Let $G$ be a finite cyclic group of order $n$. If $d$ is a positive divisor of $n$, prove that the equation $x^d = e$ has exactly $d$ distinct solutions in $G$.
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$G$ is generated by some $g\in G$. Since $d\mid n$, we can find $k$ such that $dk=n$. Now let $y=g^k$. Check that $y^j,0\leq j\leq d-1$ are pairwise distinct, so there are at least $d$ solutions to $x^d=e$. If $x\neq g^{kj}$ for all integer $0\leq j\leq d-1$, $x=g^p$ for some $p$. We write $p=kq+r$, where $1\leq r\leq k-1$. Then $x^d=g^{pd}=g^ng^{rd}=g^{rd}\neq e$. |
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