If $G \le S_n$ then $G/Z(G) \hookrightarrow S_{n^2}$

Question

Claim : If $G \le S_n$ then $G/Z(G) \hookrightarrow S_{n^2}$

$Z(G)$ means center of group $G$ and $\le $ used for subgroup, $ \hookrightarrow$ is used for embedding.

Question : How to prove the above claim?

Reference : Page no 24

Answer 1

It follows from the proof of the proposition with $H=G$, so $C_G(H)=Z(G)$. We know that $G \le {\rm Sym}(\Omega)$ with $|\Omega| = n$. There is an induced action of $G$ on $\Omega \times \Omega$.

Let $\Pi_1,\Pi_2,\ldots,\Pi_r$ be the orbits of $Z(G)$ on $\Omega \times \Omega$. Then the induced action of $G/Z(G)$ on the set $\{\Pi_1,\ldots,\Pi_r\}$ is faithful, and $r \le n^2$.