HK a subgroup, neither is included in the other's normalizer? Can anyone think of an example of the following: $H$ and $K$ are proper subgroups of $G$.  $H$ is not contained in the normalizer of $K$. $K$ is not contained in the normalizer of $H$. Nevertheless, $HK$, the set of all $hk$ where $h \in H$, and $k \in K$, is somehow a group?
 A: Consider $G = S_{4}$, of order $24$. 
Let $H$ be a $2$-Sylow subgroup, of order $8$, and $K$ a $3$-Sylow subgroup, of order $3$. 
Then $HK= G$ is a subgroup of $G$. This is because of the formula for the product of two subgroups
$$
\lvert H K \rvert = \frac{\lvert H \rvert \cdot \lvert K \rvert}{\lvert H \cap K \rvert}.
$$
However, neither $H$ normalizes $K$, nor $K$ normalizes $H$. 
To see this, note that if were the case that $H$ normalizes $K$, then $K$ would be normal in $G$, thus unique, so that $G$ would have only two elements of order $3$. However $G$ has many (actually $8$) elements of order $3$. 
A similar argument (there are $2$ elements of order $4$ in the dihedral group $H$, while $G$ has $6$) shows that $K$ does not normalize $H$.
A: To generalize the example given by Andreas, let $G$ be a finite group of order $p^a q^b$, where $p \neq q$ are primes. Suppose that $G$ does not have a normal Sylow subgroup. The $G$ contains the following example.
Let $P$ be a Sylow $p$-subgroup and $Q$ a Sylow $q$-subgroup of $G$. Now $Q$ does not normalize $P$ since otherwise $P$ would be normal. Similarly $P$ does not normalize $Q$. However, $G = PQ$ since $PQ$ has same order as $G$.
