Number of conjugacy classes in a quotient is less than or equal to number of conjugacy classes of group

Statement
Suppose $$G$$ is a group, $$H$$ is a normal subgroup, and $$K = G/H$$ is the fact about::quotient group. Then, the fact about::number of conjugacy classes of $$K$$ is less than or equal to the number of conjugacy classes of $$G$$.

When $$G$$ is a finite group, both numbers are finite and this can be thought of as a comparison of finite numbers. When $$G$$ is infinite, one or both numbers could potentially be infinite and the statement can be viewed in terms of comparisons of infinite cardinals.

Finally, the number of conjugacy classes in $$K$$ is strictly less than the number in $$G$$ if $$G$$ is finite and $$H$$ is nontrivial.

Similar facts

 * Degrees of irreducible representations of quotient group are contained in degrees of irreducible representations of group: Note that for finite groups, via the relation number of irreducible representations equals number of conjugacy classes, we already know that the number of irreducible representations of the quotient group is less than or equal to the number of irreducible representations of the whole group. The containment relation says something stronger.

Opposite facts

 * Number of conjugacy classes in a subgroup may be more than in the whole group
 * Commuting fraction in quotient group is at least as much as in whole group, which, put another way, states that the number of conjuacy classes in a quotient group is bounded from below by the number of conjugacy classes in the whole group divided by the order of the normal subgroup by which we quotiented.

Facts similar to the proof

 * Inner implies quotient-pullbackable, quotient-pullbackable implies inner

Proof
Given: A group $$G$$, a normal subgroup $$H$$, quotient $$K = G/H$$, with quotient map $$q:G \to K$$. Let $$C(G)$$ and $$C(K)$$ be the sets of conjugacy classes in $$G$$ and $$K$$ respectively.

To prove: $$|C(K)| \le |C(G)|$$, and if $$G$$ is finite and $$H$$ nontrivial, then $$|C(K)| < |C(G)|$$.

Proof: