Intersection of 2 subgroups is a subgroup?

[lockedup]

Homework Statement

H and K are subgroups of G. Prove that H$$\cap$$K is also a subgroup.

The Attempt at a Solution

For H and K to be subgroups, they both must contain G's identity. Therefore, e $$\in$$ H$$\cap$$K. Therefore, H$$\cap$$K is, at least, a trivial subgroup of G.

This was a test question. Note the past tense. The above is all that I wrote down. Since Fall Break is this week, I won't be able to see my test again until Monday. It's going to kill me. If you were my professor, would you at least give me half credit?

 

[tiny-tim]

Hi lockedup!

Sorry … maybe 1 or 2 out of 10 …

all you've proved is that it contains the identity

(and "is, at least, a trivial subgroup of G" doesn't really make sense).

If you remember what a subgroup is, always state it in the test, even if you can't see how to prove it: then at least your professor will know that you know, and can give you some credit for it.

 

[lockedup]

Well, I thought one of the trivial subgroups was the set containing just the identity (the other being the group itself)... My question is how do I prove there other other elements in H and K (specifically one element and its inverse).

 

[tiny-tim]

Well, I thought one of the trivial subgroups was the set containing just the identity (the other being the group itself)

Yes, that's right …

but why are you even mentioning trivial subgroups?

(is there something about them in the original question?)

... My question is how do I prove there other other elements in H and K (specifically one element and its inverse).

ah, you don't have to prove there are other elements …

it's enough to say that if there are other elements, then they satisfy the subgroup properties.

 

[HallsofIvy]

Well, I thought one of the trivial subgroups was the set containing just the identity (the other being the group itself)... My question is how do I prove there other other elements in H and K (specifically one element and its inverse).

What you said proves that $H\cap K$ contains a subgroup.

You do not need to "prove there are other elements". If there are no other elements in $H\cap K$ then that singleton set, {e}, is a subgroup.

But if there are other elements that e in the intersection, you still have to prove the whole thing is a subgroup. Specifically, you have to prove that the set is closed under the group operation and that, if the intersection contains member a, then it contains $a^{-1}$.

Neither of those is hard to show, but they must be shown.