Intersection of subgroups of given orders and normality Quite confused about how to solve this question..As it would have been easier to solve if the group was given to be cyclic, but no such case here
Let $H$ and $K$ be subgroups of a group $G$ of orders $14$ and $21$ respectively. If $H \cap K \neq\{e\}$, here $e$ is the identity of the group $G$, then show that $H \cap K$ is a normal subgroup of $G$. 

This is not true. The issue/trick is that there is no information about the group $G$ so I can pick it to have very few normal subgroups.
For example, we can take $G=S_n$ for some $n\geq 7$ (why?). Then $G$ has very few normal subgroups (see here), and in particular none of order $7$ (why is $7$ relevant?).
For a concrete counter-example, take $H=\langle (1,2,\ldots,7)(8,\ldots,14)\rangle$ and $K=\langle (1,2,\ldots,7)(8,\ldots,21)\rangle$.
