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I am learning about primitive $n$-th roots of unity. I came across this statements while reading and was wondering why these were true:

If $z$ is a primitive $n$-th root of unity and $n$ is even, then $z^2$ is a primitive $\frac{n}{2}$-th root of unity.

If $z$ is a primitive $n$-th root of unity, $z^{-1}$ is also primitive $n$-th root of unity.

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  • $\begingroup$ What is $(z^2)^{(n/2)}$, and is $(z^2)^k = 1$ for any $0 < k < \frac{n}{2}$? Note that $(z^{-1})^k = (z^k)^{-1}$. $\endgroup$ – Daniel Fischer Oct 7 '14 at 0:03
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We have $z^n=1$. It follows that $(z^2)^{n/2}=1$. So $z^2$ is an $(n/2)$-th root of unity.

We now show that $z^2$ is a primitive $(n/2)$-th root of unity. How we do it depends on the definition of primitive that we are using. I will use the definition that says $y$ is a primitive $m$-th root of unity if $y^m=1$, and if $1\le k\lt m$, then $y^k\ne 1$.

We argue by contradiction. If for some $k\lt \frac{n}{2}$ we $(z^2)^k=1$, then $z^{2k}=1$ and $1\le 2k\lt n$, meaning that $z$ is not a primitive $n$-th root of unity.

Showing that $z^{-1}$ is a primitive $n$-th root of unity if $z$ is follows similar lines, and is left to you.

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