Number of conjugacy classes in a subgroup may be more than in the whole group

Statement
It is possible to have a finite group $$G$$ and a subgroup $$H$$ of $$G$$ such that the fact about::number of conjugacy classes in $$H$$ is more than in $$G$$.

Similar facts

 * Commuting fraction in subgroup is at least as much as in whole group: This says that the number of conjugacy classes in the subgroup is at least as much as the number of conjugacy classes in the whole group divided by the index of the subgroup.

Opposite facts

 * Number of conjugacy classes in a subgroup of finite index is bounded by index times number of conjugacy classes in the whole group

Example of the dihedral group of degree five
The smallest pair of examples is where the group $$G$$ is $$D_{10}$$, the dihedral group of degree five and order ten, and $$H$$ is the subgroup is cyclic group:Z5. The group $$D_{10}$$ has four conjugacy classes: one of involutions, one identity element, and two conjugacy classes in the cyclic subgroup of order five. On the other hand, $$H$$ is an abelian group of order five hence has five conjugacy classes.

Other dihedral examples
More generally, for odd $$n$$, the dihedral group of order $$2n$$ and degree $$n$$ has $$(n + 3)/2$$ conjugacy classes, and the cyclic subgroup of order $$n$$ has $$n$$ conjugacy classes. For $$n \ge 5$$, the cyclic subgroup has more conjugacy classes than the whole group.

For even $$n$$, the dihedral group of order $$2n$$ has $$(n + 6)/2$$ conjugacy classes and the cyclic subgroup of order $$n$$ has $$n$$ conjugacy classes. For $$n \ge 8$$, the cyclic subgroup has more conjugacy classes than the whole group. The first example of this is cyclic group:Z8 in dihedral group:D16.