Let G acts on $\Omega$. Show that $f_{G,\Omega}(x)$ is an integer whenever x is a positive integer.

Question

Let G acts on $\Omega$, where G is a finite group and $\Omega$ is a finite set. For $g\in G$, write $a(g)$ to denote the number of orbits of <$g$> on $\Omega$. Let $f_{G,\Omega}$ be the polynomial defined by the formula

$$f_{G,\Omega}(x)=\dfrac{1}{|G|}\sum_{g\in G} x^{a(g)}. $$

Show that $f_{G,\Omega}(x)$ is an integer whenever x is a positive integer.

$\mathbf {NOTE:}$ The assertion remains true if the word "positive" is dropped.Also, observe what happens if we take $\Omega=G$ to be a cyclic group of prime order p. In that case, $f_{G,\Omega}(x)=(1/p)((p-1)x+x^p)$ , and the problem yields that $x^p-x$ is divisible by p for integers x.

This is the problem 4.8 from the "Algebra: a graduate course" of Martin Isaacs. I have searched the problem but I couldn't find anywhere (It should have been a duplicate but I didn't see where the topic is).

I have tried to take $\Omega=G$ to be any group,then for all $g\in G$, $$a(g)=\dfrac {|G|}{ord(g)}$$ (It can be seen by using "associated permutation character" of g and Cauchy-Frobenius Theorem)

and

$$f_{G,\Omega}(x)=\dfrac{1}{|G|}\sum_{g\in G} x^{\tfrac {|G|}{ord(g)}}$$

Let's say $|G|=n$, the problem implies that for all groups $G$ of order $n$

the polynomial$$\sum_{g\in G} x^{\tfrac {|G|}{ord(g)}}$$ is always zero in modulo $n$. Therefore, all elements of residue class of modulo $n$ should be a root of this polynomial and $$\sum_{g\in G} x^{\tfrac {|G|}{ord(g)}}=\prod_{i=0}^{n-1} (x-i)\tag{not sure?!} $$

But how can be the polynomial fixed while the group structure is changing with same order?

(We know that for large enough $n$ there are many different group structures with order $n$)

And is there any special name for this polynomial $(f_{G,\Omega}(x))$ satisfying this property which I couldn't prove even in the case $\Omega=G$ ?

Answer 1

if $G$ acts on a finite set $Z$ then $$\sum_{g\in G}\text{no. of $g$-fixed points in $Z$}=\sum_{z\in Z}|\text{Stab}(z)|=G\times\text{no. of $G$-orbits in $Z$}$$ is an integer divisible by $G$. Take $$Z=\text{maps}(\Omega\to\{1,2,\dots,x\})$$ with the $G$-action $(g\cdot z)(\omega):=z(g^{-1}\cdot\omega)$. Then (for any $g\in G$) the no. of $g$-fixed points in $Z$ is exactly $x^{a(g)}$, and you get your result.

(I have a feeling that this solution is too convoluted and that there is a way simpler one)