Let $G$ be a cyclic group of order $m$, and let the number $s$ be relatively prime to $m$. Prove that if $a^s = b^s$ then it follows that $a = b$. 
Let $G$ be a cyclic group of order $m$, and let the number $s$ be relatively prime to $m$. Prove that if $a^s = b^s$ then it follows that $a = b$.

Let $st+mr=1$ then we have a = $a^{st+mr}=a^{st}=b^{st}=b^{st+mr}=b$.
Is this correct?
Can someone help me please?
 A: Here's the same argument, presented using additive notation (+) rather than multiplicative notation for the group operation.  [This is conventional when a group is abelian, and indeed all cyclic groups are abelian.]  Then instead of writing $a^s = b^s$ for some integer $s$, we would instead say $sa = sb$, where the expression $sa$ is an abbreviation for $a$ added to itself $s$ times (or for the additive inverse of a similar expression when $s$ is negative).  Of course we would write the identity element as zero in this convention.
Thus $a\in G$, a cyclic group of order $m$, implies that:
$$ ma \equiv a + a + \ldots + a = 0 $$
As the OP notes in the problem statement, if $s$ is coprime to $m$, then there exist integers $t$ and $r$ such that $st + mr = 1$.  Thus:
$$ (st + mr)a = a $$
and:
$$ (st + mr)b = b $$
Therefore we can use $sa = sb$ and $ma = mb = 0$ to prove:
$$ a = (st + mr)a = t(sa) + r(ma) = t(sb) + r(mb) = b $$
As a way of persuading oneself that there is not some overlooked "trick" in using the additive notation, one can consider that $G$ must be isomorphic to $\mathbb{Z}/m\mathbb{Z}$, the group of integers modulo $m$.  In that context the above computation becomes simply a statement about residue mod $m$.
