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
Answer: (B)
Explanation: We can use Lagrange’ theorem here, which states that “The order of every subgroup of G divides the order of G”.
1<|H|<6. O(H) may be 2,3,4,5.
2 divides 6, 3 divides 6. so H may be of order 2 or 3Â which is prime, so H is always cyclic.
The order of G is not prime, so it may not be cyclic.
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