Let $G$ be a group and $H,K \le G$,then - If $H \le N_{G}(K)$ then $HK \le G$ - If $H \le K$ then $ C_{G}(K) \le C_{G}(H)$

Question

Let $G$ be a group and $H,K \le G$,then show that:

• If $H \le N_{G}(K)$ then $HK \le G$
• If $H \le K$ then $ C_{G}(K) \le C_{G}(H)$

Where $C_{G}(H)$ is the centralizer of $H$ defined as:

$${\displaystyle \mathrm {C} _{G}(H)=\{g\in G\mid gh=hg{\text{ for all }}h\in H\}.}$$

And $N_{G}(K)$ is the normalizer of $K$ defined as:

$${\displaystyle \mathrm {N} _{G}(K)=\{g\in G\mid gK=Kg\}.}$$

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For the first part I have that:

$N_G(K)$ is the normalizer of $K \le G$. It's the set of all $g$ such that $gKg^{-1} = K$.

Since $H \subset N_G(K)$, then for all $h \in H$ we know that they commute with $K$ like the $g$. Hence for all $h$ in $H$ we have that $hK = Kh, $ which means $HK = KH$.

On the other hand from this link we see that $HK \le G$,as desired.

The second part is the part that I don't know how to prove,I know that for $H \le K$ we have that $C_G(H)<N_G(H)$,but does that help?

Answer 1

Take an arbitrary element $c \in C_G(K)$. We want to show that $c \in C_G(H)$.

Note that $c \in C_G(K) \implies ck = kc, \forall k \in K$. But since $H \le K$, we have $ch = hc, \forall h \in H \implies c \in C_G(H)$.