Conjugacy classes intersecting subgroups of finite groups

Question

I figured out the first part of this years ago, but completely forget how I did it. I looked at the second, but don't think I figured it out.

This I am sure is true, but don't remember why. Suppose that G is a finite group of size $n$, and H is a normal subgroup with |G/H| = $k$. Then at least $\frac{1}{k}$ of the conjugacy classes of G are within H.

This I don't know the answer to. If H is allowed to be an arbitrary subgroup, must H intersect at least $\frac{1}{k}$ of the conjugacy classes?

An example of a simple consequence of the first statement is that if we look at $S_n$ and $A_n$ is that at least half the partitions of $n$ have an even number of even parts.

Answer 1

Let $r$ be the number of conjugacy classes in $G$. The action of $G$ on itself by conjugation gives, via the Cauchy-Frobenius formula, $r=\frac{1}{|G|}\sum |C(g)|$, where $g$ ranges over $G$. This action restricted to $H$ gives the number of $G$-conjugacy classes in $H$ as $\frac{1}{|G|}\sum |C_H(g)|$, and from the fact $|C_H(g)|\ge \frac{1}{k}|C(g)|$ the first part of your problem follows.

EDIT: As pointed out by Sergei Ivanov below, this argument also shows the second part is true as well.

Answer 2

Let me explain the "application", that "at least half the partitions of n have an even number of even parts." This fact goes back to Euler: the difference is equal to the number of partitions into distinct odd parts.

Denote by $p_o(n)$ and $p_e(n)$ the number of partitions of $n$ with odd/even number of even parts. Then $$ \sum_n (p_e(n)-p_o(n))t^n = \prod_i \frac{1}{(1+t^{2i})(1-t^{2i-1})} = \prod_i \frac{1+t^i}{1+t^{2i}} = \prod_i (1+t^{2i-1})$$ There is also a bijective proof of this - I will leave it to you to figure this out.

Answer 3

If we let $k(G)$ denote the number of conjugacy classes of the finite group $G$, then for any subgroup $H$ of $G$ (normal or not), a Theorem of P.X. Gallagher states that $[G:H]^{-1}k(H) \leq k(G) \leq [G:H]k(H)$ (this has probably been discovered and rediscovered many times). I find that the easiest way to see it is using irreducible complex characters. Of course, $k(G)$ is also the number of complex irreducible characters of $G$, and likewise for $H$. For each irreducible character $\chi$ of $G$, there is an irreducible character $\mu$ of $H$ such that $\mu$ occurs with non-zero multiplicity in the restriction of $\chi$ to $H$. By Frobenius reciprocity, $\chi$ is an irreducible constituent of the character of $G$ induced from the character $\mu$ of $H$. On the other hand, using Frobenius reciprocity again, each irreducible constituent of $\mu$ induced to $G$ must have degree at least $\mu(1).$ Thus there are (even including multiplicities) at most $[G:H]$ irreducible constituents of $\mu$ induced to $G$ Since this is true for each irreducible character of $H$, and since each irreducible character of $G$ must appear with non-zero multiplicity in at least one such character, we have $k(G) \leq [G:H]k(H).$ Going in the other direction, if $\chi$ is an irreducible character of $G$ and $\mu$ is an irreducible constituent of te restriction of $\chi$ to $H$, then we have $\chi(1) \leq [G:H]\mu(1)$, since $\chi$ occurs as a constituent of $\mu$ induced to $G$. Thus $\mu(1) \geq \frac{\chi(1)}{[G:H]}$. Hence there are at most $[G:H]$ distinct irreducible constituents of the restriction of $\chi$ to $H$. Since each irreducible character of $H$ occurs as a constituent of some such irreducible character of $G$ (consider, for example, the restriction of the regular character), we have $k(H) \leq [G:H] k(G).$