Assume . Then iff
I don't believe this will be too hard but not sure on how to start.
we want to show that the only generators of <x>, when |<x>| = ∞, are x and x^-1.
by definition <x> is generated by x, so x is obviously a generator.
since <x^-1> is a group, (x^-1)^-1 = x is in <x^-1>, so <x> is contained in <x^-1>.
similarly, since <x> is a group, and x^-1 is contained in <x>, <x^-1> is contained in <x>, hence they are equal, so x^-1
generates <x> as well.
so now we show that if a ≠ ±1 then <x^a> is not all of H. since, by the above argument, <x^a> = <x^(-a)>,
(just use "x^a" instead of "x")
we can assume without loss of generality that a > 1. the simplest way to proceed is to show that x is not in <x^a> for a > 1.
for suppose it were: then x = (x^a)^k, for some integer k. thus e = x^(ak - 1).
thus we have found a positive integer (either ak -1, or 1- ak) m, for which x^m = e.
but this means that x is of finite order, whence |<x>| is finite, a contradiction.