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Let's define the set of $r^{th}$-residues in $\mathbb{Z}_p^*$ as:

$\mathbb{Z}_p^r := \{ y \in \mathbb{Z}_p* : \exists x \in \mathbb{Z}_p^* \text{ such that } y = x^r\mod{p} \}$.

I am reading that when $p$ is prime for which $p-1=qr$ with $q$ a prime that is not a divisor of $r$, then:

  1. $\mathbb{Z}_p^r$ is an order $q$ cyclic subgroup of $\mathbb{Z}_p^*$
  2. For each $y \in \mathbb{Z}_p^*$, $y \in \mathbb{Z}_p^r$ if and only if $y^q \mod{p} = 1$

Why are this both statements true?

Nick
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Lecter
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1 Answers1

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First note that $\mathbb{Z}_p^*$ is cyclic, and so is any of its subgroups. Let $\xi$ be a generator of $\mathbb{Z}_p^*$. Then for $\xi^i,\,\xi^j\in\mathbb{Z}_p^*$, $\xi^i=\left(\xi^j\right)^r$ is equivelant to $$i\equiv jr\pmod{p-1}.$$ Let $j$ run through $\{0,\dots,q-1\}$, and we obtain $\mathbb{Z}_p^r$, which is exactly the subgroup generated by $\xi^r$. And the first statement follows.

It remains to show that if $y^q\mod p=1$, then $y\in\mathbb{Z}_p^r$. One may see here.