Recall that there are $$\frac{n!}{\prod^n\_{i = 1}i^{k\_i}k\_i!}$$ permutations in $S\_n$ which have cycle structure $(k\_1, \dots, k\_n)$, that is to say they have exactly $k\_1$ 1-cycles, $k\_2$ 2-cycles, ... and $k\_n$ n-cycles. The cycle index of $S\_m \times S\_n$ acting on the set $\{1, \dots, n\} \times \{1, \dots, n\}$ of indices of entries of an $m \times n$ matrices by row permutations and column permutations can be written as


$$Z(S\_m \times S\_n; s\_1, \dots, s\_x) := \frac{1}{m!n!} \cdot \sum\_{k\_1 + 2k\_2 + .. mk\_m = m \\ l\_1 + 2l\_2 + .. nl\_n=n} \frac{m!n!}{\prod^m\_{i = 1}i^{k\_i}k\_i! \cdot \prod^n\_{j = 1}j^{l\_j}l\_j! } \cdot \prod\_{i = 1}^m \prod\_{j = 1}^n s\_{\mathrm{lcm}(i,j)}^{\gcd(i,j) \cdot k\_i \cdot l\_j}$$


In the square case, does anyone know how to calculate the cycle index of the extension $(S\_n \times S\_n) \rtimes C\_2$ where $C\_2$ acts by transposing? I.e. extending by the permutation which sends $(i,j)$ to $(j,i)$ for all $1 \leq i,j \leq n$.


Equivalently, if $\sigma \in S\_n$ has cycle structure $(k\_1, \dots, k\_n)$ and $\rho$ has cycle structure $(l\_1, \dots, l\_n)$ then $(\sigma, \rho, 1)$ has cycle index
$$z((\sigma, \rho, 1)) := \prod\_{i = 1}^n \prod\_{j = 1}^n s\_{\mathrm{lcm}(i,j)}^{\gcd(i,j) \cdot k\_i \cdot l\_j}$$
but if $c$ is the generator of the $C\_2$ factor then what is the cycle index of $(\sigma, \rho, c)$?



Cycle index of $(S_n \times S_n) \rtimes C_2$ acting on matrix indices by row/column permutation and transposition
The question is about cycle types of elements of the wreath product $G = (S\_n \times S\_n) \ltimes C\_2 \cong S\_n \wr C\_2$ in its product action on $\{1,\ldots, n\} \times \{1,\ldots, n\}$. 


The permutation $(\sigma, \rho, c) \in G$ is conjugate, by $(\rho^{-1},\mathrm{id}\_{S\_n},1)$, to $g = (\rho\sigma, \mathrm{id}\_{S\_n}, c)$. Suppose 


$$\rho\sigma = (x\_1,\ldots, x\_a)(y\_1,\ldots, y\_b) \ldots .$$ 


Let $X = \{x\_1,\ldots, x\_a\}$ and $Y = \{y\_1,\ldots, y\_b\}$. Take all $x$ indices modulo $a$ and all $y$ indices modulo $b$. With this convention, $(x\_i,y\_j)g = (y\_j,x\_{i+1})$ and $(y\_j,x\_i)g = (x\_i,y\_{j+1})$. Hence 


$$(x\_i,y\_j)g^2 = (x\_{i+1},y\_{j+1})$$ 


and so $(x\_1,y\_1)$ is in a cycle of $g^2$ of length $\mathrm{lcm}(a,b)$. Since $g$ swaps the disjoint sets $X \times Y$ and $Y \times X$, the cycle of $g$ containing $(x\_1,y\_1)$ has length $2\mathrm{lcm}(a,b)$. Therefore $g$ acts on $X \times Y \cup Y \times X$ with cycle type 


$$\bigl((2\mathrm{lcm}(a,b))^{ab/\mathrm{lcm}(a,b)}\bigr).$$


For the action on $X \times X$ things are slightly fiddlier: if $a$ is even then there are $a/2$ cycles each of length $2a$; if $a$ is odd then the cycle type is


$$\bigl( (2a)^{(a-1)/2}, a \bigr).$$


Putting these together determines the cycle type of $g$ and hence the cycle index of $G$.


