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**letitbemww** 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.