Prove that $gcd(\mid G:H\mid,\mid G:K\mid )=1\Rightarrow G=HK$ If $G$ be a group and $H\le G,$ $K\le G$ such that $\mid G:H \mid=m$ and $\mid G:K \mid=n$ then  we know the following two relsults from group theory.

$(1)$ $lcm(m,n) \le \mid G:H\cap K \mid \le mn$ 

$(2)$ $\mid G:H \cap K \mid \le mn.$ The equality of $(2)$ holds iff
$G=HK.$ 

I know the proof of these two results. Note that $gcd(m,n)=1 \Rightarrow \mid G:H \cap K \mid =mn(by (1)) \Rightarrow G=HK(by (2)).$ But I want to prove $gcd(m,n)=1 \Rightarrow G=HK$ without using $(2)$.
Any hint is accepted.

I have seen the solutions to similar question.But it's assuming that $G$ is finite.But what happen if $G$ is infinite
Remark : 
Notice that $[G : HK] \ \mid \ [G : K] $ and $[G : HK] \ \mid \ [G : H]$ .
[It is only division of cardinality of sets, not groups.] 



So we can conclude that $[G : HK] \ \mid m$ and $[G : HK] \ \mid n$ ;
by assumption of primitivity of $m$ and $n$  ($\gcd(m,n)=1$);
we can conclude that $[G : HK] \ \mid 1$ ; 
equivalently  $[G : HK] \ \mid 1$ ; 
so we are done.
Remark : 
Notice that $[G : HK] \ \mid \ [G : K] $ and $[G : HK] \ \mid \ [G : H]$ .
[It is only division of cardinality of sets, not groups.] 



So we can conclude that $[G : HK] \ \mid m$ and $[G : HK] \ \mid n$ ;
by assumption of primitivity of $m$ and $n$  ($\gcd(m,n)=1$);
we can conclude that $[G : HK] \ \mid 1$ ; 
equivalently  $[G : HK] \ \mid 1$ ; 
so we are done.