Exercise 7.2.8

Proof.

(a)

If $x\in H,\mathit{xH}{x}^{-1}=H$, so $H\subset N_G(H)$.

$\text{∙}$

$\mathit{eH}{e}^{-1}=H$, thus $e\in N_G(H)\ne \varnothing$.

$\text{∙}$

If $x,y\in N_G(H)$, then $(\mathit{xy})H{(\mathit{xy})}^{-1}=x(\mathit{yH}{y}^{-1})x^{-1}=\mathit{xH}{x}^{-1}=H$, thus $\mathit{xy}\in N_G(H)$.

$\text{∙}$

If $x\in N_G(H)$, then $\mathit{xH}{x}^{-1}=H$, thus $\mathit{xH}=\mathit{Hx}$, and $H=x^{-1}\mathit{Hx}$: $x^{-1}\in N_G(H)$.

$N_G(H)$ is a subgroup of $G$.

(b)

For all $g\in N_G(H)$, $\mathit{gH}{g}^{-1}=H$, so $H◃N_G(H)$.

(c)

Let $N$ be a subgroup of $G$, $H\subset N\subset G$.

$H◃N⟺\forall <!--\; FUNCTION\; APPLICATION\; -->g\in N,\mathit{gH}{g}^{-1}=H⟺\forall <!--\; FUNCTION\; APPLICATION\; -->g\in N,g\in N_G(H)⟺N\subset N_G(H).$ $H$ is normal in $N_G(H)$, and every subgroup $G$ in which $H$ is normal is contained in $N_G(H)$, so $N_G(H)$ is the largest subgroup of $G$ in which $H$ is normal.

(d)

:

$\text{∙}$

If $H$ is normal in $G$, then every element of $G$ is in the normalizer of $H$ in $G$, therefore $G\subset N_G(H)$. As $N_G(H)\subset G$, $N_G(H)=G$.

$\text{∙}$

If $G=N_G(H)$, then every element $g\in G$ is in $N_G(H)$, and so satisfies $\mathit{gH}{g}^{-1}=H$, so $H$ is a normal subgroup of $G$. $$G=N_G(H)⟺H◃G.$$