This question is from group theory and involves finding the number of elements with a specific order within a finite cyclic group.
Step 1: Understanding the Question:
We have a cyclic group \(G\) with order \(|G|=25\). We need to find the number of elements \(g \in G\) such that the order of \(g\) is 5.
Step 2: Key Formula or Approach:
A fundamental theorem in group theory states that in a finite cyclic group of order \(n\), the number of elements of order \(d\) (where \(d\) is a divisor of \(n\)) is given by Euler's totient function, \( \phi(d) \).
Step 3: Detailed Explanation:
Identify n and d: The order of the group is \(n=25\). The desired order of the elements is \(d=5\).
Check the condition: We must verify that \(d\) divides \(n\). Indeed, 5 divides 25. Therefore, elements of order 5 exist.
Calculate \( \phi(d) \): We need to compute \( \phi(5) \). Euler's totient function \( \phi(k) \) counts the number of positive integers less than or equal to \(k\) that are relatively prime to \(k\).
For a prime number \(p\), \( \phi(p) = p-1 \). Since 5 is a prime number, we have:
\[ \phi(5) = 5 - 1 = 4 \]
This means there are exactly 4 elements of order 5 in any cyclic group of order 25.
Step 4: Final Answer:
There are 4 elements of order 5 in a cyclic group of order 25.