# 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?

• What are your thoughts? What have you tried? – Tobias Kildetoft Mar 11 '13 at 17:14
• What about an infinite group? – mrs Mar 11 '13 at 18:17
• This is a really nice problem! thanks for posting – user58512 Mar 11 '13 at 18:24
• @BabakS. The Prüfer $p$-group has these properties, but is not cyclic. – user1729 May 6 '13 at 13:35

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.

• Pretty nice hint, though not the complete answer. – DonAntonio Mar 11 '13 at 19:08
• I second @DonAntonio, the full answer is $G$ is cyclic of prime-power order, as shown in the other answers. – Andreas Caranti Mar 12 '13 at 16:24

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.

• This imply that $G$ is cyclic.Is this also imply that he order $G$ is prime? – rese Mar 11 '13 at 17:33

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.

• $Z(G)$ cyclic does not imply that $G$ is abelian (any nonabelian group of order $8$ is a counterexample), I'm not sure what you mean here. However, $G/Z(G)$ cyclic does imply $G$ abelian. – Mikko Korhonen Mar 11 '13 at 18:44
• @m.k.: Of course, thank you, I edited my answer. – Seirios Mar 11 '13 at 21:09
• Ok, it makes sense now. It's a nice alternative proof. – Mikko Korhonen Mar 12 '13 at 6:49