How can I see that $o(G) = p^n$ without using the fundamental theorem of finitely generated abelian groups?

Question

Let $G$ be a finite group such that for all pair of subgroups $H, K \leq G$ we have $H \subseteq K$ or $K \subseteq H$. I want to prove that there exist $p \in \mathbb{N}$ prime and $n \in \mathbb{N}$ such that $o(G) = p^n$.

I already proved that $G$ is cyclic and I know I could use the finitely generated abelian groups theorem, which would help me prove that $o(G) = p^n$, but isn't there another way? In any case, could you give me any ideas?

Answer 1

Write $G=\langle g \rangle$. Suppose that there exist distinct primes $p,q$ dividing $|G|$. Since $(p,q)=1$ it holds that $\alpha p+\beta q=1$ for some $\alpha,\beta\in\mathbb Z$. Now, $g=g^{\alpha p+\beta q}=(g^p)^\alpha(g^q)^\beta\in\langle g^p\rangle\langle g^q\rangle$ and hence $G=\langle g^p\rangle\langle g^q\rangle$. By hypothesis $\langle g^p\rangle\subseteq\langle g^q\rangle$ or $\langle g^q\rangle\subseteq\langle g^p\rangle$ and then $G=\langle g^q\rangle$ or $G=\langle g^p\rangle$. Comparing orders it follows that $|G|=\dfrac{|G|}{p}$ or $|G|=\dfrac{|G|}{q}$, which is a contradiction. Hence $|G|$ is the power of some prime.

Answer 2

One possibility is to use Cauchy's Theorem, which states that if $p$ divides the order of the group then there is a subgroup of order $p$. Now suppose that $\lvert G \rvert$ has two distinct prime divisors $p$ and $q$. Then there is at least one subgroup $H_p$ with $\lvert H_p \rvert = p$, and another subgroup $H_q$ with $\lvert H_q \rvert = q$. But clearly you cannot have $H_p$ as a subgroup of $H_q$ (or the opposite) by Lagrange's Theorem.

Thus you conclude that $\lvert G \rvert$ doesn't have two distinct prime divisors.

Answer 3

Since you already know that $G$ is cyclic, just notice that if the order $n$ of a cyclic group $G=\langle a\rangle$ has two distinct prime divisors $p,q$ then $G$ has two subgroups $H:=\langle a^{n/p}\rangle,K:=\langle a^{n/q}\rangle$ such that none of them is contained in the other one, because $|H|=p$ is not divisible by $|K|=q$, and conversely.

Answer 4

Since the family of all $G$ subgroups is finite and well ordered with respect to $\subseteq$ there is a maximum $\langle g_0\rangle=G$ among all cyclic $G$ subgroups and a maximum $\langle g_1\rangle$ among all primary $G$ subgroups. Noticeably, $G$ is cyclic. Thus $\langle g_0\rangle = \langle g_1\rangle$ since the primary $\langle g_0\rangle$ subgroups together generate $\langle g_0\rangle$ but are each contained in $\langle g_1\rangle$.