Question

Which of the following statements is false?
(I) \(\mathbb{Z}_5\times\mathbb{Z}_7\) is a cyclic group.
(II) The order of the element \((2,5)\in\mathbb{Z}_5\times\mathbb{Z}_7\) is 10.
(III) \(\text{Aut}(\mathbb{Z}_5\times\mathbb{Z}_7)\) is isomorphic to \(\mathbb{Z}_{35}\).

• A. Only (I)
• B. Only (II)
• C. Only (II) and (III)
• D. Only (I) and (III)

Hint:

Since \(\gcd(5,7)=1\), use the Chinese Remainder Theorem for (I), the lcm rule for element order in (II), and \(\text{Aut}(\mathbb{Z}_m\times\mathbb{Z}_n)\cong\text{Aut}(\mathbb{Z}_m)\times\text{Aut}(\mathbb{Z}_n)\) for coprime \(m,n\) in (III).

Answer 1

The Correct Option is C

Step 1: Reconfirm (I) using orders of elements instead of CRT.
A finite abelian group of order $n$ is cyclic iff it contains an element of order $n$. Take $(1,1) \in \mathbb{Z}_5\times\mathbb{Z}_7$. Its order is $\operatorname{lcm}(5,7) = 35$, which equals $|\mathbb{Z}_5\times\mathbb{Z}_7|$, so $(1,1)$ generates the whole group. (I) is true.

Step 2: Recompute the order of $(2,5)$ by direct repeated addition.
We need the smallest $k>0$ with $2k \equiv 0 \pmod 5$ and $5k \equiv 0 \pmod 7$. The first condition needs $k$ a multiple of $5/\gcd(2,5)=5$, the second needs $k$ a multiple of $7/\gcd(5,7)=7$. The smallest such $k$ is $\operatorname{lcm}(5,7)=35$. So the order of $(2,5)$ is $35$, not $10$; (II) is false.

Step 3: Recompute $\text{Aut}(\mathbb{Z}_5\times\mathbb{Z}_7)$ using the coprime-factor formula.
For coprime $m,n$, $\text{Aut}(\mathbb{Z}_m\times\mathbb{Z}_n)\cong\text{Aut}(\mathbb{Z}_m)\times\text{Aut}(\mathbb{Z}_n)$. Here $\text{Aut}(\mathbb{Z}_5)\cong\mathbb{Z}_4$ and $\text{Aut}(\mathbb{Z}_7)\cong\mathbb{Z}_6$, so\[\text{Aut}(\mathbb{Z}_5\times\mathbb{Z}_7) \cong \mathbb{Z}_4\times\mathbb{Z}_6\]order $24$, not $35$, and not even cyclic since $\gcd(4,6)=2\neq1$. Either way it is not $\mathbb{Z}_{35}$; (III) is false.

Step 4: Conclusion.
Only (I) is true; (II) and (III) are false.\[\boxed{\text{Only (II) and (III)}}\]