Subgroups and LaGrange's Theorem

[Obraz35]

Let H and K be finite subgroups of a group G whose orders are relatively prime. Show H and K have only the identity element in common.

By LaGrange's theorem I know that the orders of H and K must divide the order of G. I have attempted a proof by contradiction but have had no luck arriving at a contradiction. Any ideas?

Thanks.

 

[matt grime]

You are correct, and you do need Lagrange. But you're asked what the intersection of H and K is. Firstly, is HnK a group?

 

[Obraz35]

Well, I suppose it would be a group if it contained the identity, g and the inverse of g. This is what I was trying to use to find a contradiction but I'm not sure how this leads to the fact that H and K are relatively prime.

 

[Dick]

You should probably answer Matt's question without the 'if'. If the intersection of H and K is a group, then it's a subgroup of H and K. That tells you something about divisors of the order of H and K.

 

[matt grime]

Well, I suppose it would be a group if it contained the identity, g and the inverse of g. This is what I was trying to use to find a contradiction but I'm not sure how this leads to the fact that H and K are relatively prime.

You are attempting to deduce something from the fact that they are relatively prime.

H and K have coprime orders. So, by Lagrange, if L is a subgroup of H and K, then what do you know?

 

[Obraz35]

Oh, okay. I see it now. Thanks very much.