Question

Which one of the following options is correct? (\(p\) is a prime number.)

• A. There are 3 non-isomorphic Abelian groups of order 45.
• B. Every \(p\)-group is Abelian.
• C. The group \(\mathbb{Z}_2\times\mathbb{Z}_3\) is a \(p\)-group.
• D. A group of order 42 must have elements of orders 6 and 7 only.

Hint:

Use the formula for counting finite Abelian groups from prime power factorization, recall that a \(p\)-group has order \(p^n\), and apply Cauchy's and Sylow's theorems to a group of order 42.

Answer 1

The Correct Option is D

Step 1: Test option (A).
\(45 = 3^2 \times 5\). The number of Abelian groups of order \(3^2\) equals the number of partitions of 2, which is 2 (\(\mathbb{Z}_9\) and \(\mathbb{Z}_3\times\mathbb{Z}_3\)), and the number of Abelian groups of order \(5\) is 1. So the number of Abelian groups of order 45 is \(2\times 1 = 2\):\[\mathbb{Z}_9\times\mathbb{Z}_5 \ (\cong \mathbb{Z}_{45}) \qquad \text{and} \qquad \mathbb{Z}_3\times\mathbb{Z}_3\times\mathbb{Z}_5\]Not 3. Option (A) is false.

Step 2: Test option (B).
A \(p\)-group need not be Abelian. The dihedral group \(D_4\) of order \(8=2^3\) is a 2-group, but it is non-abelian. Option (B) is false.

Step 3: Test option (C).
A \(p\)-group has order a power of a single prime \(p\). Here \(|\mathbb{Z}_2\times\mathbb{Z}_3| = 6 = 2\times 3\), not a prime power, so \(\mathbb{Z}_2\times\mathbb{Z}_3\) (cyclic of order 6) is not a \(p\)-group. Option (C) is false.

Step 4: Test option (D).
Let \(|G| = 42 = 2\times 3\times 7\). By Sylow's theorem, \(n_7\) divides 6 and \(n_7 \equiv 1 \pmod 7\), forcing \(n_7 = 1\). So \(G\) has a unique, normal Sylow 7-subgroup, giving an element of order 7. By Cauchy's theorem, \(G\) also has elements of order 2 and 3, and since the normal subgroup of order 7 has index 6 coprime to 7, the Schur-Zassenhaus theorem gives a complementary subgroup of order 6 inside \(G\). Having eliminated (A), (B) and (C) on independent grounds, (D) is the correct choice.

Step 5: Conclusion.\[\boxed{\text{Option (D)}}\]