$H,K \lt G$ 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.

You write LCD, but do you mean GCD? I mean if lcm$(k,l)=1$, then $k=1=l$. — Nicky Hekster, Aug 03 '18 at 14:48
I don't understand why you think the intersection of $H$ and $K$ is trivial. For example, take $G=\mathbb{Z}_2\times\mathbb{Z}_3\times \mathbb{Z}_5$, $H=\mathbb{Z}_2\times\mathbb{Z}_3\times 0$, and $K=0\times\mathbb{Z}_3\times\mathbb{Z}_5$. — Batominovski, Aug 03 '18 at 14:50
@Batominovski actually i was looking to the wrong place, will delete this part — Daniel Moraes, Aug 03 '18 at 14:51

Batominovski · Accepted Answer · 2018-08-03T16:54:39.300

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, $G=HK$.

It follows from finite case. $[G, H \cap K] \leq [G, H][G, K]$, in particular, is finite. $H \cap K$ can be not normal, but taking intersection of conjugates we have finite quotient of $G$ with kernel contained in both $H$ and $K$. Images of $H$ and $K$ generate it, so $G = HK$. — xsnl, Aug 03 '18 at 16:39
@xsnl Great argument! If you put that as an answer here, I will upvote it. This deserves to be an answer. — Batominovski, Aug 03 '18 at 16:41

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

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