# Can these two generalizations of Polya Enumeration be combined?

Let $X$, $Y$ be sets, let $G$ be a group which acts on $X$ and let $H$, $K$ be groups which act on $Y$. Denote the set of functions from $X$ to $Y$ by $X^Y$. I will use $f$ for functions in $X^Y$ and $g, h, k$ for elements of $G$, $H$, $K$ - respectively.

As I understand it, the Polya Enumeration Theorem allows us to count the number of orbits of functions in $X^Y$ up to equivalence via the action of $G$ by:

$g \cdot f(x) = f(x ^ g)$ for all $x \in X$.

There is also a generalization of Polya Enumeration Theorem (Frank Harray and Ed Palmer call it the Power Group Enumeration Theorem) which allows us to count functions in $X^Y$ up to equivalence via the following action of the direct product $G \times H$:

$(g, h) \cdot f(x) = h \cdot f(x ^ g)$ for all $x \in X$.

The orbits of the wreath product $G \wr K = G \ltimes K^d$ (where $d = |X|$ is the degree of $G$) can be counted with respect to the following action

$(g, (k_1, \dots, k_d)) \cdot (y_1, \dots, y_d) = (k_1 \cdot y_{1^{g^{-1}}}, \dots, k_d \cdot y_{d^{g^{-1}}})$

see Enumeration under two representations of the wreath product by Palmer and Robinson. (Thanks to Max Alekseyev for pointing this out for me in a previous question). Now I want to ask whether it is possible to count the orbits of $G \times H \ltimes K^d$ with respect to the following action:

$(g, h, (k_1, \dots, k_d)) \cdot (y_1, \dots, y_d) = (k_1 \cdot (h \cdot y_{1^{g^{-1}}}), \dots, k_d\cdot (h \cdot y_{d^{g^{-1}}}))$

in other words, this is the action which has the $d$ factors of $K^d$ acting on the corresponding $d$ entries of a function in $X^Y$ whilst also having $H$ act on all factors simultaneously.

• The action looks similar to the one in Theorem 1 in the paper Enumeration of Linear Codes by Applying Methods from Algebraic Combinatorics by Fripertinger. Perhaps, their method can be adjusted for your problem. Sep 7 '18 at 17:48
• This paper is helpful for my problem and also relatively easy to read for me. Thank you again Max! Sep 11 '18 at 11:00