The first thing we should do is try to prove closure. Choose . Then there exists positive integers with and , with . Since , then . But is it also in ? Well, observe that . Also, . So then the conditions are satisfied such that , and therefore is closed.

It is pretty obvious that .

It's almost just as obvious that for any then . But let's prove it anyway. There exists integer with . Observe also that since is abelian, we can do the following: . So for any , then .

That should do it.