Let $G$ be a finite group such that for any two subgroups $H$ and $K$ either $H ⊆ K$ or $K ⊆ H$. Prove that $G$ is cyclic of prime power order. I do not understand how to approach this question. I cannot see why if any two subgroups of $G$ that are subsets of each other one way or another, it leads to $G$ being a cyclic group. Any help would be appreciated. 
 A: No need for anything as high powered as the Sylow theorems. First, we show $G$ is cyclic. Fix $x \in G$ with maximal order among elements of $G$. For any $y \in G$, we have $\langle y \rangle \subseteq \langle x \rangle$ or $\langle x \rangle \subseteq \langle y \rangle$. By hypothesis, $|y| \leqslant |x|$, so $|\langle y \rangle| \leqslant |\langle x \rangle|$. If $|\langle y \rangle| < |\langle x \rangle|$, then the only possibility is $\langle y \rangle \subseteq \langle x \rangle$. If $|\langle y \rangle| = |\langle x \rangle|$ and $\langle x \rangle \subseteq \langle y \rangle$, then in fact we must have $\langle y \rangle = \langle x \rangle$, whence $\langle y \rangle \subseteq \langle x \rangle$. Hence, every $y \in G$ belongs to $\langle x \rangle$, whence $G = \langle x \rangle$ and is therefore cyclic. 
Now, suppose $|G|$ has more than one prime divisor, say $p$ and $q$, seeking a contradiction. Then $|G|$ has an element $g$ of order $p$ and an element $h$ of order $q$, since $G$ is cyclic. Then $\langle g \rangle \cap \langle h \rangle$ has order $1$ by Lagrange's theorem, and hence cannot be all of $\langle h \rangle$ or $\langle g \rangle$, contradicting our hypothesis that either $\langle h \rangle \subseteq \langle g \rangle$ or $\langle g \rangle \subseteq \langle h \rangle$. Hence, $|G|$ must be a prime power. 
A: Another approach: First of all, the group is Abelian. If $x,y$ are two different elements, then either $x\in \langle y\rangle$ or $y\in\langle x \rangle$, and $x$ commutes with $y$.
By the fundamental theorem of finitely generated Abelian groups, $G$ is a direct sum of cyclic $p$-subgroups for some primes $p$. But if there is more than one group in the sum, then it is obvious that the subgroups are not linearly ordered. (Given $G=H\oplus K$, consider $1\oplus K$ and $H\oplus 1$).  Therefore there is only a cyclic factor of prime power order describing the entire group. 

As another alternative, consider that picking an element $x$ of maximal order forces $G\subseteq\langle x\rangle$. Otherwise $y\notin \langle x\rangle$ implies $\langle y\rangle$ properly contains $\langle x\rangle$ contradicting the maximality of the order of $x$. 
Now, if you are familiar with the fact that a cyclic group contains an element of each order dividing the group order, you can see that two different primes $p,q$ dividing the order would result in two subgroups that are impossible to linearly order. Therefore you are left with a cyclic $p$-group.
A: Suppose two distinct primes $p$ and $q$ divide the order of $G$. Then, by Cauchy's theorem, $G$ has subgroups of order $p$ and $q$, which are not contained in one another.
Therefore $|G|$ is divisible by at most one prime $p$.
Let $x_0\in G$; if $\langle x_0\rangle\ne G$, then there is $x_1\in G$, $x_1\notin\langle x_0\rangle$; by the assumption we must have $\langle x_0\rangle\subsetneq\langle x_1\rangle$.
Complete the argument using finiteness of $G$.
A: $G$ is cyclic: Because a finite group of order $n$ is cyclic iff it has at most one subgroup of order $d$ for all $d\mid n$. (ref)
$n$ is a prime power: $G$ has a unique maximal subgroup $M$ which has order $n/q$, where $q$ is a (the smallest) prime divisor of $n$. So $G\setminus M$ consists of generators. There are $\varphi(n)$ generators, so
$$\varphi(n)\geq n-\frac nq=n\cdot\frac{q-1}q.$$
But
$$\varphi(n)=n\cdot\prod_{p\mid n}\frac{p-1}p\leq n\cdot\frac{q-1}q,$$
so $n$ has only one prime divisor.
