Will the intersection of a subgroup and a normal subgroup create a normal subgroup for the main group If $G$ is a group and $H \leqslant G$ and $K \triangleleft G$, is $H \cap K \triangleleft G$?
I think the intersection of $H$ and $K$ is a subgroup, and if $K \triangleleft G$ is $H\cap K \triangleleft G$ but I couldn't prove it. This is all I have done so far:
$\forall x \in H \cap K \land \forall g \in G \Rightarrow gxg^{-1} \in H \cap K$ ?
$x \in H \cap K \Rightarrow x \in H \land x\in K$
$x \in K \land g \in G \Rightarrow gxg^{-1} \in K$
but i couldn't show it
$x \in H \land g \in G \Rightarrow gxg^{-1} \in H$
Give me a hint please
 A: No, $H\cap K$ might be not normal in $G$. For example take $G=D_4$, the dihedral group with $8$ elements. Letting $\rho$ be a rotation by angle $\frac{2\pi}{4}$ and $\epsilon$ any reflection we can take $K=\langle \rho^2,\epsilon\rangle\trianglelefteq G$ and $H=\langle \epsilon\rangle\leq G$. Then $H\cap K=\langle\epsilon\rangle$, and it is not normal in $D_4$. 
That being said, $H\cap K$ is normal in $H$. Indeed, letting $h\in H, x\in H\cap K$ we have $hxh^{-1}\in H$ because it is a product of elements in $H$, and we have $hxh^{-1}\in K$ because $K$ is normal in $G$. So indeed $hxh^{-1}\in H$. 
A: For every $a \in G$, we get:
\begin{alignat}{1}
a(H\cap K) &= \{ag\mid g\in H\cap K\} \\
&= \{ag\mid g\in H\wedge g\in K\} \\
&= \{ag\mid g\in H\}\cap \{ag\mid g\in K\} \\
&= aH\cap aK \\
\tag 1
\end{alignat}
and likewise for the right cosets. Therefore, if $K\unlhd G$:
\begin{alignat}{1}
a(H\cap K)\cap (H\cap K)a &= (aH\cap aK)\cap(H a\cap aK) \\
&= (aH\cap Ha)\cap aK \\
\tag 2
\end{alignat}
Now, $H \ntrianglelefteq G \Rightarrow \exists a'\in G\mid a'H\ne Ha' \Rightarrow a'H\cap Ha'\subsetneq a'H$, and thence, by $(2)$:
\begin{alignat}{1}
a'(H\cap K)\cap (H\cap K)a' &= (a'H\cap Ha')\cap a'K \subsetneq a'H\cap a'K=a'(H\cap K)\\
\tag 3
\end{alignat}
whence:
$$(H\cap K)a' \ne a'(H\cap K)$$
and finally $H\cap K \ntrianglelefteq G$.
A: If you take $H\leq K$ then $H\cap K=H$. So it is enough to find a chain of subgroups $H\leq K\lhd G$ such that $H\not\lhd G$.
For example, take $G=S_3\times L$ where $L$ is any non-trivial group. Then the chain $$\{id, (12)\}\leq S_3\lhd G$$ works. That is, take $K:=S_3\lhd G$ and $H:=\{id, (12)\}\leq S_3$, and the chain works as $\{id, (12)\}\not\lhd S_3$.
(I used $S_3$ because it is the smallest group containing a subgroup which is not normal. Many, many other groups work though!)
