Question

Let \( (\mathbb{Z}_n, +) \) be the group of integers modulo \(n\). Let \( G = \mathbb{Z}_{30} \oplus \mathbb{Z}_{12} \) be the external direct product of \(\mathbb{Z}_{30}\) and \(\mathbb{Z}_{12}\). Then which of the following statements is/are TRUE?

• A. \(G\) is not a cyclic group.
• B. The order of the element \((14,7)\) in \(G\) is \(98\).
• C. The order of the element \((14,7)\) in \(G\) is \(60\).
• D. There is an element in \(G\) of order \(360\).

Hint:

Use the rule that Z_m (+) Z_n is cyclic only when gcd(m,n)=1, and that the order of (a,b) is the lcm of the individual orders of a and b.

Answer 1

The Correct Option is A, C

Step 1: Check whether G is cyclic.
A basic fact about direct products of cyclic groups says \(\mathbb{Z}_m \oplus \mathbb{Z}_n\) is cyclic if and only if \(\gcd(m,n) = 1\).
Here \(m = 30\) and \(n = 12\), and \(\gcd(30, 12) = 6 \neq 1\).
So \(G = \mathbb{Z}_{30} \oplus \mathbb{Z}_{12}\) is NOT cyclic. Option (A) is TRUE.

Step 2: Recall how to find the order of an element in a direct sum.
For \(G = \mathbb{Z}_m \oplus \mathbb{Z}_n\), the order of an element \((a,b)\) is the least common multiple of the order of \(a\) in \(\mathbb{Z}_m\) and the order of \(b\) in \(\mathbb{Z}_n\). The order of \(a\) in \(\mathbb{Z}_m\) is \(\dfrac{m}{\gcd(a,m)}\).

Step 3: Compute the order of the element (14, 7).
Order of \(14\) in \(\mathbb{Z}_{30}\): \(\gcd(14, 30) = 2\), so its order is \(\dfrac{30}{2} = 15\).
Order of \(7\) in \(\mathbb{Z}_{12}\): \(\gcd(7, 12) = 1\), so its order is \(\dfrac{12}{1} = 12\).
So the order of \((14,7)\) in \(G\) is \(\text{lcm}(15, 12)\). Since \(15 = 3 \times 5\) and \(12 = 2^2 \times 3\), \(\text{lcm}(15,12) = 2^2 \times 3 \times 5 = 60\).
So the order of \((14,7)\) is \(60\), not \(98\). Option (C) is TRUE and option (B) is FALSE.

Step 4: Check whether G has an element of order 360.
The order of every element of \(G = \mathbb{Z}_{30} \oplus \mathbb{Z}_{12}\) must divide \(\text{lcm}(30, 12) = 60\) (this number is called the exponent of the group), because the order of any \((a,b)\) is \(\text{lcm}(\text{ord}(a), \text{ord}(b))\), and \(\text{ord}(a) \mid 30\), \(\text{ord}(b) \mid 12\), so their lcm always divides \(\text{lcm}(30,12) = 60\).
Since \(360 > 60\), no element of \(G\) can have order \(360\). Option (D) is FALSE.

Final Answer:
The true statements are that \(G\) is not cyclic and that the order of \((14,7)\) is \(60\).
\[ \boxed{\text{(A) and (C) are TRUE}} \]