Criterion for $a^i=a^j$ proof Let G be a group and let a be an element in G. If a has infinite order, then $$a^i=a^j$$ if and only if $i=j$
If $a$ has finite order, say $n$, then 
$$\left \langle a \right \rangle=\{e,a,a^2,...,a^n-1\}$$
and $$a^i=a^j$$ if and only if $n$ divides $i-j$.
The first part of the proof says:
If a has infinite order, then $$a^i=a^j$$ if and only if $i=j$.
I have managed to show the "only IF" part. How do I show the "IF" part? I know that if $$i=j$$ then $$a^i=a^j$$ but how do I show?
Also, as to the second part of the proof, there is an ambiguity I'd like to clarify:
Are they saying 
"If $A \implies$ $($B$ and $C$) \iff D$ or "If ($A \implies B$), and ($C \iff D$)?

Hint: If $a$ has infinite order and $a^{i}=a^{j}$ then $(a^{i})^{-1}=a^{-i}$.  Multiply $a^{i}$ and $a^{j}$ by $a^{-i}$ and see what happens...
EDIT: extra hint... you'll see that $a^{j-i}=e$ and $j\neq i$ so $j-i\neq 0$... what does it mean to have infinite order?
To clarify the ambiguity:  The statement is "Let $a$ be an element of finite order $n$.  Then $a^{i}=a^{j}$ if and only if $n|(i-j)$."  You show $A\Rightarrow B$ and $B\Rightarrow A$ where $A$ is "$a^{i}=a^{j}$" and $B$ is "$n|(i-j)$."
The IF direction would be $A\Rightarrow B$, or IF $a^{i}=a^{j}$ THEN $n|(i-j)$.
