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Let p be a prime, and let G be a non-abelian group of order . We want to find the number of conjugacy classes of G.

First we prove a lemma: Z(G) has order p.

Proof: We know that since G is a non-trivial p-group, then . Since , by Lagrange’s Theorem, .

Case 1) . We are done.

Case 2) . Then . Thus is cyclic which implies that G is abelian. (contradiction).

Case 3) . This means that the entire group G is abelian. (contradiction).

Next, let be the distinct conjugacy classes of G.

, where .

Then by the Class Equation, we have .

If , then , which means .

If , then . Since , thus . Thus we have . Since is a subgroup of , Lagrange’s Theorem forces . Thus .

By the Class Equation, we thus have , which leads us to .

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