- #1
hgj
- 15
- 0
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.
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.