Question

This number is divisible by the order of the subgroup http://arxiv.org/abs/1205.2824.

The proof is short but non-trivial. Is this fact new or is it known for a long time?

Answer 1

Here is an easy character-theoretic proof of the fact that given a subgroup $H$ of a finite group $G$ and a positive integer $k$, the number of elements $y \in G$ such that $y^k \in H$ is divisible by $|H|$. Let $\theta_k$ be the class function on $G$ defined by $\theta_k(x)$ = |{ $y \in G \mid y^k = x$ }|. It is well known that this class function is a generalized character. (In other words, it is a $\Bbb Z$-linear combination of irreducible characters.) The number of interest here is $\sum_{x \in H} \theta_k(x)$, which is equal to $|H|[(\theta_k)_H,1_H]$. This is clearly divisible by $|H|$ since the second factor is an integer because $\theta_k$ is a generalized character.

In fact, the coefficient of an irreducible character $\chi$ in $\theta_k$ is the integer I called $\nu_k(\chi)$ in my character theory book. For $k = 2$, this is the famous Frobenius-Schur indicator, whose value lies in the set {0,-1,1}. For other integers $k$, it is true that $\nu_k(\chi)$ is an integer, but there is no upper bound on its absolute value.