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. $G$ is always cyclic, but $H$ may not be cyclic.
• C. $G$ may not be cyclic, but $H$ is always cyclic.
• D. Both $G$ and $H$ may not be cyclic.

Answer 1

The correct option is C $G$ may not be cyclic, but $H$ is always cyclic.

$|G|=6$
$H$ is subgroup, so by Lagrange's theorem
$|H|=1,2,3$ or $6$ (Divisor's of 6)
Now it is given that $1<|H|<6$
or $|H|=2$ or $3$
Since $2$ and $3$ are both prime and since every group of prime order is cyclic, $H$ is surely cyclic. But order of $|G|=6$ which is not prime.
So $G$ may or may not be cyclic.
So $G$ may not be cyclic, but $H$ is always cyclic.
Option (c) is correct.