Suppose that a has infinite order Find all generators of subgroup $\left \langle a^{3} \right \rangle$
Now, since a has infinite order then so does $\left ( a^{3} \right )^{n} $for if a has finite order then $\left ( a^{3} \right )^{n}=a^{3n}=e$ But this, by definition implies that $\left | a \right |=3n$ which is a contradiction.
By the theorem: $a^{i}=a^{j}$
If element a has infinite order then all distinct powers of a are distinct group elements.
Thus, $\left ( a^{3} \right )^{k}$ are distinct group elements.
I'm pretty much stuck here.
A useful hint would certainly be helpful.