Exercise 4.3.6 (Class equation for a non-abelian group of order $15$)

Proof. Let $G$ be a non-abelian group of order $15$. Then the center $Z=Z(G)$ is a proper subgroup, and by Lagrange Theorem, $|Z|\in \{1,3,5\}$. If $|Z|=3$, then $|G∕Z|=5$ is a prime, so $G∕Z$ is cyclic, and by Exercise 3.1.36, $G$ is abelian, in contradiction with the hypothesis.

Similarly, if $|Z|=5$, then $|G∕Z|=3$, thus $G∕Z$ is cyclic and $G$ is abelian.

This shows that

Let $g_1,g_2,\dots ,g_r$ be representatives of the distinct conjugacy classes of $G$ not contained in the center $Z(G)$. The class equation (Theorem 7) gives

For every $i\in [[1,r]]$, $|G:C_G(g_i)|$ divides $15$, and $|G:C_G(g_i)|\ne 1$, otherwise $g_i\in Z$. Moreover, since $⟨g_i⟩\le C_G(g_i)$, where $g_i\ne 1$, we obtain $|C_G(g_i)>1$, thus $|G:C_G(g_i)|<15$. This shows that

So the class equation (1) is of the form

where $\lambda$ is the number of $i\in [[1,r]]$ such that $|G:C_G(g_i)|=3$ and $\mu$ is the number of $i\in [[1,r]]$ such that $|G:C_G(g_i)|=5$. Since $\lambda \ge 0$ and $\mu \ge 0$, the only solution of $14=3\lambda +5\mu$ is $\lambda =3,\mu =1$, which gives

There is at most one possible class equation for $G$. □