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.

  • $\begingroup$ An arbitrary element of $\langle a^3 \rangle$ is of the form $a^{3k}$, where $k \in \mathbb Z$. If $a^{3k}$ generates $\langle a^3 \rangle$ then some power of $a^{3k}$ must equal $a^3$. What does this imply? $\endgroup$ – Bungo Apr 17 '16 at 3:38
  • $\begingroup$ @Bungo This implies k=1? $\endgroup$ – Mathematicing Apr 17 '16 at 3:39
  • $\begingroup$ Is that the only solution? $\endgroup$ – Bungo Apr 17 '16 at 3:40
  • $\begingroup$ @Bungo It isn't but I am thinking how to determine k=-1 $\endgroup$ – Mathematicing Apr 17 '16 at 3:41
  • 1
    $\begingroup$ If $a^{3k}$ generates $\langle a^3 \rangle$ then some power of $a^{3k}$ must equal $a^3$. So for some integer $n$ we must have $(a^{3k})^n = a^{3kn} = a^3$. Now use the fact that distinct powers of $a$ are distinct group elements. $\endgroup$ – Bungo Apr 17 '16 at 3:43

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