## 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 $\boxed{n=p^2+p-1}$.