Conjugacy Classes of non-abelian group of order p^3

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Let p be a prime, and let G be a non-abelian group of order p^3. 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 Z(G)\neq 1. Since Z(G)\trianglelefteq G, by Lagrange’s Theorem, |Z(G)|=p,p^2,\text{or }p^3.

Case 1) |Z(G)|=p. We are done.

Case 2) |Z(G)|=p^2. Then |G/Z(G)|=p^3/p=p. Thus G/Z(G) is cyclic which implies that G is abelian. (contradiction).

Case 3) |Z(G)|=p^3. This means that the entire group G is abelian. (contradiction).

Next, let O(x_1),\dots, O(x_n) be the distinct conjugacy classes of G.

O(x_i)=\{gx_i g^{-1}:g\in G\}, where C_G(x_i)=\{g\in G:gx_i=x_ig\}.

Then by the Class Equation, we have \displaystyle p^3=|G|=\sum_{i=1}^n [G:C_G(x_i)].

If x_i\in Z(G), then C_G(x_i)=G, which means [G:C_G(x_i)]=1.

If x_i \notin Z(G), then C_G(x_i)\neq G. Since x_i\in C_G(x_i), thus Z(G)\subsetneq C_G(x_i). Thus we have p=|Z(G)|<|C_G(x_i)|<|G|=p^3. Since C_G(x_i) is a subgroup of G, Lagrange’s Theorem forces |C_G(x_i)|=p^2. Thus [G:C_G(x_i)]=p^3/p^2=p.

By the Class Equation, we thus have p^3=p+(n-p)p, which leads us to \boxed{n=p^2+p-1}.

Author: mathtuition88

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