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Lesson 9 of 30

Find the true period of a unit

Prove order divisibility and compute periods of powers.

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Powers of two modulo seven are 2, 4, 1, then repeat. Euler’s theorem supplies exponent six, but the first return to one occurs at three. This least positive exponent is the multiplicative order. It is defined for a unit, since invertibility ensures a finite multiplicative cycle returning to one. A nonunit can instead enter a pattern that never includes one.

Predict

Predict the order of four modulo seven. Since four is 2² and two has order three, its powers also cycle with length three. But squaring an element of order six would produce order three. The answer depends on the gcd between the exponent and the original order, not on a universal rule that squaring halves the period.

Experiment

Inspect powers of three modulo seven.

Exponent123456
Residue326451

The order is six. Read off the orders of 3² and 3³ from the table, then verify the formula d/gcd(d, k). Use the table to solve 3^x≡5 modulo seven, stating the full exponent class rather than one exponent.

Understand

If d is the least positive exponent with a^d=1, division k=qd+r reduces any return exponent k to a^r=1 with 0≤r< d. Minimality forces r=0. Thus d divides every return exponent, including φ(m) by Euler’s theorem. For a^k, the return condition is d|kt. Dividing by gcd(d, k) makes the remaining factors coprime, so the least positive t is d/gcd(d, k). To certify d=N when a^N=1, it suffices to check a^(N/q)≠1 for every prime divisor q of N. Any proper divisor d of N would divide one of those exponents.

Apply

Orders also combine constructively. If commuting elements have coprime orders r and s, their product has order rs. A power equal to one puts a^k=b^(−k) in both cyclic subgroups; its order divides both r and s, hence it is one. Therefore both r and s divide k. This lemma allows prime-power pieces of orders to be assembled into a larger order, a key step in proving primitive roots exist modulo primes. Commutation is essential; arbitrary products in nonabelian groups need not obey the same conclusion.

Finding an order efficiently begins with a known multiple, such as φ(n) for a unit. Factor that multiple and test whether dividing it by a prime still gives exponent one. Whenever it does, replace the bound by the smaller exponent and repeat. The process terminates at the exact order because any remaining proper divisor would omit at least one prime factor and would have been detected. This method depends on knowing a factorization of the bound; the elementary theorem does not guarantee that obtaining that factorization is cheap.

Continue

An order certificate should include a return exponent and tests excluding all proper divisors through its prime factors. A trace of several nonidentity powers is not enough if possible divisors remain unchecked. Computing orders can be difficult when the surrounding group size is not factored. The next lesson uses field root counts to prove existence of a generator and explains precisely which composite moduli admit primitive roots.

Return to the number theory unit to choose a targeted retry. Continue with prove when one residue generates all units.

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Practice and reasoning

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Find the order of 2 modulo 7.

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

If a has order 18, find the order of a⁶.

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

If a has order d and a^k=1, prove d|k.

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Prove ord(a^k)=d/gcd(d, k).

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Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.

Suppose a^N=1. Prove ord(a)=N iff a^(N/q)≠1 for every prime q dividing N.

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For commuting elements of coprime orders r, s, prove their product has order rs.

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Reflect and continue

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