Let $H,K\subset G$ be group and subgroups, if $H$ is a subset of the normalizer of $K$ in $G$, why is $H\cap K$ normal in H? Let $H,K\subset G$ be group and subgroups, if $H$ is a subset of the normalizer of $K$ in $G$, why is $H\cap K$ normal in H?

 A: The normalizer $N_{G}(K)=\{g \in G\, | \, gKg^{-1}=K\}$ and we are given that $H \subset N_{G}(K)$.
We need to show that $h(H \cap K)h^{-1} \subseteq H \cap K$ for all $h \in H$. 
Let $x\in H \cap K$. Consider $hxh^{-1}$. Since $x \in K$ and $h \in H \subset N_{G}(K)$ so $hxh^{-1}=y$ for some $y \in K$. Thus $hxh^{-1} \in K$ and by closure under the operation, $hxh^{-1} \in H$ (since all the elements are in $H$ and $H$ is a subgroup). Thus $hxh^{-1} \in H \cap K$. 
This shows that $$\forall \, h \in H \quad  h(H \cap K)h^{-1} \subseteq H \cap K.$$
Thus $H \cap K \unlhd H$. 
A: Let $g$ be any element of $H\cap K$. We want to show that given any $h\in H$, the element $hgh^{-1}$ is in $H\cap K$. 


*

*Since $g\in H$ by definition, and $h\in H$ by assumption, we know that $hgh^{-1}\in H$ since $H$ is a subgroup.

*Since $g\in K$ by definition, and $H$ is in the normalizer of $K$, we know that $hgh^{-1}\in K$.


Thus $hgh^{-1}$ is in both $H$ and $K$, so $H\cap K$ is normal in $H$.
A: $H\cap K$ is clearly a subgroup of $H$. We know that $H\subset N(K)$, where $N(K)$ is the normalizer of $K$ in $G$ which means that $\forall h\in H$, we have $hKh^{-1}=K$. 
Now to check that $H\cap K$ is normal in $H$, it suffices to show that $\forall h\in H$, $h(H\cap K)h^{-1}=H\cap K$ which is obviously the case.
A: A short proof:
$K$ is normal in $N_G(K)$, whch means that for any $x$ in the normaliser, $\;xK=Kx$.
Now, if $H \subset N_G(K)$, we have in particular $xK=Kx$ for all $x\in H$. We have to prove that
$$x(H\cap K)=(H\cap K)x\quad\forall x\in H$$
Note that we also have $xH=Hx$ since $x\in H$, and
$x(H\cap K)=(xH)\cap(xK)$, similarly $(H\cap K)x=(Hx)\cap(Kx)$, so it comes down to the obvious 
$$(A=A')\wedge(B=B')\implies A\cap B=A'\cap B'.$$
