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To solve this problem, we need to calculate how many generators the cyclic group \( G \) of order 11 has.

### Key Points:
- A **cyclic group** of order \( n \) (in this case \( n = 11 \)) is a group where each element can be written as powers of a single element called the **generator**.
- The number of generators of a cyclic group of order \( n \) is equal to the number of elements in \( \{1, 2, ..., n-1\} \) that are **coprime** with \( n \). This is given by **Euler’s Totient Function**, \( \phi(n) \).
  
For a prime number \( p \), \( \phi(p) = p - 1 \), because every number from \( 1 \) to \( p-1 \) is coprime to \( p \).

Since 11 is prime:
\[
\phi(11) = 11 - 1 = 10
\]

Thus, the cyclic group \( G \) of order 11 has **10 generators**.

### Correct answer: (d) 10 generators.

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Here are 5 related questions you might find useful:
1. What is a cyclic group?
2. How do you calculate Euler’s Totient Function for a composite number?
3. What are generators in group theory?
4. How do generators of cyclic groups behave under modular arithmetic?
5. How can we find all generators of a cyclic group manually?

**Tip:** For any prime \( p \), the number of generators in its cyclic group is always \( p - 1 \).

If the cyclic group G contains 11 distinct elements, then it has (a) 2 generators (b) 7 generators (c) 9 generators (d) 10 generators

Learn how to determine the number of generators for a cyclic group of order 11 using Euler's Totient Function. This problem explores key concepts in group theory and number theory. Suitable for undergraduate-level mathematics students.

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