Let $\displaystyle H=<x>$.

Assume $\displaystyle |x|=\infty$. Then $\displaystyle H=<x^a>$ iff $\displaystyle a=\pm 1$

I don't believe this will be too hard but not sure on how to start.

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- Sep 30th 2011, 07:10 PMdwsmithH=<x>
Let $\displaystyle H=<x>$.

Assume $\displaystyle |x|=\infty$. Then $\displaystyle H=<x^a>$ iff $\displaystyle a=\pm 1$

I don't believe this will be too hard but not sure on how to start. - Sep 30th 2011, 07:58 PMDevenoRe: H=<x>
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.