# Prove every Sylow Subgroup of $G$ is Cyclic

I'm working on the following problem. Let $$G$$ be a finite group such that for each $$n \mid |G|$$, we have that $$G$$ contains at most $$1$$ subgroup of order $$n$$. Prove that every Sylow $$p$$ subgroup of $$G$$ is cyclic. Then prove that $$G$$ is abelian and cyclic itself.

So I can see immediately that for each prime $$p \mid |G|$$, $$G$$ has a unique (hence normal) Sylow subgroup $$P \lhd G$$. It suffices to find a single $$g \in G$$ with $$|g|=|P|$$.

The first Sylow theorem along with the uniqueness hypothesis here also says that $$P$$ contains $$p$$ subgroups of each order $$1,p,p^2,\dots,p^{k-1}$$ where $$|P|=p^k$$. Hence $$P$$ contains the unique subgroups of $$G$$ of these orders. However, I don't see how to prove there exists $$g \in G$$ with $$|g|=|P|$$.

Take $$g\in P$$ of maximal order, say $$p^i$$. If $$i then there is some element $$h\in P\setminus\langle g\rangle$$. It has order $$p^j$$, and by maximality of $$i$$ we must have $$j\leq i$$. Then $$\langle h\rangle$$ is a subgroup of $$G$$ of order $$p^j$$. However, $$\langle g\rangle$$ also contains a subgroup of order $$p^j$$, which contradicts the uniqueness of a subgroup of such order.
Since, $$G$$ has unique subgroup of each order then every sylow-$$P$$ subgroup of G is unique. Let, $$P$$ is not cyclic. Let, $$h$$ be the element of maximal order in $$P$$. Consider, $$H = $$ since, G is non cyclic then $$•(h) <|P|$$.hence, $$|H|<|P|$$ Take, $$S= P\setminus H$$ , since S is non empty, S contains an element $$q$$ of order $$p^t$$ Since, $$h$$ is element of maximal order & if $$•(h) = p^m$$ then m≥t. Being, Cyclic $$H$$ contains an element of order $$p^t$$.
Hence, We found two subgroups of order $$p^t$$ one is contained in H & one is outside of $$H$$.
It gives a contradiction to given condition. Hence, $$P\setminus H$$ is empty =>$$P$$ is cyclic & generated by $$h$$.