$H,K \lt G$ with $HK=G$ $G$ is a group and $H \lt G$ and $K \lt G$, with $|G|=n$ then If $GCD( (G:H) ,(G:K))=1$ then $G=HK$
Any thoughts?
Edit by Batominovski: To prevent this thread from being closed or getting unnecessary downvotes, the OP has made an attempt to solve the problem.  But the attempt was wrong, so it was removed.
 A: I assume that $G$ is a finite group (due to the tag "finite-groups" and the quote "$|G|=n$").  Note  (for example, from here) that
$$|HK|\,|H\cap K|=|H|\,|K|\,.$$
That is, 
$$\frac{|HK|}{|H|}=\frac{|K|}{|H\cap K|}=[K:H\cap K]\in\mathbb{Z}\,.$$
Similarly, 
$$\frac{|HK|}{|K|}=\frac{|H|}{|H\cap K|}=[H:H\cap K]\in\mathbb{Z}\,.$$
That is, $|H|$ and $|K|$ both divides $|HK|$.  This mean $\text{lcm}\big(|H|,|K|\big)$ divides $|HK|$, or
$$|HK|\geq \text{lcm}\big(|H|,|K|\big)\,.$$
Consequently,
$$\frac{|G|}{|HK|}\leq \frac{|G|}{\text{lcm}\big(|H|,|K|\big)}=\gcd\left(\frac{|G|}{|H|},\frac{|G|}{|K|}\right)\,.$$
From the given condition 
$$\gcd\left(\frac{|G|}{|H|},\frac{|G|}{|K|}\right)=\gcd\big([G:H],[G:K]\big)=1\,,$$
we conclude that
$$\frac{|G|}{|HK|}\leq 1\text{ or }|G|\leq |HK|\,.$$
Since $HK\subseteq G$ and $G$ is a finite set, we deduce that $G=HK$.

P.S.  We indeed have the following proposition.  See a comment by xsnl below for a sketch of a proof.

Proposition.  Let $H$ and $K$ be subgroups of an arbitrary (finite or infinite) group $G$ such that $[G:H]$ and $[G:K]$ are finite.  Suppose further that $\gcd\big([G:H],[G:K]\big)=1$.  Then, $G=HK$.

