TITLE: Inclusion of subgroups implies the group is cyclic
QUESTION [5 upvotes]: Let $G$ be a finite group such that for any two subgroups $H_{1}$ and $H_{2}$ of $G$ we have $H_{1} \subseteq H_{2}$ or $H_{2} \subseteq H_{1}$. Why this implies $G$ is a cyclic group?
Ah. I think this works: Suppose $g \in G$ then $G = \langle g \rangle$. Suppose otherwise, then we can find $z$ such that $z \not \in \langle g  \rangle$. Now let $H_{1}=\langle z \rangle$ and $H_{2}=\langle g \rangle$. By assumption we have $H_{1} \subseteq H_{2}$ or $H_{2} \subseteq H_{1}$. In both cases we get the contradiction $z \in \langle g \rangle$.

REPLY [2 votes]: Another (marginally different) proof: Consider a maximal cyclic subgroup $H$ of $G$ (one exists since $G$ is finite).  If $g \in G$ then the assumption implies either that $\langle g \rangle \subset H$ or else that $H \subset \langle g \rangle$.  In the latter case, we have equality, by maximality of $H$, and so in all cases $g \in H$.  Thus $H = G$,
and so $G$ is cyclic.