Conjugacy class size and simple group For a finite group $G$, it is denoted by $N(G)$ the set of conjugacy class sizes. Let $G$ be a finite group and $H$ be a finite non-abelian simple group. Is it possible that $N(G)=N(H)$? In fact, I guess that $G$ should be simple but I cannot prove it. Can you help me?

Let $H = A_5$, the finite simple group of order $60$. Let $G = A_5 \times C_2$ the direct product of $A_5$ with a cyclic group of order $2$. Then the set of conjugacy class sizes of $H$, i.e. $N(H)$ is equal to $\{1,12,15,20\}$. The set $N(G)$ of conjugacy class sizes of $G$ is also equal to $\{1,12,15,20\}$. So $N(G)=N(H)$, and the answer to your question is yes, it is possible to have $N(G)=N(H)$ without $G$ being simple.
