# Subgroups-Stuck on a question

1. Sep 18, 2005

### hgj

Here's the question:

Prove that a nonempty subset H of a group G is a subgroup if for all x and y in H, the element xy^(-1) is also in H.

We have a theorem that says if G is a group and H is a nonempty subset of G, then H is a subgroup of G iff:
(1) H is closed
(2) if h is in H, then the inverse of h in G lies in H.

I know I need to use this theorem, and I have two ideas about how to go about it:
First, I think if I can prove y^(-1) is in H, then the two parts of the theorem follow from that. Unfortunately, I can't figure out how to do this.
My second thought is to prove the two parts separately. If I do this, my proof for closure is:
Assume xy is not in H. Then x is not in H or y is not in H. But this is a contradiction, so xy must be in H. Then closure is satisfied.
After this, though, I get stuck again when I try to prove part (2) about the inverse.

2. Sep 18, 2005

### AKG

If for all x and y in H, xy-1 is also in H, then what happens if x = y? Using that, you will have determined the existence of another special element in H. Using that element and the knowledge that xy-1 is in H for every x, y in H, what more can you prove?

For closure, you want to prove that for all x in H, if y is in H then so is xy. If you have proven that (if y is in H then so is y-1) and you use the given fact (if x and y are in H, so is xy-1) then you can prove that xy is in H because it is true so long as:

x(y-1)-1 is in H which is true using the given assumption if
x is in H and y-1 is in H. The first is true by assumption, and the second is true because a) y is in H and b) you should have proved inverses are in H.