Exercise 2.2.6 (Conditions for $H \subseteq N_G(H)$, $H \subseteq C_G(H)$.)

Proof. Let $H$ be a subgroup of the group $G$.

(a)

Let $g$ be any element of $H$. Since $H$ is a subgroup of $G$, for all $h\in H$, $\mathit{gh}{g}^{-1}\in H$, so $$\mathit{gH}{g}^{-1}\subset H.$$

Since $g^{-1}\in H$, we have similarly $g^{-1}\mathit{Hg}\subset H$. Therefore $H\subset \mathit{gH}{g}^{-1}$, so

$$\mathit{gH}{g}^{-1}=H.$$

This shows that $g\in N_G(H)$, for every $g\in H$, thus

$$H\le N_G(H).$$

This is not necessarily true if $H$ is not a subgroup. As a counterexample, consider the group $G=S_3$ and $H=\{(12),(13)\}$. Since

$$(13)(12){(13)}^{-1}=(32)\notin H,$$

this shows that $(13)\notin N_G(H)$, so

$$H⊈N_G(H).$$

(b)

Suppose that $H$ is an abelian subgroup of $G$. Let $h$ be any element of $H$. Then for any $k\in H$, $\mathit{hk}=\mathit{kh}$, so $h\in C_G(H)$. This shows that $$H\le C_G(H)\text{(if }H\text{ is abelian)}.$$

Conversely, suppose that $H\le C_G(H)$. If $h,k$ are elements of $H$, then $k\in H\subseteq C_G(H)$, so $k\in C_G(H)$, where $h\in H$, thus $\mathit{hk}=\mathit{kh}$. This shows that $H$ is abelian.

$H\le C_G(H)$ if and only if $H$ is abelian.