# 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 Quotient group (?). Then, the 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.

## 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:

Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 The quotient map from $G$ to $K$ sends conjugate elements to conjugate elements, hence induces a set map $q_1:C(G) \to C(K)$
2 The set map $q_1$ from Step (1) is surjective.
3 If $H$ is nontrivial, $q_1$ is not injective.
4 $|C(K)| \le |C(G)|$ and inequality is strict if $G$ is finite and $H$ is nontrivial.