# Set of normal forms for for $C_2\wr S_3$

I am working with the subgroup of $$SL(3,\mathbb{Z})$$ with generating set $$\omega=\begin{bmatrix} -1& 0& 0\\ 0& 1& 0\\ 0& 0&1 \end{bmatrix}r=\begin{bmatrix} 0& 1& 0\\ 0& 0&1 \\ 1& 0&0 \end{bmatrix} \text{ and } s=\begin{bmatrix} 0& 1&0 \\ 1&0 &0 \\ 0&0 &1 \end{bmatrix}$$ With this generating set, the group is more easily seen as $$C_2\wr S_3$$, which is another name for $$C_2^3\rtimes_\rho S_3$$ with the natural action of $$S_3$$.

Is there a nice set of normal forms this group with this generating set?

• See this question and its references. – Dietrich Burde Aug 14 at 18:35
• Your title has $C_2\wr C_3$ (which I would assume to be the regular representation). But you talk about a different wreath product, $C_2\wr_{\Omega} S_3$, where $\Omega=\{1,2,3\}$ with the natural action of $S_3$ on $\Omega$. – Arturo Magidin Aug 14 at 18:56
• Again: do you mean $C_2\wr_{\rho} S_3$ in the title, rather than $C_2\wr C_3$? – Arturo Magidin Aug 14 at 20:55
• You assert in the body that the matrices generate $C_2\wr_{\rho} S_3$; this is different from $C_2\wr C_3$. The latter has order $2^3\times 3 = 24$; the former has order $2^3\times 6 = 48$. So if you meant the latter, then your assertion in the body is false and you should fix it. – Arturo Magidin Aug 15 at 21:52
• Thanks, I was misreading your comments. I have fixed the title. – Prototank Aug 16 at 19:41

## 1 Answer

I think I've figured out a set of normal forms. The group acts on the set of (componentwise signed) permutations of (1,2,3). For example, there's an element of $$G$$ taking the vector $$\begin{bmatrix}1 \\ 2 \\ 3\end{bmatrix}$$ to $$\begin{bmatrix}-3 \\ 1 \\ -2\end{bmatrix}$$. So if we can determine the permutation first, and then place signs on the permutation, this will account for the $$48$$ elements of the group. Here's what I got:

$$\underbrace{\omega^i (r\omega r^{-1})^j(r^2\omega r^{-2})^k}_{\text{Determines \{\pm 1\}^3}}\underbrace{r^ls^m}_{\text{Dihedral form of S_3}}$$