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?
Show 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.
The order of the element \((14,7)\) in \(G\) is \(98\).
The order of the element \((14,7)\) in \(G\) is \(60\).
There is an element in \(G\) of order \(360\).
Show Solution
The Correct Option isA, C
Solution and Explanation
Step 1: Break $G$ into a product of prime power cyclic groups.
$\mathbb{Z}_{30} \cong \mathbb{Z}_2 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_5$ (since $30 = 2 \times 3 \times 5$, pairwise coprime), and $\mathbb{Z}_{12} \cong \mathbb{Z}_4 \oplus \mathbb{Z}_3$ (since $12 = 4 \times 3$).
So $G \cong \mathbb{Z}_2 \oplus \mathbb{Z}_3 \oplus \mathbb{Z}_5 \oplus \mathbb{Z}_4 \oplus \mathbb{Z}_3$.
Step 2: Use this decomposition to test cyclicity.
$G$ is cyclic exactly when it decomposes into cyclic groups of pairwise coprime order with no prime repeated. Here $\mathbb{Z}_3$ appears twice (once from the $30$ part, once from the $12$ part), so the prime $3$ shows up twice, and a genuinely cyclic group cannot be built this way.
This repeated factor means $G$ is not cyclic, matching statement (A) being TRUE.
Step 3: Find the order of $(14,7)$ using the division rule directly.
In $\mathbb{Z}_{30}$, the order of an element $a$ is $30$ divided by $\gcd(a,30)$. With $a=14$: $\gcd(14,30)=2$, so the order is $30/2 = 15$.
In $\mathbb{Z}_{12}$, the order of $b=7$ is $12/\gcd(7,12) = 12/1 = 12$, since $7$ and $12$ share no common factor.
The order of the pair $(14,7)$ in the direct sum is the smallest number that is a multiple of both individual orders, that is $\text{lcm}(15,12)$. Writing $15=3\cdot5$ and $12=2^2\cdot3$, the lcm takes the highest power of each prime: $2^2 \cdot 3 \cdot 5 = 60$.
So the order of $(14,7)$ is $60$, confirming (C) is TRUE and (B) (claiming 98) is FALSE.
Step 4: Bound the largest possible order in $G$ to test option (D).
Since every element of $\mathbb{Z}_{30}$ has order dividing $30$ and every element of $\mathbb{Z}_{12}$ has order dividing $12$, any element $(a,b)$ of $G$ has order $\text{lcm}(\text{ord}(a),\text{ord}(b))$ which must divide $\text{lcm}(30,12) = 60$.
So no element of $G$ can have order bigger than $60$, ruling out an element of order $360$. Option (D) is FALSE.
Final Answer:
$$\boxed{\text{Only (A) } G \text{ is not cyclic, and (C) the order of } (14,7) \text{ is } 60 \text{, are TRUE}}$$
