$n$th power of two groups' members Let $G$ be a group with $a,b\in G$ and $a^n=b^n\not=1$. What is the least property that $G$ must have to ensure that $a=b$?

If $G$ is finite then gcd($|G|,n)=1$ will do. By Bézout's Theorem one can find integers $k, l \in \mathbb{Z}$, such that $k|G|+ln=1$. Hence $a=a^1=a^{k|G|+ln}=a^{ln}=(a^n)^l=(b^n)^l=b^{k|G|+ln}=b$. Here we are using the fact that $g^{|G|}=1$ for all $g \in G$. This follows from Lagrange's Theorem.
In a similar manner, one can prove another condition ($G$ not necessarily finite): if both $a$ and $b$ have finite order and these orders are relatively prime to $n$, then $a^n=b^n$ implies $a=b$.
