Question:medium

Find the number of automorphisms of the cyclic group $\mathbb{Z}_n$ for $n = 30$.

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Number of automorphisms of $\mathbb{Z}_n$ equals $\varphi(n)$ because automorphisms correspond to choosing generators.
Updated On: Feb 15, 2026
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Correct Answer: 8

Solution and Explanation

Step 1: Understanding the Topic
The question asks for the number of automorphisms of the cyclic group $\mathbb{Z}_{30}$. An automorphism is an isomorphism from a group to itself. For a cyclic group $\mathbb{Z}_n$, an automorphism is determined by mapping a generator (like 1) to another generator of the group.
Step 2: Key Formula or Approach
The number of automorphisms of $\mathbb{Z}_n$, denoted $|\text{Aut}(\mathbb{Z}_n)|$, is equal to the number of generators of $\mathbb{Z}_n$. The generators of $\mathbb{Z}_n$ are the integers $k$ such that $1 \le k<n$ and $\gcd(k, n) = 1$. The number of such integers is given by Euler's totient function, $\varphi(n)$.
So, we need to calculate $\varphi(30)$.
Step 3: Detailed Calculation
A. Find the prime factorization of n=30:
\[ 30 = 2 \times 3 \times 5 \] B. Apply the formula for Euler's totient function:
The function $\varphi(n)$ is multiplicative, so $\varphi(abc) = \varphi(a)\varphi(b)\varphi(c)$ if a, b, c are pairwise coprime. For a prime $p$, $\varphi(p) = p-1$. \[ \varphi(30) = \varphi(2) \times \varphi(3) \times \varphi(5) \] \[ = (2-1) \times (3-1) \times (5-1) \] \[ = 1 \times 2 \times 4 = 8 \] Alternatively, using the formula $ \varphi(n) = n \prod_{p|n} (1 - 1/p) $: \[ \varphi(30) = 30 \left(1 - \frac{1}{2}\right) \left(1 - \frac{1}{3}\right) \left(1 - \frac{1}{5}\right) \] \[ = 30 \times \frac{1}{2} \times \frac{2}{3} \times \frac{4}{5} \] \[ = \frac{30 \times 1 \times 2 \times 4}{2 \times 3 \times 5} = \frac{240}{30} = 8 \] Step 4: Final Answer
The number of automorphisms of the cyclic group $\mathbb{Z}_{30}$ is 8. \[ \boxed{8} \]
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