Suppose $H$ is a subgroup of a group $G$. I can prove that if $K$ is a normal subgroup of $G$, then $H\cap K$ is a normal subgroup of $H$.
My question is whether $H\cap K$ a normal subgroup of $G$? If it is not, can we give a non-trivial counterexample?
What if for an arbitrary subgroup $J$, my guess is that $H\cap J $ may not be normal, but I am struggling to give a counterexample.
Many thanks for the help!
What if $G$ is a group and $K=G$ and $H$ is any subgroup which is not normal in $G$?
Edit: The OP wanted a non-trivial example. It is known that $A_n$ (the collection of even permutations) is the only proper normal subgroup of $S_n$ for $n\geq 5$. See Proving that $A_n$ is the only proper nontrivial normal subgroup of $S_n$, $n\geq 5$
Thus you can simply choose any proper subgroup $H$ of $A_n$ and it will not be normal in $G=S_n$ when you intersect it with $A_n=K$.
Consider $D_8$ which is Dihedral group of order 8. Then $<s>$ is subgroup of $D_8$ and $<s,sr^2>$ is a normal subgroup of $D_8$ (because, $|<s,sr^2>| = 4$). Now, $<s> \cap <s,sr^2> = <s>$ is a normal subgroup of $<s,sr^2>$ (because, $|<s>|=2$), BUT $<s>$ is not a normal subgroup of $D_8$.
For, let $s \in <s>$ and $r^3 \in D_8$. Then $r^3s(r^3)^{-1}=r^3sr=rrrsr=rrs(r^{-1})r=rrs=rs(r^{-1})=s(r^{-1})(r^{-1})=sr^2 \notin <s>$.
Now, counter example for your other argument about J:
Consider $G=S_4$, $H=\{e,(1 2)\}$ and $J=S_3$ on the symbols $1$,$2$,$3$. Then both $H$ and $J$ are subgroups of $G$ and none of them is a normal subgroup of $G$. Also note that $H \cap J = H$.
Let $G_0$ be any group with non-normal subgroup $H_0$. Let $H=H_0\times G_0$, $G=G_0\times G_0$, and $K=G_0\times\{1\}$
