= Solution
The <Sylow theorems> also give $n_5\mid14$ and $n_5\equiv1\pmod5$. Of $1,2,7,14$, only $1$ is congruent to one modulo five. Thus $H$ is a <normal subgroup> of $G$ as well.
By <Lagrange's theorem>, $H\cap K=\{1\}$, since its <group order> divides both $5$ and $7$. For $h\in H$ and $k\in K$, the <group commutator> $hkh^{-1}k^{-1}$ lies in $H$ by the <normal subgroup> property of $H$, and in $K$ by the <normal subgroup> property of $K$. It is therefore the <identity element>, so all elements of $H$ commute with all elements of $K$. This is the fact that <normal subgroups of coprime order commute>.
The <subgroup> $HK$ consequently has $|H||K|=35$ elements: the expression $hk$ is unique because $H\cap K=\{1\}$. Since $H$ and $K$ have <prime> orders, they are <cyclic groups>. Write $H=\langle h\rangle$ and $K=\langle k\rangle$. The <product of commuting elements of coprime order> has order $35$, so $hk$ generates $HK$. Finally, $HK$ is a <normal subgroup> because both its factors are <normal subgroups>; equivalently, it has <subgroup index> two. Therefore
$$
\boxed{HK\cong C_{35},\qquad HK\triangleleft G.}
$$
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