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 $p - 1$.