Lecture 22: Properties of Order

# Summary

Today we continued our discussion of the order of an integer a modulo m. We discussed many arithmetic properties of order, including its relationship to $\phi(m)$ as well as how one can predict the order of a power of an integer based on the order of the integer itself. We also discussed primitive roots more deeply, counting the number of primitive roots when they exist.

# More Properties of Order

Last class period we finished with the following

Lemma: For integers m and a with $(a,m) = 1$, an integer n satisfies $a^n \equiv 1 \mod{m}$ if and only if $\mbox{ord}_m(a) \mid n$.

One of the important consequences of this result is the following

Corollary: For integers m and a with $(a,m) = 1$, the order of a mod m is a divisor of $\phi(m)$.

Proof: From the last lemma we know that whenever $a^n \equiv 1 \mod{m}$ then we have $\mbox{ord}_m(a) \mid n$. Since Euler's Theorem tells us that $a^{\phi(m)} \equiv 1 \mod{m}$, this gives us the desired result. $\square$

#### Example: Calculating an order mod 11

Suppose that you want to calculate $\mbox{ord}_{11}(2)$. Normally we'd need to calculate $2^j \mod{11}$ for all j in the range $1 \leq j \leq 10$, but according to the last corollary we don't need to hit all these j: it's enough to try out those j which are divisors of 10. So let's do it:

(1)
\begin{split} 2^1&\equiv 2 \not\equiv 1 \mod{11}\\ 2^2&\equiv 4 \not\equiv 1 \mod{11}\\ 2^5&\equiv 32 \equiv -1 \mod{11}\\ 2^{10}&\equiv (2^5)^2 \equiv (-1)^2 \equiv 1 \mod{11}. \end{split}

Hence we see that $\ord_9(2) = 10$. $\square$

#### Example: Calculating an order mod 47

Let's use the same idea to calculate $\mbox{ord}_{47}(2)$. To do this, we need to know that the prime factorization of $\phi(47)$ is $\phi(47) = 2\cdot 23$. Then we only need to check the value of $2^j \mod{47}$ when $j \in \{1,2,23,47\}$.

(2)
\begin{split} 2^1 &\equiv 1 \notequiv 1 \mod{47}\\ 2^2 &\equiv 4 \notequiv 1 \mod{47}\\ 2^{23} &\equiv 2^{16}2^{4}2^{2}2^{1} \equiv 18 \cdot 16 \cdot 4\cdot 2 \equiv 1 \mod{47}. \end{split}

Hence we see that $\mbox{ord}_{47}(2) = 23$. Notice that this means that 2 is not a primitive root for this prime number. $\square$

# Primitive Roots as Generators

This is not the only useful corollary to come out of our lemma.

Corollary: For integers m and a with $(a,m) = 1$, then $a^i \equiv a^j \mod{m}$ if and only if $i \equiv j \mod{\mbox{ord}_m(a)}$.

Proof: Suppose first that $i \equiv j \mod{\mbox{ord}_m(a)}$. This tells us that

(3)
\begin{align} i = j+k\cdot \mbox{ord}_{m}(a) \end{align}

for some integer k. Hence we get

(4)
\begin{align} a^i \equiv a^{j+k\cdot \mbox{\tiny{ord}}_m(a)} \equiv a^j \left(a^{\mbox{\tiny{ord}}_m(a)}\right)^k = a^j (1)^k \equiv a^j \mod{m}. \end{align}

On the other hand, suppose that we know $a^i \equiv a^j \mod{m}$. Without loss of generality, assume additionally that $i>j$. Then we have

(5)
\begin{align} a^j a^{i-j} \equiv a^i \equiv a^j \mod{m}. \end{align}

Now since $(a,m) = 1$ we know $(a^j,m) = 1$, and hence we can "cancel" the appearance of $a^j$ from both sides of the above equation to conclude

(6)
\begin{align} a^{i-j} \equiv 1 \mod{m}. \end{align}

But our previous corollary says that this is only possible if $\mbox{ord}_m(a) \mid i-j$, which is what we wanted to prove. $\square$

This lemma doesn't wind up being especially useful for computing the order of a given element, since typically one doesn't have access to two integers i and j so that $a^i \equiv a^j \mod{m}$. It is, however, quite useful from a theoretical standpoint, as we're abou to see.

Lemma: Suppose that a is a primitive root mod m. Then the set $\{a,a^2,\cdots,a^{\phi(m)}\}$ is a complete set of reduced residues mod m.

Recall that a residue r is said to be reduced mod m if $(r,m) = 1$. Hence the content of the above corollary is that for any number n which has $(n,m) =1$ there is some exponent $1 \leq k \leq \phi(m)$ so that

(7)
\begin{align} a^k \equiv n \mod{m}. \end{align}

Proof: Certainly since $(a,m) = 1$ we know that $(a^k,m) =1$ for all $k \geq 1$. Hence we only need to show that if i and j are integers between 1 and $\phi(m)$, then $a^i \not\equiv a^j \mod{m}$. To do this, suppose instead that we had distinct i and j between 1 and $\phi(m)$ such that $a^i \equiv a^j \mod{m}$. According to the previous result, this would imply that $\mbox{ord}_m(a) \mid i-j$. But since $\mbox{ord}_m(a) = \phi(m)$, this means $\phi(m) \mid i-j$. This, however, is impossible because $1 \leq |i-j| \leq \phi(m)-1$. $\square$

This theorem is really quite powerful, because it tells us that if we can get a hold of a primitive root mod m, then we can use this element to express all other reduced residues. This is especially useful in light of the following

Lemma: Suppose that m and a are integers satisfying $(a,m) = 1$. Then we have

$\displaystyle \mbox{ord}_m(a^i) = \frac{\mbox{ord}_m(a)}{(\mbox{ord}_m(a),i)}$

Proof: Let's write d in place of $(\mbox{ord}_m(a),i)$. We'll also write $\mbox{ord}_m(a) = d \cdot b$ and $i = d\cdot k$. Notice that when we do this, we have $(b,k) = 1$. Now to compute the order of $a^i$, we need to find the smallest exponent which send this element to 1 mod m.

To start, notice that we have

(8)
\begin{align} (a^i)^b \equiv (a^i)^{\frac{\mbox{\tiny{ord}}_m(a)}{d}} \equiv (a)^{\frac{i\cdot \mbox{\tiny{ord}}_m(a)}{d}} \equiv \left(a^{\mbox{\tiny{ord}}_m(a)}\right)^{\frac{i}{d}} \equiv \left(a^{\mbox{\tiny{ord}}_m(a)}\right)^{k} \equiv 1^k \equiv 1. \end{align}

This implies that we have

(9)
\begin{align} \mbox{ord}_m(a^i) \mid b. \end{align}

On the other hand, we know that

(10)
\begin{align} (a)^{i \cdot \mbox{\tiny{ord}}_m(a^i)} \equiv (a^i)^{\mbox{\tiny{ord}}_m(a)} \equiv 1 \mod{m} \end{align}

by definition. Hence we know that $\mbox{ord}_m(a) \mid i \mbox{ord}_m(a^i).$ Considering $bd = \mbox{ord}_m(a)$ and $kd = i$, this means we get $bd \mid (kd)\mbox{ord}_m(a^i).$ Canceling the "d" on both sides then gives $b \mid k \mbox{ord}_m(a^i)$, and since $(b,k) = 1$ we conclude that

(11)
\begin{align} b \mid \mbox{ord}_m(a^i). \end{align}

Putting together Equations (9) and (11) gives $b = \mbox{ord}_m(a^i)$. $\square$

#### Example: Computing orders modulo 11

We already know that 2 is a primitive roots mod 11, so let's use this fact to compute the order of other elements mod 11.

$j$ $2^j \mod{11}$ $\mbox{gcd}(\mbox{ord}_{11}(2),j)$ $\mbox{ord}_{11}(2^j)$
1 2 1 10
2 4 2 5
3 8 1 10
4 $2\cdot 8 \equiv 5$ 2 5
5 $2\cdot 5 \equiv 10$ 5 2
6 $2\cdot 10 \equiv 9$ 2 5
7 $2\cdot 9 \equiv 7$ 1 10
8 $2\cdot 7 \equiv 3$ 2 5
9 $2\cdot 3 \equiv 6$ 1 10
10 $2\cdot 6 \equiv 1$ 10 1

# Counting Primitive Roots

A nice consequence of the previous result is that we can count primitive roots — at least when they exist.

Corollary: Suppose that a primitive root exists mod m. Then there are $\phi(\phi(m))$ many primitive roots.

Proof: Let a be a primitive root. We already know that all reduced residues take the form $a^k$ where $1 \leq k \leq \phi(m)$, and the previous result tells us that

(12)
\begin{align} \mbox{ord}_m(a^k) = \frac{\mbox{ord}_m(a)}{(\mbox{ord}_m(a),k)} = \frac{\phi(m)}{(\phi(m),k)}. \end{align}

Hence we see that $a^k$ has order equal to $\phi(m)$ precisely when $(\phi(m),k) = 1$. By the definition of the $\phi$ function, there are precisely $\phi(\phi(m))$ many choices for k that satisfy this criterion. $\square$

One of the questions that was asked in class was: why do we need to assume there is a primitive root to prove this theorem? The answer is that in order to write an equation like

(13)
\begin{align} \mbox{ord}_m(a^k) = \frac{\phi(m)}{(\phi(m),k)} \end{align}

we need to know that there really is an element a that has $\mbox{ord}_m(a) = \phi(m)$. It is worth nothing that not all m have a primitive root.

#### Example: A Modulus without Primitive Root

Let $m = 12$. Then the reduced residues are $\{1,5,7, 11\}$. Notice that $1^1 \equiv 1 \mod{12}$, and that

• $5^2 \equiv 25 \equiv 1 \mod{12}$
• $7^2 \equiv 49 \equiv 1 \mod{12}$
• $11^2 \equiv (-1)^2 \equiv 1 \mod{12}$

Hence we have

$a$ $\mbox{ord}_{12}(a)$
1 1
5 2
7 2
9 2

You can see that in this case there is no primitive root. $\square$