The exercise is as follows:
Prove that the following polynomial takes integer values when evaluated at integer points: $$f(x) = \frac{1}{|G|} \sum_m a(m) x^{\frac{|G|}{m}}$$ where $a(m) = |\{g \in G \mid o(g) = m\}|$
He suggests using that the function $\theta_n(g) = n^{m(g)}$ is a character of $G$, where $m(g) = \langle \chi_{\langle g \rangle}, 1_{\langle g \rangle} \rangle$ and $\chi = (1_H)^G$, for some subgroup $H$ of $G$.
This makes it obvious to me that $f(x)$ is going to be some inner product of some character and $\theta_x$. That said, it’s unclear why $m(g) = |G|/m$ for some $m$, and, most importantly, that, for every $m$ which has an element in $G$ of that order, we can find some $g$ that makes it work.
$m(g)$ is the number of orbits of $\langle g \rangle$ under the multiplication action on the cosets of $H$ in $G$, so I have to find a proper $H$, which I couldn’t do…
I’m also unsure about what to do with $a(m)$. The inclusion-exclusion principle yields that it’s a $\mathbb{Z}$-linear combination of $R_n(1)$, with $R_n(1) = |\{x \in G \mid x^n = 1\}|$, as $n$ varies through the divisors of $m$. That said, I don’t know if this helps at all…
Does anyone have any hints?
Thanks in advance!
Best Answer
I don't know why my answer was deleted, but here is another try: The claim follows by choosing $H=1$ in Exercise 5.17. (If this answer is deleted again, I won't try again, and step back from MSE in general.)
EDIT (according to the comment of the OP): I believe, it is a general theorem that a polynomial is integer-valued on integers if it is integer-valued on the positive integers. You can express such a polynomial uniquely as an integral linear combination of binomial "polynomials" $\binom{x}{k}$.