Proving G=HK: A Finite Group Theorem

[happyg1]

Homework Statement

Let H and K be subgroups of a finite group G with coprime indices. Prove that G=HK

Homework Equations

From a theorem we have, If |G| and |G:K| are finite and coprime, we have:
|G intersect K|=|G|*|G:K|

|G| indicates the index of G over H, not the order here...a notational point that hurts my head.

The Attempt at a Solution

I used the thoerem and I got
|G intersect K|=|G|*|G:K|

but since G and H are of coprime index, (H intersect K=1),
So that I get

|G|=|G|*|G:K|

if I let |G|=p and |G:K|=q then |G|=pq

That's where I am and I don't think I'm headed in the right direction.

pointers and clarification will be greatly appreciated.
CC

 

[NateTG]

Well, if you determine that |G:(HK)| =1 you're golden, right?

BTW:

but since G and H are of coprime index, (H intersect K=1),

Isn't generally true:
G = \mathbb{Z}_{12}
H=\{0,3,6,9\}
K=\{0,2,4,6,8,10\}
then
|G<img src="/styles/physicsforums/xenforo/smilies/arghh.png" class="smilie" loading="lazy" alt=":H" title="Gah! :H" data-shortname=":H" />|=3
and
|G:K|=2
are coprime, but
|G \cap K|\neq 1
and
|G<img src="/styles/physicsforums/xenforo/smilies/arghh.png" class="smilie" loading="lazy" alt=":H" title="Gah! :H" data-shortname=":H" />| \times |G:K| \neq |G|