Question

The number of distinct subgroup of Z₉₉₉ is_____.

Answer 1

Correct Answer: 8

To determine the number of distinct subgroups of \( \mathbb{Z}_{999} \), we first consider its structure. The group \( \mathbb{Z}_{999} \) is cyclic because it is an additive group of integers modulo 999. A cyclic group of order \( n \) has subgroups for every divisor of \( n \).

First, factorize 999: \( 999 = 3^3 \times 37 \).

We find the divisors of 999 by considering all combinations of powers of its prime factors:

• \(1 = 3^0 \times 37^0\)
• \(3 = 3^1 \times 37^0\)
• \(9 = 3^2 \times 37^0\)
• \(27 = 3^3 \times 37^0\)
• \(37 = 3^0 \times 37^1\)
• \(111 = 3^1 \times 37^1\)
• \(333 = 3^2 \times 37^1\)
• \(999 = 3^3 \times 37^1\)

Thus, the divisors of 999 are: \( \{1, 3, 9, 27, 37, 111, 333, 999\} \).

The number of distinct subgroups of a cyclic group is equal to the number of its divisors. Thus, there are 8 divisors, which correspond to 8 distinct subgroups.

Finally, confirm the solution falls within the given range: The expected range is 8 to 8, and the computed number of distinct subgroups is indeed 8.

Conclusion: The number of distinct subgroups of \( \mathbb{Z}_{999} \) is 8.