# Subgroups of Relatively Prime Index

1. Nov 30, 2008

### e(ho0n3

The problem statement, all variables and given/known data
Let H and K be subgroups of G of finite index such that [G] and [G:K] are relatively prime. Prove that G = HK.

The attempt at a solution
All I know is that [G intersect K] = [G] [G:K]. What would be nice is if [GK] = [G] [G:K] / [G intersect K], for then I would be done. Anywho, I must somehow show that [GK] = 1 or prove that G = HK directly. Any tips?

2. Nov 30, 2008

### morphism

3. Nov 30, 2008

### e(ho0n3

Funny. The poster knows the part of the proof I do not and I know the part of the proof that the poster does not. Do you know what counting principle the poster is talking about? In any case, the poster states that "it will come down to saying that |HK|=c|G| for some multiple c", but then the poster is assuming that G is finite.

4. Nov 30, 2008

### morphism

If HK is finite, then

$$|HK| = \frac{|H||K|}{|H \cap K|}.$$

This is the counting principle the was poster was referring to.

Of course it won't do us much good here, because G isn't finite (something I missed when I looked at your post). The problem is more difficult without this assumption; here's a hint: use the fact that [G intersect K]=[G][G:K] to deduce that [G:K]=[H intersect K]. Then show that this implies that G=HK (look at the cosets of K in G and the cosets of H intersect K in H).

5. Nov 30, 2008

### e(ho0n3

If [G:K]=[H intersect K], there must be some kind of bijection between the cosets of K in G and the cosets of H intersect K in H. However, I'm unable to figure out what that bijection could possibly be. Hmm...this is harder than I thought.

6. Nov 30, 2008

### morphism

I think you should try to finish this off by yourself. It's one of those problems that'll give you a massive headache until you finally notice the right way to go about doing them. This is a valuable educational experience! Good luck.

7. Nov 30, 2008

### e(ho0n3

OK. I'll try. Thanks for your help.