Let $H, K \leq G. $ Let $G' = [H,K]$ be a subgroup of $G$ which generated by all $[h,k]$ with $ h \in H , k \in K $. Show that:
$1)$ $H\subseteq C_{G}(K)$ iff $[H,K]=1$
$2)$ $H\subseteq N_{G}(K)$ iff $[H,K]\subseteq K$
$3)$ If $H,K$ normal subgroups of $G$ and $H \cap K = 1$ then $H\subseteq C_{G}(K)$.
My effort:
$1)$ $[H,K]=1$ so all $[h,k]=1$ so all the elements of $H,K$ commute.
$H \subset C_{G}(K)$ so if $h \in H$ then $h \in C_{G}(K)$ so $hk=kh$ for all $k \in K$ which is correct since elements of $H,K$ commute.
(Is that ok?)
$2)$
$3)$ $[H,K] \subset K$
$[H,K]^{-1} = [K,H] \subset H$ so $[H,K] \subset H$. So $[H,K] \subset H \cap K =1$ so all the elements of $H,K$ commute and we have the same as part $1$.
(Is that ok?)
1)seems fine, I would just add a conclusion such as "hence [h,k]=e $\forall h\in H,k\in K$".
For 2), assume $H\subset N_G(K)$. Let $h\in H , k\in K$ , we must show $[h,k]\in K$.
By assumption , $h\in N_G(K)$ so $hKh^{-1}=K$ so $hkh^{-1} = k' \in K$ so $hkh^{-1}k^{-1} = k'k^{-1} \in K$ since $K $ is a subgroup and thus closed to multiplication.
So indeed $[H,K]\subset K$.
Now assume $[H,K]\subset K$. Let $h\in H$. we want to show that $h\in N_G(K)$ , that is $hKh^{-1}=K$. But, for any $k \in K$ , $[h,k]=hkh^{-1}k^{-1} \in K$ by assumption, so $hkh^{-1}\in Kk = K$ so indeed $hKh^{-1}=K$ and we have $h\in N_G(K)$.
For 3) , by 1) we need to show that $[H,K] = 1$. So let $h\in H$ and $k\in K$. then $hkh^{-1} \in K$ since $K$ is normal. So $[h,k] =hkh^{-1}k^{-1}\in K $. On the other hand, $kh^{-1}k^{-1}\in H$ since $H$ is normal, and aain $[h,k] =hkh^{-1}k^{-1}\in H $. So we get $[h,k]\in H\cap K$. By assumption , $H\cap K=1$ and we are done.