Does $a^n=b^n \implies a=b$ in an abelian group? Does $a^n=b^n \implies a=b$ in an abelian group?
My intuition tells me this mug not always be the case. Under what conditions is this true?
Thanks!
 A: No. If $\,A\,$ is any (abelian or not) group of order $\,m\,$ , then $\,x^m=y^m=1\;,\;\forall\,x,y,\in A\,$
For the other part (now with abelian groups): try taking $\,m\in\Bbb N\;,\;\;gcd(m,|G|)=1\,\ldots$
A: Let $(A,+)$ be an abelian group (written additively: thus instead of $x^n$ we write $nx$...)  For $n \in \mathbb{Z}^+$ we define
$A[n] = \{x \in A \ | \ nx = 0\}$, the n-torsion subgroup, and
$A[\operatorname{tors}] = \bigcup_{n \in \mathbb{Z}^+} A[n]$, the torsion subgroup.
Why are these relevant?  Because for $a,b \in A$, if $na  =nb$, then $n(a-b) = 0$ so $a-b \in A[n]$.  Conversely, if there is a nonzero $b \in A[n]$, then $n0 = nb = 0$. Thus:
$\bullet$ For a fixed positive integer $n$ and all $a,b \in A$, $na = nb \implies a = b$ holds iff $A[n] = 0$.  
$\bullet$ For an abelian group $A$, for all $a,b \in A$ and all $n \in \mathbb{Z}^+$, 
$na = nb \implies a= b$ holds iff $A[\operatorname{tors}] = 0$.  
Another, yet cleaner, way to express this is that for any abelian group and $n \in \mathbb{Z}^+$, multiplication by $n$ is an endomorphism of $A$ (i.e., a group homomorphism from $A$ to itself).  Then $A[n]$ is precisely the kernel of this endomorphism.  The OP is asking when the endomorphism is injective, and the answer is...if and only if the kernel $A[n]$ is trivial.  
A: No. (1)(1) = (-1)(-1).
For an arbitrary $n$,
if $w= e^{2\pi i/n}$,
$w^n = 1^n$.
