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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?

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Welcome to Mathoverflow! –  HJRW Jun 2 '12 at 10:12

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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.

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Thank you, Marty! Your proof is shorter than ours but less elementary. However, I would be happy if someone provides a reference proving that the fact is known. Actually, the paper cited in the question contains a more general fact (Corollary 5). <i>Suppose that $H$ is a subgroup of a group $G$ and $W$ is a subgroup (or a subset) of a free group $F$. Then the number of homomorphisms $f\colon F\to G$ such that $f(W)\subseteq H$ is divisible by $|H|$.</i> (Taking $F=\Bbb Z$, we obtain the statement from the question.) Does there exist a short character-theoretic proof for this too? –  Anton Klyachko Jun 4 '12 at 23:41
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It seems to be easy to prove via character theory that if |W| = 1, then number of homomorphisms f such that f(W) <= H is a multiple of |H|. I don't see a proof along these lines if W has cardinality exceeding 1. –  Marty Isaacs Jun 5 '12 at 22:12
    
Is this because the number of tuples $(g_1,\dots,g_n)$ such that $w(g_1,\dots,g_n)=x$ is a generalised character $\theta(x)$ for ANY word $w\in F$ or this is not so easy? –  Anton Klyachko Jun 6 '12 at 17:59

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