Groups such that inclusion on collection of all its subgroup is a total order Characterize the Groups with the following property: 
Suppose G is any group such that for any two subgroups, H and K either H $\subseteq$K or K $\subseteq$ H. Now what can we tell about cardinality, also about abelianess(Information about commutator subgroup) of the group?
 A: This is not a complete answer to the question, but I'll show that such groups are abelian and torsion (i.e., each element has finite order).
Abelianity:
Suppose $G$ is a group with this property, and we show that it is abelian. Take $a$ and $b$ be two elements of the group $G$. Let $H=\langle a \rangle$ and $K=\langle b\rangle$. Either $H$ contains $K$ or vice versa, meaning that $b$ is a power of $a$ or vice versa.
Suppose WLOG that $b$ is a power of $a$, writing $b=a^k$ gives:
$$
ab=aa^k=a^{k+1}=a^k a=ba
$$
Because $a$ and $b$ were arbitrary, we conclude that $G$ is abelian.
I now claim that each element of $G$ has finite order:
 Take any $a\in G$, and let $H=\langle a^2 \rangle$, $K=\langle a^3\rangle$. Either $H$ contains $K$ or vice versa, meaning that $a^2$ is a power of $a^3$ or vice versa. Thus we find some integer $k$ such that one of the following is true:
$$
a^{2k}=(a^2)^k=a^3 \;\;or\;\; a^{3k}=(a^3)^k=a^2
$$
thus $a^{2k-3}=id$ or $a^{3k-2}=id$ (where $id$ is the group identity), and both of the exponents are non-zero (as $k$ is an integer). Thus $a$ has finite order, where $a\in G$ was chosen arbitrarily.
A: Here is an extension of the answer by Miel Sharf, proving that such a group is either a finite cyclic group of prime power order or a Prüfer group.
First, we deal with the case where $G$ is finite. In this case, $G$ has a unique maximal subgroup, and hence is generated by any element not in this subgroup, so it is cyclic. It must also have prime power order, or it would have two non-trivial subgroups intersecting trivially (by picking them of distinct prime orders).
Next, we note that the property of subgroups being totally ordered is inherited by subgroups, so we conclude that there are no elements of infinite order, as that would give an infinite cyclic subgroup, and infinite cyclic groups do not satisfy the requirements. It also shows that all finite subgroups are cyclic of prime power order for some fixed prime, so in fact all elements have prime power order.
Finally, we assume that $G$ is infinite and wish to show that it is a Prüfer group, i.e. that for each $x\in G$ there are precisely $p$ solutions in $G$ to the equation $y^p = x$. To see this, we prove that this is in fact the case for any $x$ in a cyclic $p$-group which is not a generator (and then it follows for $G$ by noting that $\langle x\rangle$ sits inside a larger cyclic $p$-group in $G$).
So let $H$ be a cyclic $p$-group and $x\in H$ be an element which is not a generator. If $y$ is a generator for $H$ then $x = y^{pn}$ for some $n$, and we see that the elements $z$ with $z^p = x$ are precisely those of the form $y^{nk}$ where $k\equiv 1\pmod{|x|}$. That there are precisely $p$ solutions to this follows since $y^{n(1 + m|x|)} \neq y^{n(1+m'|x|)}$ when $1\leq m < m'\leq p-1$ while $y^{n(1+ p|x|)} = y^n$.
