The question in the image asks:

Determine the number of elements of order five in $Z_{125}$. Explain!

To solve this, let’s break down the problem.

Solution:

1. Understanding $Z_{125}$: $Z_{125}$ represents the set of integers modulo 125, i.e., the group $\{ 0, 1, 2, \ldots, 124 \}$ under addition modulo 125.

2. Element Order in $Z_{125}$: The order of an element $a$ in $Z_{125}$ is the smallest positive integer $k$ such that $k \cdot a \equiv 0 \mod 125$.

3. Prime Factorization: Since 125 can be factored as $5^3$, the order of any element in $Z_{125}$ must divide 125.

4. Finding Elements of Order 5: Elements of order 5 are those that satisfy:

   • $k \cdot a \equiv 0 \mod 125$ for $k = 5$.

   This means $a$ should be such that the least common multiple of 5 and 125 is 125. Thus, the order of each element $a$ needs to be 5, and $a$ must be of the form $125/5 = 25$.

5. Counting the Elements: The elements of order 5 in $Z_{125}$ will correspond to those integers that form a subgroup of order 5. Each cyclic group with an element of order 5 has exactly 4 elements of this order.

Thus, there are 4 elements of order 5 in $Z_{125}$.

Would you like more detailed steps or have further questions?

Related Questions:

1. How do we find elements of a specific order in other groups like $Z_{64}$?
2. What are the properties of cyclic groups that help in counting elements of certain orders?
3. How can we determine the order of elements in direct product groups?
4. Why do groups like $Z_{n}$ always have elements whose orders divide $n$?
5. How does the structure of $Z_{n}$ change when $n$ is prime compared to when it’s a composite number?

Tip:

In cyclic groups, the number of elements of a certain order can often be found using the divisors of the group's order and understanding which elements satisfy these divisors.