Question

Let \( G \) be a group of order 39 such that it has exactly one subgroup of order 3 and exactly one subgroup of order 13. Then, which one of the following statements is TRUE?

• A. \( G \) is necessarily cyclic
• B. \( G \) is abelian but need not be cyclic
• C. \( G \) need not be abelian
• D. \( G \) has 13 elements of order 13

Hint:

A group of order \( pq \), where \( p \) and \( q \) are coprime, is always cyclic.

Answer 1

The Correct Option is A

Let \( G \) be a group of order 39. By Lagrange's Theorem, the orders of the elements of \( G \) must divide 39, so the possible orders of elements are 1, 3, 13, or 39. The subgroup of order 3 is unique, and the subgroup of order 13 is also unique. Since the orders of these subgroups are coprime (3 and 13 are relatively prime), by the Chinese Remainder Theorem, the group \( G \) is isomorphic to the direct product of these two subgroups, which are cyclic. Therefore, \( G \) must be cyclic, as the direct product of two cyclic groups of coprime orders is cyclic. Thus, the correct answer is that \( G \) is necessarily cyclic.