Let G be a group of order 6, and H be a subgroup of G such that 1 < |H| < 6.…
2021
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.
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Correct answer: B
Key insight: use Lagrange's theorem and the classification of groups of order 6.
Possible orders for a proper nontrivial subgroup H are divisors of 6 strictly between 1 and 6, so |H| = 2 or |H| = 3.
Any group of prime order is cyclic, so H is necessarily cyclic.
Groups of order 6 are either cyclic (isomorphic to C6) or non-abelian (isomorphic to S3). Hence G may or may not be cyclic.
Conclusion: The correct statement is "G may not be cyclic, but H is always cyclic."