Question

Let \( G \) be a group of order \(6\), and \( H \) be a subgroup of \( G \) such that \( 1 < |H| < 6 \). Which one of the following options is correct?

• A. Both \( G \) and \( H \) are always cyclic.
• B. \( G \) may not be cyclic, but \( H \) is always cyclic.
• C. \( G \) is always cyclic, but \( H \) may not be cyclic.
• D. Both \( G \) and \( H \) may not be cyclic.

Hint:

All groups of prime order are cyclic, regardless of the structure of the parent group.

Answer 1

The Correct Option is B

Step 1: Possible groups of order 6.
Up to isomorphism, there are two groups of order 6: the cyclic group \( \mathbb{Z}_6 \), which is cyclic, and the symmetric group \( S_3 \), which is non-cyclic. Hence, \( G \) need not be cyclic.

Step 2: Possible orders of subgroup \( H \).
By Lagrange's Theorem, the order of \( H \) must divide 6. Since \( 1 < |H| < 6 \), we have \( |H| = 2 \) or \( |H| = 3 \).

Step 3: Cyclicity of \( H \).
Any group of prime order is cyclic. Therefore, every subgroup of order 2 or 3 is cyclic.

Step 4: Conclusion.
Thus, \( G \) may not be cyclic, but \( H \) is always cyclic.