5
$\begingroup$

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 $z \in \langle g \rangle$.

  • 0
    @Dan Petersen: I think you mean't we don't get this if $\langle g \rangle \subseteq \langle z \rangle$ right? how do we fix this?2011-03-11
  • 1
    Hint: Under the given conditions, the group must have exactly one maximal proper subgroup. Now consider an element not contained in this subgroup.2011-03-11

3 Answers 3