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

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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.

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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.