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?

According to Lagrange's Theorem, the order of any subgroup $H$ of a finite group $G$ must divide the order of $G$.

Given $|G| = 6$ and $1 < |H| < 6$, the possible values for $|H|$ are the proper divisors of $6$, which are $2$ and $3$.

• Every group of prime order is cyclic. Since both $2$ and $3$ are prime numbers, any subgroup $H$ of order $2$ or $3$ must be cyclic. Thus, $H$ is always cyclic.

• A group of order $6$ can be isomorphic to either the cyclic group $\mathbb{Z}_6$ (which is cyclic) or the symmetric group $S_3$ (which is non-abelian and therefore not cyclic). Thus, $G$ may not be cyclic.

Therefore, $G$ may not be cyclic, but $H$ is always cyclic, which corresponds to option (B).