If a is a group element and a has infinite order, prove that $a^m\neq a^n$ when $m \neq n$ 
If $a$ is a group element and $a$ has infinite order, prove that $a^m\neq a^n$ when $m \neq n$. (Gallian, Contemporary Abstract Algebra, Exercise 19, Chapter 3.)

To prove that $m \neq n$ implies $a^m\neq a^n$. Taking contrapositive of this i get
Given when $a^m=a^n$ prove $m=n$
Let $m \neq n$. So $m-n \neq 0$ Now $a^m=a^n$ implies $a^{m-n}=e$ Thus we get finite order of a which is contradiction. SO m=n. IS THIS RIGHT ?
Thanks
 A: Though your solution is correct, some remarks may be helpful to ameliorate it.
About the logical structure of your argument: you want to prove the implication $P\Rightarrow Q$. You prefer to prove the contrapositive $\neg Q\Rightarrow \neg P$. This you do by contradiction, showing that $\neg Q\wedge P$ leads to a contradiction. But this is the same as showing right away by contradiction the implication $P\Rightarrow Q$! 
About the statement itself: showing that $m\neq n$ implies $a^m\neq a^n$ smells like injectivity of a map. Indeed, defining a map
$$
f\colon \mathbf Z\rightarrow G
$$
by $f(a)=a^m$ for all $m\in\mathbf Z$, where $G$ is the group containing the element $a$, the statement you have to prove is that $f$ is injective. This you can do by contradiction. Note that $f$ is a morphism of groups. If $f$ is not injective, you have probably seen that then the kernel of $f$ is nonzero. It follows that there is a nonzero element $n\in\mathbf Z$ such that
$$
e=f(n)=a^n.
$$
Contradiction, since $a$ was supposed to be of infinite order.
In fact, the argument is exactly the same as yours, except that it is more economical from a mathematical point of view. Your argument reproves a statement that you have probably already seen: a noninjective morphism of groups has a nonzero kernel. 
