# Intersection of 2 subgroups is a subgroup?

1. Oct 14, 2009

### lockedup

1. The problem statement, all variables and given/known data
H and K are subgroups of G. Prove that H$$\cap$$K is also a subgroup.

3. 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?

2. Oct 14, 2009

### 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.

3. Oct 14, 2009

### 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).

4. Oct 14, 2009

### tiny-tim

Yes, that's right

but why are you even mentioning trivial subgroups?

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

5. Oct 14, 2009

### HallsofIvy

Staff Emeritus
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.