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.



Conjugacy classes intersecting subgroups of finite groups
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.


