Question

If G is a cyclic group of order 12, then the order of Aut(G) is:

• A. 1
• B. 5
• C. 4
• D. 77

Hint:

Remembering the relationship \( |\text{Aut}(\mathbb{Z}_n)| = \phi(n) \) is a major shortcut for this type of problem. An automorphism of \(\mathbb{Z}_n\) is determined by where it sends the generator 1. It must send 1 to another generator, and the generators of \(\mathbb{Z}_n\) are precisely the integers \(k\) such that \(1 \le k<n\) and gcd(k,n)=1. The number of such integers is \( \phi(n) \).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The question asks for the order of the automorphism group of a cyclic group of order 12. An automorphism is an isomorphism from a group to itself. The set of all automorphisms of a group G, denoted Aut(G), forms a group under composition.

Step 2: Key Formula or Approach:
For a cyclic group of order n, \( G \cong \mathbb{Z}_n \), the automorphism group Aut(G) is isomorphic to the group of units modulo n, \( U(n) \). The order of this group is given by Euler's totient function, \( \phi(n) \). \[ |\text{Aut}(\mathbb{Z}_n)| = |U(n)| = \phi(n) \] We need to calculate \( \phi(12) \).

Step 3: Detailed Explanation:
Euler's totient function, \( \phi(n) \), counts the number of positive integers up to a given integer \(n\) that are relatively prime to \(n\). We can calculate \( \phi(12) \) in two ways: 1. By counting: List the integers from 1 to 12 and find those with a greatest common divisor of 1 with 12. The numbers are: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11. The ones relatively prime to 12 are {1, 5, 7, 11}. There are 4 such numbers. So, \( \phi(12) = 4 \). 2. Using the formula: First, find the prime factorization of 12: \( 12 = 2^2 \times 3^1 \). The formula for \( \phi(n) \) is \( \phi(n) = n \prod_{p|n, p \text{ is prime}} (1 - \frac{1}{p}) \). \[ \phi(12) = 12 \left(1 - \frac{1}{2}\right) \left(1 - \frac{1}{3}\right) = 12 \times \frac{1}{2} \times \frac{2}{3} = 4 \]
Step 4: Final Answer:
The order of Aut(G) is 4.