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?
Best Answer
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.