G is a cyclic group of cardinality of a power of a prime number

Question

Let $G$ be a finite group for which for every subgroups $H,K$ of it we have $H\subseteq K$ or $K\subseteq H$. Prove that $G$ is a cyclic group and its cardinal is a power of a prime number.

Answer 1

There are two things we need to show about $G$, and they are more or less independent.

$|G|$ is a power of some prime.

Assume for contradiction that $|G| = pqm$ where $p \neq q$ are prime and $m\in \Bbb N$ (i.e. assume there is more than one prime factor present in $|G|$, and let $p$ and $q$ be two of them). In that case, by the Sylow theorems, there are subgroups $H, K\subseteq G$ with $|H| = p^i$ and $|K| = q^j$ for some $i, j \geq 1$. By Lagrange's theorem none of those is a subgroup of the other, which contradicts the assuptions on $G$.

$G$ is cyclic

Assume for contradiction that $G$ is not cyclic. That specifically means that there are $g_1, g_2 \in G$ such that there is no $g$ with $g_1, g_2 \in \langle g\rangle$. Specifically, this means that $g_1 \notin \langle g_2\rangle$ and $g_2 \notin \langle g_1\rangle$. In other words, if $H = \langle g_1\rangle$ and $K = \langle g_2\rangle$, then $g_1\notin K$ means $H\not\subseteq K$, and similarily $K \not \subseteq H$, which again contradicts the assumptions on $G$.

Answer 2

If $\lvert G\rvert$ has two prime divisors $p$ and $q$, there exist two elements with orders $p$ and $q$ by Cauchy's theorem, and none of them is contained in the other. Thus $\lvert G\rvert$ must be a prime power $p^n$.

Let $a\in G$ be an element with maximal order. For any other element $b\in G$, either $b\in\langle a\rangle$ or $a\in\langle b\rangle$ by hypothesis. The latter case implies $\langle a\rangle=\langle b\rangle$ by the maximality of the order of $a$. In both cases we can assert that $b\in\langle a\rangle$, thereby proving $a$ is a generator of $G$.