Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
Presumably, given the natural action of $S_n$ on $(\mathbb Z/2\mathbb Z)^n$ of permuting the elements... – Thomas Andrews Jul 26 '12 at 18:02
To define a semidirect product $H\rtimes K$ one must give $H$ and $K$ and an action of $K$ on $H$. Presumably, here you are wanting the action to be as Thomas Andrews suggested. However, there is nothing in your post to say that the action isn't the trivial action and so we would just have $H\times K$. Indeed, you could make up some silly action like switching the first two elements and keeping everything else fixed if the permutation is odd, and keeping everything fixed otherwise (using the fact that $S_n\rightarrow\mathbb{Z}/2\mathbb{Z}\leq \operatorname{Aut}((\mathbb{Z}/2\mathbb{Z})^n)$). – user1729 Jul 26 '12 at 19:43
up vote 1 down vote accepted

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.

share|cite|improve this answer
Why do you write $e_i^2 = 1, e_i e_j = e_j e_i$? Couldn't you just write $e_i \in \mathbb{Z} / 2 \mathbb{Z}$? Or, if you want to have representations, $e_i \in \{0,1\}$? That somewhat seems simpler to me. – Albert May 24 '13 at 13:01
@Albert: the question asked for a presentation. “$e_i \in \mathbb{Z}/2\mathbb{Z}$” is not a presentation. – Jack Schmidt May 24 '13 at 13:39

If $H= \langle X | R \rangle$ and $K= \langle Y |S \rangle$, then $H \rtimes_{\phi} K = \langle X,Y | R,S, (khk^{-1}= \phi(k) \cdot h,h \in X, k \in Y) \rangle$.

share|cite|improve this answer

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.