i,i+1\}$.  The fist line is the presentation of the normal subgroup ${\left(\mathbb{Z}/2\mathbb{Z}\right)}^n$, the second line is the presentation of its “complement”, $S_n$, and the third line describes how the complement acts on the normal subgroup.
Notice how $s_i e_{i-1} s_i = e_i$ lets us "bump" a basis vector up. Starting with $e_1$, we get $e_2 = s_2 e_1 s_2$, $e_3 = s_3 e_2 s_3 = s_3 s_2 e_1 s_2 s_3$, etc. We don't actually need all those generators $e_i$; just $e_1$ will do. If we set $s_1 = e_1$, and remove some superfluous relations, we get the standard Coxeter presentation after requiring $(s_1 s_2)^4 = 1$.
Note that Qiaochu's standard presentation uses the other standard ordering of the generators. In the OP's ordering which I preserve, $s_1$ is the special generator, and in Qiaochu's ordering, it is $s_n$. If the OP is not wedded to a particular name, then I might suggest considering $s_0$ as well.
