Let G be a group of order 6, and H be a subgroup of G such that 1

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.

Accepted Answer

B. G may not be cyclic, but H is always cyclic. A group of order 6 can be either Z_6 (cyclic) or S_3 (non-cyclic, the symmetric group on 3 elements). So G may not be cyclic. However, by Lagrange's theorem |H| must divide 6 and satisfy 1 < |H| < 6, leaving only |H| = 2 or |H| = 3. Both Z_2 and Z_3 are cyclic (every group of prime order is cyclic), so H is always cyclic regardless of what G is.

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.