Primitive Roots (I) Example: Consider U32. For any element a ∈U32 ...

Primitive Roots (I) Example: Consider U 32 . For any element a ∈ U 32 , ord 32a|ϕ (32) = 16. But (±a ± 16)2 ≡ a 2 (mod 32), so € a mod32 €

1 3 € 5 7 15 13 11 9 17 19 21 23 31 29 27 25 a 2 mod 32 1 9 25 17

€

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. € what we observed in this last example In fact, 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 2e 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 2e ), € € € € €

and therefore also ord 2e a k = 2. But by the Order Theorem, k

2€= ord 2e a =

ord 2e a (k, ord 2e a)

=

2e−1 (k, 2

e−1

)

,

so (k, 2 e−1 ) = 2e−2 . Thus, the only element of order 2 in € U 2e is a

2e−2

2e−2

. However, since a is odd, −a

must

e−2

€ €

be a second element (incongruent to a 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 €

ϕ (m )/2 € ϕ ( s) ϕ (t )/ 2 € a ≡ 1 (mod n) € €≡ (a )

aϕ (m )/2 ≡ (a ϕ (t ) )ϕ (s )/ 2 ≡ 1 (mod n) so that by the CRT, aϕ (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? € need a rather To answer this question, we first € 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]. € €

Proof Recall that if write the prime factorizations of k and l in the form k = p1d1 p2d2 L p dr r , l = p1e1 p2e2 L prer where di , ei ≥ 0, then €

(k,l ) = p1δ1 pδ22 L prδr , [k,l] = p1ε1 pε22 L pεrr

€

where δi = min(di, ei ) and εi = max(di, ei ). Let s be €the product of those prime power factors of k for which di = δi , and let u be the product of the € remaining prime€power factors (for which di = εi ). Similarly, let v be the product of those prime power € factors of l for which ei = εi , and let t be the product of the remaining prime power factors € (for which ei = δi ). 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]. € €

€ € s t Now put c = a b ; we claim € €that c is the desired element satisfying ord mc = [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 zb tw€≡ a szbtz +tw ≡ a sz (mod m)

€

so that ord ma sz = ord mb tw. By the Order Theorem, ord ma ord mb k l = , or = , (sz,ord ma) (tw, ord mb) (sz,k) (tw,l)

€

which we can write as €

su tv u v = , or = . (sz,su) (tw,tv) (z,u) (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. € particular, the argument € in the last paragraph In € can be used to show that u|ord mc . An entirely similar argument implies that v|ordmc . But again, (u,v) = 1, so uv|ordmc . On the other hand,

€ €

c €

uv

s t uv

≡ (a b )

€ tv u k v l u ≡ (a su )v (b ) ≡ (a ) (b ) ≡ 1 (modm) €

€ whence ord mc|uv. Therefore, ord mc = uv = [k,l]. // €

Let us illustrate the procedure outlined in the proof of the lemma by means of an example: € €

Example: 215 ≡ 1 (mod100), so ord10021 = 5. Also, 432 ≡ 49 (mod100) and 434 ≡ 1 (mod100), so ord10043 = 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 = 211 ⋅ 431 ≡ 3 € (mod100). Thus, ord1003 = [5,4] = 20.

€

Proposition λ (m) = max ord ma. In particular, €

€ €

a∈Um

there exists an element in U m of order λ (m). Proof€Let a ∈ U m be such that k = ordma is largest possible. By definition of λ (m), € we must then have € k ≤ λ(m). Also, if b is another element in U m and l =€ ordmb does not divide € k, then [k,l] > k, but by the lemma we can find € a c ∈ U m so that ord mc = [k,l], and this violates the maximality € of the order of a. Thus, the order of every element in U m must divide k k. This means for all x ∈ U m € that x ≡ 1 (modm) € whence λ (m) ≤ k . Thus λ (m) = k = ordma. // € € € € € €