How do I proceed with this proof about order of elements in a group G? Let a and b be elements of a group G. Let ord(a)=m and ord(b)=n.
Prove if m and n are relatively prime, then no power of a can be equal to any power of b (except for e).
My work so far:
Suppose m and n are relatively prime. Then gcd(m,n)=1.
Then 1=mx+ny for some integers x,y. 
NTS a^m cannot equal b^n, unless m=n=e.
My concerns: I would like to prove this using gcd if possible. I'm not sure how to proceed this proof and how to show what I need, that is a^m cannot equal b^n unless m=n=e. Any suggestions or hints? \
My ideas: I'm thinking of exploiting properties of gcd to somehow get two equations. I'm not very familiar with the properties of gcd so it would be helpful to have some explanation on this and how to use it for my proof.
Sorry I don't know any lexicon. 
 A: By Lagrange's Theorem, the order of an element of a group must divide the order of the group, so the order of any power of $a$ must divide the order of $\langle a\rangle = $ord($a)=m$ and the order of any power of $b$ must divide the order of $\langle b\rangle = $ord($b)=n$.
Suppose there is an element $x$ in $G$ that is both a power of $a$ and a power of $b$. We want to show that $x$ is the identity.
Since $x$ is a power of $a$, $x \in \langle a \rangle$, so ord($x$) divides $m$.
Since $x$ is a power of $b$, $x \in \langle b \rangle$, so ord($x$) divides $n$.
However, $\gcd(m,n)=1 \implies $ord($x)=1 \implies x=1$, which means that if an element is both a power of $a$ and a power of $b$, then it is the identity element of $G$. 
A: Let $\exists\ p,q \in \Bbb{Z}$ such that $a^p=b^q$, then $|a^p|=|b^q|$ i.e. $\frac{m}{(p,m)}=\frac{n}{(q,n)}$ then as $(m,n)=1$ so only common divisor they both have is $1$ so we can only have $\frac{m}{(p,m)}=\frac{n}{(q,n)}=1$ which happens when $p=m,q=n$ but that is not what we want. 
A: Let $a^k = b^l.$ Then $a^{kn} = b^{ln} = e$ which forces $m$ to divide $kn$ since $m$ is the order of $a.$ But we know that $gcd(m,n) = 1.$ Hence, we have $m$ divides $k.$
Therefore, the first equation becomes $$e = a^k = b^l,$$ which forces $n$ to divide $l$ and we get the required result.
A: This is 10.E.2 in Pinter.

Let a and b be elements of a group G.
Let ord(a) = m and ord(b) = n.
If m and n are relatively prime, then no power of a can be equal
  to any power of b (except for e).
(HINT: Use 10.D.2)

As directed in the hint, we'll be using 10.D.2:

Let a be any element of finite order of a group G.
The order of a^k is a divisor (factor) of the order of a.

Let's begin.
ord(a) = m
ord(b) = n

Contrary to the claim to be proved, let's assume that a power of a is equal to a power of b:
a^k = b^j

Then
ord(a^k) = ord(b^j)

By 10.D.2:
ord(a^k)|m   In words:   ord(a^k) is a factor of m

ord(b^j)|n   In words:   ord(b^j) is a factor of n

However, since m and n are relatively prime, they do not have common factors and so 
ord(a^k) and ord(b^j) cannot be equal to each other.
Thus:
ord(a^k) ≠ ord(b^j)

