TITLE: Does $HK = G$ imply that $KH = G$ when $H,K$ aren't subgroups?
QUESTION [6 upvotes]: Let $G$ be a finite group and let $H$ and $K$ be two "subsets" of $G$ such that $HK = G$. Does that imply that $KH = G$?
Since $H$ and $K$ are not subgroups we cannot use the formula $o(HK) = \frac{o(H)o(K)}{o(H\cap K)}$.
I believe this is false and I am looking for counterexamples.
Thank you.

REPLY [11 votes]: Hopefully I didn't make any stupid mistake :)
Let $G=S_3$ and 
$$H= \{ e, (1,2) \} \\
K=\{ e, (1,3), (1,2,3) \}$$
Then 
$$HK= S_3$$
but
$$(1,3)(1,2)=(1,2,3)=(1,2,3)e$$
showing that $KH$ can have at most 5 distinct elements.