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

Answer 1

The Correct Option is A

To determine whether the group \( G \) of order 39 is cyclic, we can analyze the properties and the structure of the group based on the given information. 

The order of the group \( G \) is 39. The order of a group is the total number of elements in the group. The prime factorization of 39 is:

\(39 = 3 \cdot 13\)

By the Sylow theorems, the number of subgroups of a given prime order \( p \) in a group is congruent to 1 modulo \( p \) and also divides the order of the group. Let's apply this to both prime factors:

1. For the prime \( p = 3 \):
   • The number of subgroups of order 3 must divide 39 and must be congruent to 1 modulo 3. The possible divisors of 39 are 1, 3, 13, and 39. Out of these, only 1 satisfies both conditions (i.e., \( 1 \equiv 1 \mod 3 \)). Therefore, there is exactly one subgroup of order 3.
2. For the prime \( p = 13 \):
   • Similarly, the number of subgroups of order 13 must divide 39 and must be congruent to 1 modulo 13. The divisors of 39 are the same, and only 1 satisfies \( 1 \equiv 1 \mod 13 \). Thus, there is exactly one subgroup of order 13.

Having exactly one subgroup of each prime order suggests that these subgroups are normal in \( G \). In general, if a group has exactly one Sylow \( p \)-subgroup for each prime \( p \) dividing the group's order, then these subgroups must be normal.

Consequently, \( G \) is the internal direct product of these normal subgroups. Since both subgroups are cyclic of prime order, they are generated by their respective orders' elements. Thus, \( G \) is generated by these elements:

• A cyclic group of order 3 generated by an element \( a \).
• A cyclic group of order 13 generated by an element \( b \).

The group \( G \) can then be expressed as the external direct product of cyclic groups of orders 3 and 13. This product group is cyclic if and only if the orders, 3 and 13, are coprime. Since they are coprime, \( G \) is cyclic:

\(G \cong \mathbb{Z}_{39} \cong \mathbb{Z}_{3} \times \mathbb{Z}_{13}\)

Therefore, the correct answer is: G is necessarily cyclic.

With this, we can confidently conclude that 'G is necessarily cyclic' is the true statement.