$H,K \lt G$ with $HK=G$

Question

$G$ 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.

Answer 1

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, $G=HK$.