Summary
Today we proved the (majority of) the primitive root theorem, and so we are now able to determine exactly which moduli m have primitive roots. Indeed, this means we can count precisely how many primitive roots a number m has just by looking at a factorization of m.
Integers with Primitive Roots
Last class period we were able to show that there are a handful of integers which do not have primitive roots. Our two big classes of such moduli were
- powers of 2 that are at least 8
- products mn so that $(m,n) = 1$ and so that $m,n>2$.
Today we'll aim to show that the remaining integers — i.e., the integers $1,24$ together with integers of the form $p^e$ or $2p^e$, where p is an odd prime — all *do* have primitive roots. We already know that primitive roots exist modulo p, so we'll start our investigation by verifying that primitive roots exist modulo $p^2$.
Theorem: A modulus m has primitive roots if and only if $m = 1,2,4,p^e \mbox{ or } 2p^e$, where p is an odd prime.
Example: When do primitive roots exist?
From the previous theorem we know that there are no primitive roots modulo 65, since $65 = 3\cdot 13$ and hence doesn't match any of the "right" forms from our theorem. On the other hand, we do know that $242 = 2 \cdot 11^2$ does have primitive roots. $\square$
Proposition: If p is an odd prime, then there exists a primitive root mod $p^2$. In fact, if $r$ is a primitive root modulo p, then r is a primitive root modulo $p^2$ if and only if
$r^{p-1} \nequiv 1 \mod{p^2}$.
In the case that $r^{p-1} \equiv 1 \mod{p^2}$, then the number $r+p$ is a primitive root modulo $p^2$.
Notice that this proposition doesn't just say there are primitive roots mod $p^2$, but gives us a test we can run to actually product primitive roots modulo $p^2$ quickly.
Proof: Suppose that r is a primitive root mod p; we already know such an element exists. Now we know that if $n = \mbox{ord}_{p^2}(r)$, then we have $n \mid \phi(p^2) = p(p-1)$. We also know that $r^n \equiv 1 \mod{p^2}$ implies $r^n \equiv 1 \mod{p}$, and so we get $p-1 = \mbox{ord}_p(r) \mid n$. Together, this means that we have either $n = p-1$ or $n = p(p-1)$. If the latter case is true, then we're finished: such an r is a primitive root mod $p^2$.
Otherwise, consider the element $r+p$. Since it is congruence to r mod p, it is still a primitive root mod p. Hence its order is either $p-1$ or $p(p-1)$. We'll show that the former is impossible, and so we will be able to conclude that $r+p$ is a primitive root mod $p^2$. To see that $r+p$ can't have order $p-1$, we compute:
(1)Now the "tail" of this expression is clearly divisible by $p^2$, and so modulo $p^2$ we have
(2)We are already assuming that the order of r mod $p^2$ is $p-1$, and hence we get $r^{p-1} \equiv 1 \mod{p^2}$. Also, we know that $pr^{p-2} \not\equiv 0 \mod{p^2}$ since $(r,p) = 1$. Therefore the previous equation says
(3)Hence we get that $\mbox{ord}_{p^2}(r+p) \neq p-1$, and so we must have that $r+p$ is a primitive root mod $p^2$. $\square$
Example: Finding a primitive root mod 121
From the book we know that 2 is a primitive root modulo 11 (note that we could prove that 2 is a primitive root by calculating $2^1,2^2$ and $2^5$ mod 11 and showing that none of these quantities is $1 \mod{11}$). Hence to find a primitive root modulo 121, we need to determine the value of $2^{10} \mod{121}$. From the result above, if this quantity isn't 1, then we'll know 2 is a primitive root; on the other hand, if $2^{10} \equiv 1 \mod{121}$, then we'll know that 2 isn't a primitive root mod 121, but that $2+11 = 13$ is. Using successive squaring we were able to see that
(4)and so we know that 2 is a primitive root mod 121. $\square$
This might seem like we've only made slight progress, moving from primitive roots mod p to primitive roots mod $p^2$. As it turns out, though, this firs step is all one needs to do: primitive roots mod $p^2$ are always primitive roots mod $p^m$.
Proposition: Any primitive root mod $p^2$ is a primitive root mod $p^e$, where p is an odd prime.
Proof: Let r be a primitive root mod $p^2$, and write n for $\mbox{ord}_{p^e}(r)$. As before, we know that
(5)Moreover we have that
(6)Since we know that $\mbox{ord}_{p^2}(r) = p(p-1)$ (since r is a primitive root mod $p^2$), Equations (5) and (6) together imply
(7)Now we would like to show that $n = p^{e-1}(p-1)$, so we need to show that no choice of $k<e-1$ is possible in the above equation. To do this, we'll show that
(8)in the next lemma. In particular, this equation will force $\mbox{ord}_{p^e}(r) \neq p^k(p-1)$ for $k<e-1$, and hence we will be forced into the condition $\mbox{ord}_{p^e}(r) = p^{e-1}(p-1)$; i.e., r will be a primitive root. $\square$
Lemma: Suppose that r is a primitive root mod $p^2$. Then
$r^{p^{e-2}(p-1)} \not\equiv 1 \mod{p^e}$.
Proof: We'll prove this result by induction. The case $e=2$ is already taken care of since we know that
(9)(since r is a primitive root, the smallest exponent which sends r to 1 is $\phi(p^2) = p(p-1)$). So assume that we know the result holds for $e-1$.
Now we know that
(10)and hence we have
(11)notice that $p \nmid c$ since otherwise we'd have $r^{p^{e-3}(p-1)} \equiv 1 \mod{p^e-1}$, contrary to the induction hypothesis. Now we'll raise both sides of Equation (11) to the pth power. We get
(12)Notice that all the first two terms are the only which survive when we consider this equation mod $p^e$. For instance, we know that $\left(\begin{array}{cc}p\\2\end{array}\right)(cp^{e-2})^2 = \frac{p(p-1)}{2}c^2p^{2e-4}$. Since p is odd, the term $\frac{p(p-1)}{2}$ is a multiple of p, and so this term is divisible by $p^{2e-3}$. Since we know $e \geq 3$, this means $2e-3 \geq e$. Hence the third term of this expansion is 0 modulo $p^e$. Any other term has a factor of the form $(p^{e-2})^k$ for $k \geq 3$, and since $ke-2k \geq e$ for such k, we know these terms are also congruent to 0 modulo $p^e$.
Hence we have
(13)Now since $p \nmid c$ we get $cp^{e-1} \not\equiv 0 \mod{p^e}$, and so we have
(14)as desired. $\square$
Having proven this result, we then stated the following
Proposition: If r is an odd primitive root modulo $p^2$, then it is also a primitive root modulo $2p^e$ for every e.