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Lesson 10 of 30

Prove when one residue generates all units

Establish primitive roots for primes and their permitted composite extensions.

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Three generates every nonzero class modulo seven: its six successive powers are 3, 2, 6, 4, 5, 1. Such a unit is called a primitive root. Modulo eight there are four units, but each squares to one, so none generates all four. The existence question therefore needs a theorem, not a presumption that every finite multiplicative system resembles the prime example.

Predict

Predict why the prime modulus matters. Over a field, a polynomial of degree d has at most d roots. The composite residue ring modulo eight violates that conclusion for X²−1, which has four roots. This root-count distinction will force cyclicity in the prime case. Counting a few powers can find a generator for a small modulus, but it cannot prove that one exists for every prime.

Experiment

Compare two unit systems.

ModulusUnit countLargest orderPrimitive root?
7663 works
842None

For modulus eleven, test a candidate using only exponents ten divided by its prime divisors two and five. Explain why passing those tests certifies full order rather than merely suggesting it.

Understand

Let M be the least common multiple of the orders of all units in F_p. For each prime power dividing M, choose an element whose order contains that full prime power and raise it to remove the other factors. Multiplying these commuting elements of pairwise coprime orders produces order M. Every unit satisfies x^M=1, so the polynomial root bound gives p−1≤M. Fermat makes every order divide p−1, hence M divides p−1. Therefore M=p−1 and a primitive root exists. The same proof works for any finite subgroup of a field’s multiplicative group. Once a generator exists, its generating powers are exactly the exponents coprime to the group order.

Apply

For odd p, choose a primitive root g modulo p with g^(p−1) not congruent to one modulo p²; if necessary replace g by g+p, using the binomial correction proved in practice. If g^(p−1)=1+pc with p∤c, repeated binomial expansion shows v_p(g^((p−1)p^j)−1)=j+1. The leading term gains one factor p at each step, while all later terms gain at least two because p is odd. Thus g has order (p−1)p^(k−1) modulo p^k, the number of units. Choosing an odd representative also gives a primitive root modulo 2p^k by CRT.

Continue

The complete list is 1, 2, 4, p^k and 2p^k for odd prime p and k≥1, with the trivial modulus-one unit convention stated explicitly. The positive cases above include direct checks for two and four. For exclusion, two distinct odd prime factors, or a factor four together with an odd prime, give by CRT a quotient whose two unit factors both have even order; no element can have the product order. A modulus divisible by eight maps onto the noncyclic unit group modulo eight, whereas every quotient of a cyclic group is cyclic. These obstructions exhaust the other factorizations. This classification depends on the proved lifting and CRT arguments, not merely a table of small moduli.

Return to the number theory unit to choose a targeted retry. Continue with count divisors and units through prime powers.

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Practice and reasoning

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Find the order of 3 modulo 7.

Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.

How many primitive roots modulo 11 are there?

Integers, fractions and finite decimals are accepted. Expressions such as sqrt(2) are not parsed by this checker.

Why does a nonzero degree-d polynomial over a field have at most d roots?

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Outline the constructive group argument proving U(p) cyclic.

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Prove no primitive root exists modulo 8.

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For odd p, why can a primitive root g mod p be replaced by g+p if g^(p−1)≡1 mod p²?

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Reflect and continue

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