# Wreath Product and Generalized Symmetric Group

I'm trying to understand the generalized symmetric group $$G(m,1,n)$$. Two simplest cases are the symmetric group $$G(1,1,n)$$, and the hyperoctahedral group $$G(2,1,n)$$. The symmetric group is learned in general abstract algebra course, so it is easy. The hyperoctahedral group the signed symmetric group of permutations $$w$$ either of the set ⁠$$\{-n,-n+1,\cdots,-1,1,2,\cdots,n\}$$ such that $$w(i)=-w(-i)$$ for all $$i$$, which is a little more complicated but still not difficult to understand.

The generalized symmetric group is the wreath product $$G(m,1,n):=\mathbb{Z}_{m}\wr S_{n}$$ of the cyclic group of order $$m$$ and the symmetric group of order $$n$$. The definition of wreath product looks quite more intricate. In Wikipedia, it gives the generalized symmetric group as an example:

The base of this wreath product is the $$n$$-fold direct product $$\mathbb{Z}_m^n = \mathbb{Z}_m \cdots \mathbb{Z}_m$$ of copies of $$\mathbb{Z}_m$$ where the action $$\varphi:S_n \to \text{Aut}(\mathbb{Z}_m^n)$$ of the symmetric group of degree $$n$$ is given by $$\varphi(\sigma)(\alpha_1,..., \alpha_n) := (\alpha_{\sigma(1)},..., \alpha_{\sigma(n)})$$.

It makes me more confused. I can understand how $$S_n$$ acts on $$\mathbb{Z}_m^2$$, but I don't understand the relation between the above interpretation and the definition of $$G(1,1,n)$$ or $$G(2,1,n)$$ I state at first. Could someone please illustrate $$\mathbb{Z}_{m}\wr S_{n}$$ in a simpler and clearer way? Or describing $$G(3,1,n)=\mathbb{Z}_{3}\wr S_{n}$$ also may help!

• You also see a useful subgroup of $\Bbb{Z}_3\wr S_8$ in a 2x2x2 Rubik's cube. The eight corner pieces can all be turned into three positions (in place), and then you can permute them any which way you want. Do keep in mind that not all the elements of $\Bbb{Z}_3^8$ can be obtained by legal sequences of moves (only a subgroup of index three), but anyway. Commented Aug 6 at 6:44

The wreath product is defined as a semidirect product; what Wikipedia is making explicit is what the action is in the semidirect product. In any case, there's a simpler and more intuitive way to understand $$\mathbb{Z}_m \wr S_n$$, because it has a particularly nice faithful $$n$$-dimensional matrix representation: namely, it is the group of generalized permutation matrices where the entries can be $$m^{th}$$ roots of unity. For example,
$$\begin{bmatrix} 0 & \zeta_5 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & \zeta_5^2 \end{bmatrix}$$
is a typical element of $$\mathbb{Z}_5 \wr S_3$$, where $$\zeta_5 = e^{\frac{2\pi i}{5}} \in \mathbb{C}^{\times}$$ is a primitive $$5$$-th root of unity. The group operation is just matrix multiplication. It's a nice exercise to check that this thing is actually the wreath product that it's supposed to be. From here you can do some exercises to check your understanding like classifying conjugacy classes (it will be some generalization of cycle decomposition that takes the roots of unity into account somehow), etc.
The elements of $$\mathbb{Z}_m\wr{S}_{n}$$ are permutation matrices with nonzero entries $$x_1,x_2,\dots,x_k$$, where $$x_i\in\mathbb{Z}_m$$. More generally, for all finite group $$G$$, the elements of $$G\wr{S}_{n}$$ are permutation matrices with nonzero entries replaced by elements in $$G$$.
• This is a good way to think about the wreath product. But as you use matrices, it may be worth clearing up that you use the group operation when multiplying matrix entries. Particularly as your example has $G=\Bbb{Z}_m$, which is an additive group. May be replacing it with $C_m$, or $\mu_m$ (=the group of $m$th complex roots of unity would diminish the chance of a misunderstanding? Commented Aug 6 at 6:39