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If $n>2$, and $\zeta$ is a primitive nth root of unity, and we have an integer $a$ where $\zeta^a=1$, can we simply say that $\zeta^a=\zeta$? Or is there some other way to prove that $n|a$?

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If $\zeta\ne1$ but $\zeta^a=1$ then $\zeta^a\ne\zeta$. –  anon Oct 12 '11 at 2:52
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No, that's not true. If $\zeta^a=1$ and $1\ne\zeta$ (which is true because $\zeta$ is primitive and $1$ is not), then they can't be equal, i.e. $\zeta^a\ne\zeta$.

Here's most of the work: If the order of a root of unity $\zeta$ is $m$ and we know that $\zeta^a=1$, then consider $\zeta^{a-b}$, where $b$ is the greatest multiple of $m$ less than or equal to $a$. Since $b=km$ for some $k$, we have $\zeta^b=(\zeta^m)^k=1$, so we know that $\zeta^{a-b}=\zeta^a=1$. But if $a-b\ne0$ this means that there is a power of $\zeta$ that evaluates to $1$ and is less than $a$ (why?). Recall the definition of "order" carefully...

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Perhaps what OP is missing is the connection between "primitive" and "order". Perhaps OP has forgotten that saying $\zeta$ is a primitive $n$th root of unity is saying that $\zeta^n=1$ and that there is no $r$, $0\lt r\lt n$, such that $\zeta^r=1$ - in your language, that $\zeta$ has order $n$. –  Gerry Myerson Oct 12 '11 at 3:18
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