TITLE: Prove that $G$ must be cyclic
QUESTION [7 upvotes]: Let $G$ be a finite group, assume that for any pair of subgroup $H, K$ of $G$ either $H\subseteq K$ or $K\subseteq H$. Prove that $G$ must be cyclic. 


Suppose $a$ is maximum in $G$. Let $\langle a\rangle =H$ and $\langle b\rangle =K$. If $H\subseteq K$, then it is a contraction since $a$ is maximum. If $K\subseteq H$, then $b=a^n\in H$ where $n\in \mathbb{N}$, hence $H$ is cyclic which implies that $G$ is cyclic.

Can anyone check the proof valid or not? thanks

REPLY [5 votes]: Perhaps one rewrite the proof along the following lines.
Let $a$ be an element of $G$ of maximal order, and let $H$ be the subgroup of $G$ generated by $a$.  We show that $H$ is all of $G$.
Suppose to the contrary that there is a $b\in G$ such that $b$ is not in $H$, and let $K$ be the subgroup of $G$ generated by $b$. It is clear that $K$ is not a subset of $H$. So $H$ is a subset of $K$, necessarily proper. This contradicts the maximality of the order of $a$.