Finite group with condition that $H\subseteq K $ or $K\subseteq H$ on its subgroups 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?
 A: 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.
A: Suppose $G$ is not cyclic. Take an element $x \in G$ of maximal order. Since $\langle x \rangle \neq G$ there is an $y \in G \setminus \langle x \rangle$. Now neither $\langle x \rangle \subseteq \langle y \rangle$ nor $\langle y \rangle \subseteq \langle x \rangle$ holds.
A: Hint: Choose an element $g\in G$ of largest order. Let $x$ be any element in $G$ . We find that $\langle g\rangle\leq \langle x\rangle$ or $\langle x\rangle \leq \langle g \rangle$. Since $g$ has the largest order we find that $\langle x \rangle \leq \langle g\rangle$. What does this imply ?
If $p,q$ are two primes that divide the order of $G$, then we know by Cauchy's theorem that there exists 2 elements $x,y$ of order $p,q$ respectively. Since $\langle x\rangle\leq\langle y\rangle$ or $\langle y\rangle\leq \langle x\rangle$. By Lagrange's theorem, we get $p|q$ or $q|p$. Thus, $p=q$ because $p,q$ are primes.
