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While studying primitive roots, I came across the following lemma:

Lemma: Let $p$ and $q$ be primes and suppose that $q^\alpha\mid p-1$, where $\alpha\geq 1$. Then there are precisely $q^\alpha - q^{\alpha -1}$ residue classes $a\pmod p$ of order $q^\alpha$.

However, I also know that given a $d\mid p-1$, $x^d\equiv 1\pmod p$ has $d$ solutions ($p$ is a prime). So, is the lemma in agreement with the previous statement? Because according to the statement shouldn't it be $q^\alpha$ residue classes instead of $q^\alpha - q^{\alpha -1}$ residue classes?

Am I missing something? Where am I going wrong?

Michael Hardy
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Apurv
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  • Is $\alpha$ the greatest number with $q^{\alpha}|p-1$ ? – Peter Jun 10 '15 at 17:45
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    I changed a(\text{mod }p) to a\pmod p. That is standard. Notice that if there is more than one character after \pmod, then you need braces, as in a\pmod{43}, so that you see $a\pmod{43}$ and not $a\pmod43$. ${}\qquad{}$ – Michael Hardy Jun 10 '15 at 17:52

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There are $q^\alpha$ residue classes of order less than or equal to $q^\alpha$. There are $q^{\alpha-1}$ of order strictly less than $q^{\alpha}$. So there are $q^\alpha-q^{\alpha-1}$ of order exactly $q^\alpha$.

Added There are indeed $q^\alpha$ solutions of the congruence $x^{q^\alpha}\equiv 1\pmod{p}$. They all have order that divides $q^\alpha$. But they do not all have order $q^{\alpha}$. For concreteness, suppose from now on that $\alpha=4$.

Note that $x=1$ is a solution of the congruence $x^{q^4}\equiv 1\pmod{p}$, and $1$ has order $1$. And among the solutions of $x^{q^4}\equiv 1\pmod{p}$, there are the $q$ solutions of $x^q\equiv 1\pmod{p}$. Of these, one has order $1$, and the rest have order $p$. So being a solution of $x^{q^4}\equiv 1\pmod{p}$ does not make $x$ have order $4$.

Among the $q^4$ solutions of $x^{q^4}\equiv 1\pmod{p}$ there are the $q^3$ solutions of $x^{q^3}\equiv 1\pmod{p}$. There all have order a divisor of $q^3$. The rest of the solutions have order $q^4$, giving $q^4-q^3$ elements of order exactly $q^4$.

André Nicolas
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  • But doesn't the statement say that there are $q^\alpha$ solutions to $x^{q^\alpha}\equiv 1\pmod p$? – Apurv Jun 10 '15 at 17:58
  • It certainly does. One of the solutions is $x=1$, which certainly does not have order $q^\alpha$. More generally, take for example $\alpha=5$. Among the $q^5$ solutions of $x^{q^5}\equiv 1\pmod{p}$, there are the $q^2$ solutions of $x^{q^2}\equiv 1\pmod{p}$. These solutions do not have order $q^5$. They have order $\le q^2$. – André Nicolas Jun 10 '15 at 18:02
  • Okay, Thanks, you clarified my doubt. +1 – Apurv Jun 10 '15 at 18:04
  • You are welcome. – André Nicolas Jun 10 '15 at 18:04