Subgroups of Relatively Prime Index

[e(ho0n3]

Homework Statement
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?

 

[morphism]

Look here: https://www.physicsforums.com/showthread.php?t=260722.

By the way, I'd be careful about saying things like [GK], because we don't know a priori that HK is a subgroup of G.

 

[e(ho0n3]

Look here: https://www.physicsforums.com/showthread.php?t=260722.

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.

 

[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).

 

[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.

 

[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.

 

[e(ho0n3]

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