Prove G=HK When G, H, K are Finite Subgroups of G

[fk378]

Homework Statement

Let G be a group and let H,K be subgroups of G.
Assume that G is finite and that the indices |G| and |G:K| are relatively prime. Show that G=HK.

Hint: Show that |G(intersect)K| is divisible by both |G| and |G:K| and then use the counting principle for |HK|.

The Attempt at a Solution

First off, why do the indices have to be relatively prime?
I don't know how to show that |G(intersect)K| is divisible by both |G| and |G:K|, but I do know that if I assume those, I know how to use the counting principle because ultimately it will come down to saying that |HK|=c|G| for some multiple c, and c must = 1 otherwise it says that for c>1, |HK|>|G| and that is not possible.

EDIT:
Is the intersection of the left coset of H and the left coset of K disjoint? Since they are both equivalence classes they would have to either be disjoint or equal, no? So then |G(intersect)K| would consist of both xH and xK for some x in G...?

 

[morphism]

If the indices were not prime then it's easy to come up with examples where G does not equal HK.

As to showing that |G$\cap$K| is divisible by both |G| and |G:K|, here's a hint: H$\cap$K is a subgroup of H, K and G.

 

[fk378]

I understand that H(union)K is a subgroup of H, K and G. But I don't understand how the numbers would work. How do we know that |G(union)K| is definitely a multiple of both |G| and |G:K|?

 

[fk378]

I just added this "edit" into my original question:

Is the intersection of the left coset of H and the left coset of K disjoint? Since they are both equivalence classes they would have to either be disjoint or equal, no? So then |G(intersect)K| would consist of both xH and xK for some x in G...?

 

[morphism]

Just write down what |G|, |G:K| and |G$\cap$K| are. It will also help to think about what |H$\cap$K| and |K$\cap$K| are.

Is the intersection of the left coset of H and the left coset of K disjoint? Since they are both equivalence classes they would have to either be disjoint or equal, no? So then |G(intersect)K| would consist of both xH and xK for some x in G...?

"The" left coset of H? I think you need to review your definitions. A coset is not an equivalence class; it's a set.