The question is :

Suppose $G$ is a $p$ group ($|G|=p^n$)and $A\leq B$ or $B\leq A$ for any $A,B\leq G$ then, Prove that $G$ is cyclic.

Suppose $A,B$ are two subgroups of $G$ of same order (say $p^m$ for some $m<n$)

As we have $A\leq B$ or $B\leq A$, in any case we can not have $A$ properly contained in $B$ (or the other way). Thus, we should have $A=B$.

I mean no two subgroups having same order can be different.

Thus, I see any subgroup is normal (As $|gAg^{-1}|=|A|$ for any $g\in G$).

Any subgroup is normal... This does not even say that $G$ is abelian :( :(..

Highly disappointed with this... But, as i think this is the only "considerable conclusion" i have concluded from given data I think i will include in the question..

I would be thankful if someone can suggest me an approach/ give me a hint to approach the problem.... 

Thank You 

EDIT : Just after this Question it is sited that : 

> 
> "This Statement is also true without the assumption that $G$ is a p-group"
> 
> 
> 

As now, I have understood the proof for p-groups, I would like to work on the general statement.

I would be thankful if some one can give hint for general statement "preferably an extension of this proof"

Thank you :)