Question

Let G be a finite group. Then G is necessarily a cyclic group if the order of G is

• A. 4
• B. 7
• C. 6
• D. 10

Answer 1

The Correct Option is B

To determine when a finite group \( G \) is necessarily a cyclic group, we should consider some key properties of group theory.

A group \( G \) is called cyclic if there exists an element \( a \) in \( G \) such that every element of \( G \) can be expressed as \( a^n \) for some integer \( n \). Essentially, a cyclic group is generated by a single element.

A fundamental result in group theory states that if the order of a group (the number of elements in the group) is a prime number, then the group is cyclic. This is because, in such a group, every non-identity element must generate the entire group.

Let's examine each option:

• Option 1: Order = 4
  The number 4 is not a prime number (it is \( 2^2 \)). Groups of order 4, such as the Klein four-group, exist which are not cyclic since they are not generated by a single element. Hence, a group of order 4 is not necessarily cyclic.
• Option 2: Order = 7
  The number 7 is a prime number. By the aforementioned theory, any group with a prime order is cyclic. Therefore, any group of order 7 is necessarily cyclic.
• Option 3: Order = 6
  The number 6 is not a prime number (it is \( 2 \times 3 \)). Groups of order 6, such as the symmetric group \( S_3 \), can exist which are not cyclic. Thus, a group of order 6 is not necessarily cyclic.
• Option 4: Order = 10
  The number 10 is not a prime number (it is \( 2 \times 5 \)). Groups of order 10, for example, the dihedral group \( D_5 \), exist which are not cyclic. Therefore, a group of order 10 is not necessarily cyclic.

Based on the analysis, the correct answer is that a group \( G \) is necessarily cyclic if the order of \( G \) is \( 7 \), a prime number.