If $[G:H]$ and $[G:K]$ are relatively prime, then $G=HK$ I'm struggling to proof that if $H$ and $K$ are subgroups of  finite index of a group $G$ such that $[G:H]$ and $[G:K]$ are relatively prime, then $G=HK$. I don't know why I can't answer it, because this question seems easy. I'm stuck maybe because I've studied so far just Lagrange's theorem and some of its consequences. But I think we don't need much more, because this is the material covered so far by the Hungerford's book.
I need help.
Thanks.
 A: We have the following fact:


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*If $H$ and $K$ are subgroups of a group $G$, then $[H:H\cap K]\leq [G:K]$. If $[G:K]$ is finite, then $[H:H\cap K]=[G:K]$ if and only if $G=KH$. This is proposition $4.8$ of chapter I in Hungerford's Algebra. 


It is easy to prove that $[G:H][H:H\cap K]=[G:K][K:H\cap K]$, from this conclude that $[G:H]\mid [K:H\cap K]$; since $[G:H]$ and $[G:K]$ are relatively prime, and combinig this with $(1)$ we get $[G:H]=[K:H\cap K]$, so...
A: Try proving that [$G$ : $H\cap K$] $=$ [$G$ : $H$][$G$ : $K$].
And then see how you can use the fact that $|HK| = \frac{|H||K|}{|H\cap K|}$
A: If $[G:H]$ and $[G:K]$ are coprime, $[G:H∩K]=[G:H][G:K]$ (see here for a proof)
Observe that a coset of $H\cap K$ is just an intersection of a left coset of $H$ and a left coset of $K$.
Thus, every intersection of cosets of $H$ and $K$ are non-empty and make a coset of $H∩K$. In particular,
$$xK∩H≠\emptyset $$ $∀x∈G$. So if $g\in G$, then there exists a $h\in H$ such that $gK=hK$
So for every there is an element of $H$ in every left coset of $K$. Hence
$$G=\bigcup_{g\in G} gK=\bigcup_{h\in H} hK=HK$$
Notation: $[G:H]$ is the index of subgroup $H$ in group $G$
