Exercise 7.3.12

Proof.

(a)

Let $k\in G$. Then $$\mathit{kN}{k}^{-1}=k\left(\underset{g\in G}{\bigcap }\mathit{gH}{g}^{-1}\right)k^{-1}=\underset{g\in G}{\bigcap }(\mathit{kg})H{(\mathit{kg})}^{-1}=\underset{u\in G}{\bigcap }\mathit{uH}{u}^{-1}=N,$$

thus $N◃G$.

(b)

$H=\mathit{eH}{e}^{-1}\supset \underset{g\in G}{\bigcap <!--\; FUNCTION\; APPLICATION\; -->}\mathit{gH}{g}^{-1}=N$, so $N\subset H\subset G$.

If any subgroup $M$ of $H$ is normal in G, then for all $g\in G$, $\mathit{gM}{g}^{-1}=M$, therefore $M=\underset{g\in G}{\bigcap <!--\; FUNCTION\; APPLICATION\; -->}\mathit{gM}{g}^{-1}\subset \underset{g\in G}{\bigcap }\mathit{gH}{g}^{-1}=N$.

Conclusion: ${\mathrm{Core}}_G(H)=\underset{g\in G}{\bigcap <!--\; FUNCTION\; APPLICATION\; -->}\mathit{gH}{g}^{-1}$ is the largest subgroup of $H$ normal in $G$.