Lecture 22: Properties of Order


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:

\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 $j \in \{1,2,23,47\}$.

\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

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

for some integer k. Hence we get

\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

\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

\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

\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

\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

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

On the other hand, we know that

\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

\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

\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

\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. $\square$

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