Math Problem Statement

If G is a finite cyclic group of order n generated by a then prove that a^i is a generator of G if gcd ( i,n)=1

Solution

To prove that aia^i is a generator of the cyclic group GG (which is of order nn) if and only if gcd(i,n)=1\gcd(i, n) = 1, we proceed step-by-step.

Given:

  • GG is a finite cyclic group of order nn, so G=aG = \langle a \rangle, meaning every element of GG can be written as some power of aa, specifically a0,a1,a2,,an1a^0, a^1, a^2, \ldots, a^{n-1}.
  • gcd(i,n)=1\gcd(i, n) = 1, meaning that ii and nn are coprime (i.e., their greatest common divisor is 1).

We need to show that aia^i is a generator of GG, which means that the powers of aia^i must cover every element in GG.

Proof:

  1. Order of aia^i: The element aia^i is an element of GG, and we want to find the order of aia^i. The order of an element aia^i is the smallest integer kk such that: (ai)k=e,(a^i)^k = e, where ee is the identity element of GG. Since aia^i is a power of aa, we have: (ai)k=aik.(a^i)^k = a^{ik}. Thus, we need aik=ea^{ik} = e, and since an=ea^n = e (because aa generates a group of order nn), this implies: ik0(modn).ik \equiv 0 \pmod{n}. This means that kk must satisfy ikik being divisible by nn. In other words, we need kk to be the smallest integer such that ikik is a multiple of nn. Since gcd(i,n)=1\gcd(i, n) = 1, there exists a unique solution for kk, which is k=nk = n (because ii and nn are coprime, and multiplying ii by nn gives a multiple of nn).

    Therefore, the order of aia^i is nn.

  2. Conclusion: Since the order of aia^i is nn, it follows that aia^i generates all the elements of GG. Specifically, the elements of GG can be written as: {(ai)k:k=0,1,2,,n1}={aik:k=0,1,2,,n1}.\{(a^i)^k : k = 0, 1, 2, \ldots, n-1\} = \{a^{ik} : k = 0, 1, 2, \ldots, n-1\}. Since kk runs from 00 to n1n-1, and the powers aika^{ik} cover all distinct elements of GG, aia^i is indeed a generator of GG.

Thus, we have shown that if gcd(i,n)=1\gcd(i, n) = 1, then aia^i is a generator of GG.


Would you like further clarification on any part of this proof, or have any related questions?

Here are some further questions you might find interesting:

  1. How do we prove that if gcd(i,n)=d\gcd(i, n) = d, then the order of aia^i is nd\frac{n}{d}?
  2. Can we prove a similar result for infinite cyclic groups, like Z\mathbb{Z}?
  3. What is the relationship between generators of cyclic groups and primitive roots in modular arithmetic?
  4. How does the structure of a cyclic group relate to the divisor function in number theory?
  5. What is the order of an element in a direct product of cyclic groups, and how can this relate to generators?

Tip: When dealing with generators in cyclic groups, always check if the exponent and the group's order are coprime to determine if the element is a generator!

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Math Problem Analysis

Mathematical Concepts

Group Theory
Cyclic Groups
Number Theory

Formulas

gcd(i, n) = 1
(a^i)^k = e

Theorems

Cyclic Group Generator Theorem
Properties of gcd in Cyclic Groups

Suitable Grade Level

Undergraduate