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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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Welcome to math.SE. I'm afraid that it is not considered polite here to tell other users to do something. Your question does not show that you have thought about the problem. Please explain what you've tried so far, and where you are stuck. – Zev Chonoles Apr 8 '12 at 21:40
The (strong) converse is also true and useful for proving the existence of primitive roots. (Amazingly, Wikipedia does not contain this proof. Or did I miss it?) – lhf Apr 8 '12 at 21:57
(There is no need to say in the title that the question is a question, nor that it is abstract algebra, for that's what the tags are for!) – Mariano Suárez-Alvarez Apr 9 '12 at 0:08

$G$ 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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Alright if I need to try it out, this is what I tried to do. $x^{d} = e$ if and only if $a^{ld} = e$ if and only if $ld$ is a multiple of $n$. I got this far. Im not sure how to go about the rest of the problem. – user28615 Apr 9 '12 at 0:07
since d divides n you can pick such an $k$ so that $n$ EQUALS $kd$. take your generator $a$ and consider the subgroup generated by $a^k$. show $a^k$ has order $d$, which will give you $d$ solutions. then use the division algorithm to show there aren't any others. – David Wheeler Apr 9 '12 at 12:39

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