$|G:H|=2$, $K \leq G$ with an element not in $H$. Prove $HK=G$ I know it has something to do with normal subgroups and cosets partitioning, but I don't know how to deal with cosets of $HK$.

Let $x \in K$ with $x \notin H$. Since index$[G:H]=2$, we have $G = H \cup Hx$. Now let $g \in G$ be arbitrary. Then either $g \in H \subseteq HK$. Or $g \in Hx \subseteq HK$. So either way, $g \in HK$. Hence $G$ lies in the subset $HK$. But of course $HK \subseteq G$. It follows that $G=HK$.
