a mod a 2 mod

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1 Primitive Roots (I) Example: Consider U 32. For any element a U 32, ord 32 a ϕ(32) = 16. But (±a ±16) 2 a 2 (mod 32), so a mod a 2 mod This shows that 15, 17, 31 have order 2; that 7, 9, 23, and 25 have order 4; and that the other eight elements of U 32 (excluding 1) have order 8. In particular, no element has order 16, so there is no primitive root mod 32. In fact, what we observed in this last example extends to most powers of 2. Proposition There is no primitive root mod 2 e for any e 3. Proof Suppose e 3 and that a is a primitive root mod 2 e. Then since ϕ(2 e ) = 2 e 1, U 2 e can be represented by the powers a k of a for 1 k 2 e 1. One of these powers must satisfy a k 1 (mod 2 e ),

2 and therefore also ord 2 ea k = 2. But by the Order Theorem, 2 = ord 2 ea k = ord 2 ea (k,ord 2 ea) = 2 e 1 (k,2 e 1 ), so (k,2 e 1 ) = 2 e 2. Thus, the only element of order 2 in U 2 e is a 2e 2. However, since a is odd, a 2e 2 must be a second element (incongruent to a 2e 2 ) of order 2, contradiction. // This leads to an obvious question: for which moduli m do there exist primitive roots? It turns out that the nonexistence of primitive roots is rather common: Proposition If m can be expressed as the product of two relatively prime numbers greater than 2, then there is no primitive root mod m. Proof Write m = st where s,t 2 and (s,t) = 1. Then since both ϕ(s) and ϕ(t) are even, we find that for any a relatively prime to m, and a ϕ(m)/2 (a ϕ( s) ) ϕ(t )/ 2 1 (mod n)

3 a ϕ(m)/2 (a ϕ(t ) ) ϕ(s )/ 2 1 (mod n) so that by the CRT, a ϕ(m)/2 1 (modm). This means that there is no element of U m of order ϕ(m), so there can be no primitive root mod m. // The set of all moduli for which there is a primitive root was first determined by Gauss. The argument we present here is based on application of Lagrange s Theorem and a function first studied by Carmichael in the 1920s, the minimal universal exponent function: λ(m) = smallest positive integer for which a λ(m) 1(mod m) holds for all a U m For instance, we saw earlier that λ(32) = 8. Also, if there is a primitive root mod m then if λ(m) = ϕ(m). Notice that the definition of λ(m) does not necessarily imply that the converse of this statement must be true. But is it true nonetheless? To answer this question, we first need a rather technical Lemma Suppose a,b U m have orders k and l, respectively. Then there must be an element in U m of order [k,l].

4 Proof Recall that if write the prime factorizations of k and l in the form where d i,e i 0, then k = p 1 d 1 d 2 Lp r d r, l = p 1 e 1 e 2 Lp r e r (k,l) = p 1 δ 1 δ 2 Lp r δ r, [k,l] = p 1 ε 1 ε 2 L p r ε r where δ i = min(d i,e i ) and ε i = max(d i,e i ). Let s be the product of those prime power factors of k for which d i = δ i, and let u be the product of the remaining prime power factors (those for which d i = ε i ). Similarly, let v be the product of those prime power factors of l for which e i = ε i, and let t be the product of the remaining prime power factors (those for which e i = δ i ). Then k = su, l = tv, and (s,u) = (t,v) = 1 as well as (s,t) = (u,v) = 1. Moreover, st = (k,l) and uv = [k,l]. Now put c = a s b t ; we claim that c is the desired element satisfying ord m c = [k,l]. Given an integer z, define w to be the standard residue of z mod l. Then, if z is chosen so that c z 1 (modm), it follows that b tw c z b tw a sz b tz+tw a sz (mod m)

5 so that ord m a sz = ord m b tw. By the Order Theorem, ord m a (sz,ord m a) = ord m b (tw,ord m b), or k (sz,k) = l (tw,l), which we can write as su (sz,su) = tv (tw,tv), or u (z,u) = v (w,v). But then u (w,v) = v (z,u), whence u v (z,u). Since (u,v) = 1, we deduce that u (z,u). But this forces u = (z,u) and we conclude that u z. In particular, the argument in the last paragraph can be used to show that u ord m c. An entirely similar argument implies that v ord m c. But again, (u,v) = 1, so uv ord m c. On the other hand, c uv (a s b t ) uv (a su ) v (b tv ) u (a k ) v (b l ) u 1 (modm) whence ord m c uv. Therefore, ord m c = uv = [k,l]. // Let us illustrate the procedure outlined in the proof of the lemma by means of an example:

6 Example: (mod100), so ord = 5. Also, (mod100) and (mod100), so ord = 4. As a = 21, b = 43, we have k = 5, l = 4. But (k,l) = 1, so s = 1, u = 5, and t = 1, v = 4, and c = (mod100). Thus, ord = [5,4] = 20. Proposition λ(m) = max ord m a. In particular, a U m there exists an element in U m of order λ(m). Proof Let a U m be such that k = ord m a is the maximum possible order of all elements of U m. By definition of λ(m), we must then have k λ(m). Also, if b is another element in U m and l = ord m b does not divide k, then [k,l] > k. But by the lemma we can find a c U m so that ord m c = [k,l], and this violates the maximality of the order of a. Thus, the order of every element in U m must divide k. This means that x k 1 (modm) for all x U m whence λ(m) k. Thus λ(m) = k = ord m a. //