Examples of groups satisfying that its number of conjugacy classes is less that the number of conjugacy classes of centain subgroup If $G$ is a group define $r(G)$ to be the number of conjugacy classes of $G$. I'm trying to construct simple examples of groups $G$ with at least one subgroup $H$ such that $r(H)>r(G)$. I won't find simple examples of this, so I'd appreciate some hints.
 A: Let $D_n$ denote the dihedral group of order $2n$ (I'm specifying, since the notation is never consistent); the subgroup of rotations $C_n \subset D_n$ has $n$ conjugacy classes, so we count the conjugacy classes of $D_n$.
If $n$ is odd, then reflections form a single conjugacy class, and the rotations pair up, excepting the identity, so there are $\frac{n+3}{2}$ conjugacy classes. This is less than $n$ when $n \ge 5$.
If $n$ is even, then reflections form two conjugacy classes, and the rotations pair up, excepting the identity and $180^\circ$ rotation, so there are $\frac{n+6}{2}$ conjugacy classes. This is less than $n$ when $n \ge 8$.
A: If you can find "large" abelian subgroups, this will work -- they just need to have more elements than the full group has conjugacy classes.
It turns out this is kind of hard for symmetric groups, at least if I'm being lazy. The symmetric group $S_{n^2}$ of degree $n^2$ has an abelian subgroup of order $n^n$ ($n$ copies of the cyclic group of order $n$). Consulting OEIS, it turns out $S_{16}$ has only $231$ conjugacy classes, while $4^4 = 256$, yielding an example. A similar thing happens with $S_{25}$, which has $1958$ conjugacy classes and an abelian subgroup of order $5^5 = 3125$. It would be interesting to know whether this happens for all $S_{n^2}$ where $n \geq 4$, and I suspect it does, but I have no idea.
