TITLE: Finite group with condition that $H\subseteq K $ or $K\subseteq H$ on its subgroups
QUESTION [8 upvotes]: Let $G$ be a finite group such that for every two subgroups $H$ and $K$ we have $H\subseteq K $ or $K\subseteq H$.
Is $G$ is necessarily cyclic?
Is the order of $G$ is power of prime number?

REPLY [6 votes]: If there exist two different primes $p,q$ dividing $|G|$, then neither $\langle x \rangle \subset \langle y \rangle$ nor $\langle y \rangle \subset \langle x \rangle$ if $\text{ord}(x)=p$ and $\text{ord}(y)=q$. Therefore, $G$ is a $p$-group. 
In particular, $Z(G) \neq \{1\}$, so by induction, $G/Z(G)$ is cyclic. Therefore $G$ is an abelian $p$-group, so $G$ is a direct product of cyclic $p$-groups; if there are more than one such cyclic group, then you can find easily two subgroups contradicting the property.