$HK$ is a subgroup where $H \subseteq N_G(K)$ 
Let $H, K$ be subgroups of $G$.  Let $HK=\{hk: h \in H, k \in K\}$ and $H \subseteq N_G(K)$.

I'm a bit stuck on something.  It makes me a bit nervous that $H \subseteq N_G(K)$.  Does this mean that $HK=KH$ or for every $h \in H$ and $k \in K$ there is an $h' \in H$ such that $hk=kh'$.  I saw this, but in this question it has that $K$ is a subgroup.
 A: $N_G(K)$ is the normalizer in $G$ of the subgroup $K$.  It's the set of all $g$ such that $gKg^{-1} = K$.
So since $H \subset N_G(K)$, all of $h \in H$ also commute with $K$ like the $g$ above. So $hK = Kh, \ \forall h \in H$  Take the union over $H$ to show that $HK = KH$.
QED
A: Does not
$N_G(K) = \{g \in G \mid g^{-1} K g = K \}? \tag{1}$
Indeed it does; now for $h_1, h_2 \in H$ and $k_1, k_2 \in K$ we wish to establish that
$h_1k_1h_2k_2 \in HK, \tag{2}$
so we follow (1) and note that since
$h_2 \in H \subseteq N_G(K) \tag{3}$
we have
$h_2^{-1}Kh_2 = K, \tag{4}$
which readily implies
$h_2^{-1}k_1h_2 \in K, \tag{5}$
whence there exists $k_3 \in K$ such that
$h_2^{-1}k_1h_2 = k_3, \tag{6}$
or
$k_1h_2 = h_2k_3, \tag{7}$
whence
$h_1k_1h_2k_2 = h_1h_2k_3k_2 \in HK, \tag{8}$
which shows that $HK$ is multiplicatively closed.  If $H$ and $K$ are finite, so that $HK$ is also finite, we are done, since a finite multiplicatively closed subset of any group is a subgroup.  In the event that $HK$ is infinite, we need to establish that the identity $e \in G$ is in $HK$ and that $(hk)^{-1} \in HK$ for $h \in H, \, k \in K$.  Well, $e \in HK$ is virually self-evident, since $H, \, K$ being subgroups implies $e \in H, \, K$; if $hk \in HK$, then $(hk)^{-1} = k^{-1}h^{-1}$; but again, since
$h^{-1} \in H \subseteq N_G(K), \tag{9}$
we find that
$hk^{-1}h^{-1} = (h^{-1})^{-1}k^{-1}h^{-1} \in K, \tag{10}$
and thus we must have $k_4 \in K$ with
$hk^{-1}h^{-1} = k_4, \tag{11}$
whence
$k^{-1}h^{-1} = h^{-1}k_4 \in HK, \tag{12}$
showing that indeed $(hk)^{-1} \in HK$.  This completes the proof that $HK$ is a subgroup of $G$ under the given hypothesis.  QED
Hope this helps.  Cheerio,
and as always, 
Fiat Lux
