# Determining the conjugacy classes of a wreath product $G \wr S_n$

If $$G$$ is a finite group and its conjugacy classes are known, can the conjugacy classes of the wreath product $$G \wr S_n \cong G^n \rtimes S_n$$ be determined?

• Yes, this is well known. It is described in detail in James and Kerber's book on the representation theory of symmetric groups. – Christian Gaetz Oct 9 '18 at 15:59
• Did you mean $G\wr S_n\cong G^n\rtimes S_n$? – Alex B. Oct 9 '18 at 16:14
• Alex: I mean for $S_n$ to be acting on $n$ copies of $G$, in the natural way. Have I gotten the notation the wrong way round? – Chris Russell Oct 9 '18 at 16:19
• In that case Chris, the notation is the wrong way round, and Alex's is what you meant ( and I think the group dealt with in James-Kerber as mentioned by Christian). – Geoff Robinson Oct 9 '18 at 16:25

This is indeed handled in Section 4.2 of James and Kerber's book on Representations of the Symmetric group.

We also considered this problem in Section 2 our paper describing a computer algorithm for computing conjugacy class representatives in permutation groups.

J. Cannon and D. Holt, Computing conjugacy class representatives in permutation groups, J. Algebra 300 (2006), 213--222.

I can cut and paste the main bit of theory you need. The problem easily reduces to finding those classes that map onto a specific element of $$S_n$$.

Let $$A$$ be any group, let $$P$$ be a permutation group acting on the set $$\{ 1\ldots d\}$$, and let $$W := A \wr P$$. Then $$W$$ is a semidirect product of $$A^d$$ by $$P$$, and elements of $$W$$ have the form $$(g,x)$$ with $$g \in P$$, $$x \in A^d$$, and $$x = (x_1,\ldots,x_d)$$ with $$x_i \in A$$. In general, for $$x \in A^d$$, we denote the $$i$$-th component of $$x$$ by $$x_i$$. The action of $$P$$ on $$A^d$$ in $$W$$ is given by: $$(g,1)^{-1}(1,(x_1,\ldots,x_d))(g,1) = (1,(x_{1^{g^{-1}}},\ldots, x_{d^{g^{-1}}})).$$

Hence, for $$g \in P$$, $$x,z\in A^d$$, we have $$$$\label{conjeqn} (1,z)^{-1}(g,x)(1,z) = (g,y), \ \mathrm{where}\ y_i = z^{-1}_{i^{g^{-1}}}x_iz_i \ \mathrm{for}\ 1 \le i \le d.$$$$

$$\textbf{Theorem}$$ With the above notation, let $$x^{(1)},\ldots,x^{(k)}$$ be representatives of the conjugacy classes of $$A$$. Fix an element $$g \in P$$, and let $$r_1,r_2,\ldots,r_s$$ be representatives of the cycles of $$g$$ in its action on $$\{ 1\ldots d\}$$. Then $$R := \{\,(g,x) \mid x_i = 1 \ \mathrm{for}\ i \not\in \{r_1,\ldots,r_s\},\, x_i \in \{x^{(1)},\ldots,x^{(k)}\}\ \mathrm{for}\ i \in \{r_1,\ldots,r_s\}\,\}$$ is a set of representatives of the $$A^d$$-classes of elements of $$W$$ of the form $$(g,x)$$ for $$x \in A^d$$.

The size of the $$A^d$$-class $$R_{gx}$$ containing $$(g,x)$$ is the product of the sizes of the classes of the $$x_i$$ in $$A$$ with $$i \in \{r_1,\ldots,r_d\}$$ with the lengths of the cycles of $$g$$ that contain $$r_1,\ldots,r_d$$.

The above describes the $$A^d$$-classes of $$W$$; i.e. the orbits under conjugation by the subgroup $$A^d$$ of $$W$$. To get the classes of $$W$$ itself, first of all we only consider those $$R_{gx}$$ as $$g$$ ranges over a set of class representatives of $$P$$. Then, for each such $$g \in W$$, the $$A^d$$-class representatives $$R_{gx}$$ are fused under the action of $$C_P(g)$$. This action is the same as the action of $$C_P(g)$$ on the cycles of $$g$$, so is easy to calculate.

Then the size of a $$W$$-class $$R_{gx}$$ is the product of the size of the class of $$g$$ in $$P$$ with the sum of the the sizes of the $$A^d$$-classes $$R_{gx'}$$ that lie in the orbit under the action of $$C_P(g)$$.

• Thank you Derek, I see how an element $(g, x)$ is conjugate to something in the set $R$ you define. If I write $$R_g := \{\,(g,x) \mid x_i = 1 \ \mathrm{for}\ i \not\in \{r_1,\ldots,r_s\},\, x_i \in \{x^{(1)},\ldots,x^{(k)}\}\ \mathrm{for}\ i \in \{r_1,\ldots,r_s\}\,\}$$ is there a way to describe a subset of $$\bigcup_{g \in G}R_g$$ which contains only one representative from each conjugacy class? Also is there a way to determine how many elements are in a each conjugacy class? – Chris Russell Oct 10 '18 at 9:32
• My answer was incomplete because it only described the $A^d$-classes. Calculating the $G$-classes from thsi information is not difficult - I have added some further explanation. – Derek Holt Oct 10 '18 at 10:48
• I think I get the idea of how it works now but I'm not sure what $G$ is in the second last paragraph of your updated answer, which is confusing me. In one place I'm interpreting that $A^d$ is a subgroup of $G$ but in another $g \in G$ where I thought $g$ is an element of $P$. – Chris Russell Oct 10 '18 at 11:29
• Sorry $G$ should be $W$. I don't have a $G$. I have $W = A^d \rtimes P$. Of course $P=S_n$ in your question. – Derek Holt Oct 10 '18 at 12:10