If $A\leq B$ or $B\leq A$ for any $A,B\leq G$ then $G$ is a cyclic group 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 :)
 A: Note that $G$ is necessarily a $p$-group since if $p,q\mid |G|$ where $p,q$ were distinct primes, then $\langle x\rangle\leqslant\langle y\rangle$ or $
\langle y\rangle\leqslant\langle x\rangle$ would be impossible for $|x|=p$ and $|y|=q$. But, then note that since $G$ is a $p$-group it has a subgroup of every order dividing $|G|$. But, from the assumption, it can have at most one subgroup of every order dividing $|G|$. Thus, $G$ has precisely one subgroup of every order dividing $|G|$, and so it must be cyclic. So, the $G$ (finite) that satisfy your condition are cyclic $p$-groups.

Theorem: If $G$ is a finite group of order $n$ such that for every $d\mid n$, $G$ has at most one subgroup of order $d$, then $G$ is cyclic.

Proof: Partition $G$ into classes $C_d$ according to the order $d$ of its elements. Evidently then, if $G$ has $s_d$ number of cyclic subgroups of order $d$, then $\#(C_d)=\varphi(d)s_d$. But, note that 
$$|G|=\sum_{d\mid n}\#(C_d)=\sum_{d\mid n}s_d\varphi(d)$$
Since 
$$\sum_{d\mid n}\varphi(d)=n$$
it follows from the fact that $s_d\leqslant 1$, that $s_d=1$ for all $d$. In particular, $s_n=1$, so there exists a cyclic subgroup of $G$ of order $n$. So, $G$ is cyclic.
EDIT: Prompted by Mikko's comment below, I think it's nice that with a little oomph, we can actually classify all groups, finite or infinite, that satisfy the property in the title. 
Namely, let $G$ be any group that satisfies the titular property. Note that $G$ is necessarily abelian. Indeed, if $g,h\in G$ then either $\langle g\rangle\leqslant\langle h\rangle$ or $\langle h\rangle\leqslant \langle g\rangle$ and so evidently $g$ and $h$ commute. If $G$ is an f.g. group satisfying the titular property, it must evidently be torsion. Indeed, since $G$ is abelian it suffices to show that it has rank $0$. But, since the titular property is obvious hereditary, and $\mathbb{Z}$ doesn't satisfy the titular property, it follows that $\text{rank}(G)=0$. Thus, $G=\mathbb{Z}/p^n\mathbb{Z}$ for some $n$ and some prime $p$.
Now, let $G$ be any group satisfying the titular property. The above now shows that $G$ is abelian and every finitely generated subgroup of $G$ is $\mathbb{Z}/p^n\mathbb{Z}$. Recall that $G=\varinjlim F$, where $F$ varies over the diagram of finitely generated subgroups. Now, the above shows that each finitely generated subgroup is of the form $\mathbb{Z}/p^n\mathbb{Z}$ and that they are all contained in one another. By the infinitude of $G$, we may actually see that $G$ contains $\mathbb{Z}/p^n\mathbb{Z}$ for all $n$, and so the diagram of finitely generated subgroups is 
$$\mathbb{Z}/p\mathbb{Z}\to\mathbb{Z}/p^2\mathbb{Z}\to\cdots$$
thus, 
$$G=\varinjlim F=\varinjlim \mathbb{Z}/p^n\mathbb{Z}\cong \mathbb{Z}(p^\infty)$$
(where $\mathbb{Z}(p^\infty)=\varinjlim\mathbb{Z}/p^n\mathbb{Z}$ is the Prufer $p$-group). Thus, the only groups $G$ which satisfy the titular properties are:


*

*The groups $\mathbb{Z}/p^n\mathbb{Z}$ for $p$ a prime.

*The Prufer $p$-group $\mathbb{Z}(p^\infty)$ for $p$ a prime.

A: I am not sure I understand your very first line. (Addendum: if you mean that in a cyclic group any two subgroups are comparable, this is only true in cyclic $p$-groups. Check the cyclic group of order $6$ to see why your argument fails.)
As to the rest, if $G$ has only one maximal subgroup $M$, then clearly any $x \in G \setminus M$ generates $G$, as $\langle x \rangle \not\le M$.
Now if $A$ and $B$ are two maximal subgroups of $G$, one can be contained in the other only if they are equal.
A: The beginning of your question is false, I think. Consider $\def\Z{\mathbb Z}\Z_6$ and $A=\{0,2,4\} \leq \Z_6$ and $B=\{0,3\}\leq \Z_6$ - none of $A,B$ is a subgroup of the other.
