Prove that if $i$ and $j$ are integers, the $\langle a^i \rangle = \langle a^j \rangle$ iff $i=\pm j$ Let $G$ be a group and let $a \in G$ be an element of infinite order. Prove that if $i$ and $j$ are integers, the $\langle a^i \rangle = \langle a^j \rangle$ iff $i=\pm j$.
My attempt,
Since a is an element of infinite order $a^i = a^j$ iff i = j, therefore $\langle a^i \rangle = \langle a^j \rangle$. Since j is coprime to j-1, $C_n = \langle a^j \rangle = \langle a^{j-1} \rangle = \langle a^{-j} \rangle$ mod J.
 A: One direction is almost trivial. For the other, note that
$$\bigl\langle a^i\bigr\rangle:=\bigl\{a^{ik}:k\in\Bbb Z\bigr\}$$ and $$\bigl\langle a^j\bigr\rangle:=\bigl\{a^{jk}:k\in\Bbb Z\bigr\}.$$
If $\bigl\langle a^i\bigr\rangle=\bigl\langle a^j\bigr\rangle,$ then, $a^i\in\bigl\langle a^j\bigr\rangle$ so since $a$ has infinite order, then $i$ is an integer multiple of $j$. Similarly we have that $j$ is an integer multiple of $i$. The conclusion follows fairly readily after that.
A: Note that you have an isomorphism from $\mathbb{Z}$ onto $\langle a\rangle $ given by
$$
n\longmapsto a^n.
$$
So it boils down to 
$$
i\mathbb{Z}=j\mathbb{Z}\quad\Leftrightarrow\quad |i|=|j|.
$$
The implication $\Leftarrow$ is trivial.
Now if $i\mathbb{Z}=j\mathbb{Z}$, you have
$$
i\in j\mathbb{Z}\quad \Rightarrow \quad j|i
$$
and likewise $i|j$.
So the implication $\Rightarrow$ follows from the fundamental theorem of arithmetic.
Note: if you don't want to use the fundamental theorem of arithmetic, simply wite $i=mj$ and $j=ni$. If $i=0$ then $j=0$. If $i\neq 0$ then $i=mni$ implies  $mn=1$ which implies in turn $m=n=1$ of $m=n=-1$.
A: *

*For the easier direction, it is enough to observe that $a^{-i}=(a^i)^{-1}$, and that, for any $b\in G$, we have $\langle b\rangle=\langle b^{-1}\rangle$ (why?).

*Conversely, suppose that $S:=\langle a^i\rangle=\langle a^j\rangle$. Using the above observation, we can assume w.l.o.g. that both $i,j>0$. By symmetry we can also assume that $i\le j$. Now consider the remainder $r$ when $j$ is divided by $i$, and conclude that $a^r\in S$.

