Let A and B be groups. Prove that the following sets are subgroups of the direct product A x B:
{(a, 1) | a is in A}
{(1, b) | b is in B}
{(a, a) | a is in A} where we assume B = A
To prove that a set is a subgroup, you need to prove it is closed under identity, the group action, and the inverse. I also know that the direct product of groups is defined as when (a1, b1) * (a2, b2) = (a1*a2, b1*b2)
I think I get the identity parts (saying that (e, 1), (1, e) or (e, e) is the identity) but it's the other two that I'm confused about.
Here is my guess for the first one: (e, 1) is the identity since (a, 1)*(e, 1) = (a, 1).
Let x, y be in A. Then (x, 1)*(y, 1) = (xy, 1) which is in A x B (but I'm not sure if this is actually true). This shows it is closed under multiplication. And (a^-1, 1) is the inverse because (a^-1, 1)*(a,1) = (e, 1). I guess I'm just the most confused about proving the group action is closed.
Ok so for the second one would it be that (1, e) is the identity because (1, b)*(1, e) = (1, b). Let x, y be in B. Then (1, x)*(1, y) = (1, xy) which is in A x B. Therefore this is closed under multiplication. Also, (1, b^-1) is the inverse because (1, b^-1)*(1, b) = (1, e).
For the third one, I would have (e, e) is the identity because (a, a)*(e, e) = (a, a). Let x, y be in A. Then (x, x)*(y, y) = (xy, xy) which is in A x B = A x A. Therefore this is closed under multiplication. Also, (a^-1, a^-1) is the inverse because (a^-1, a^-1)*(a, a) = (e, e).
Is that enough to say that they are subgroups of AxB? I just feel like I'm not using the direct product enough or something. It seems too simple...
for closure, all you need to do is take 2 "generic" elements of each subset, and prove their product is again in the subset.
for example, two typical elements of the 3rd subset are (a1,a1) and (a2,a2). their product is (a1*a1, a2*a2). is this in the subset {(a,a) | a in A}?
well, clearly the two components are the same, so it just remains to show that a1*a1 and a2*a2 are both in A. however, A is a group, so.....