Prove every Sylow Subgroup of $G$ is Cyclic

Question

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 $|g|=|P|$.

Answer 1

Take $g\in P$ of maximal order, say $p^i$. If $i<k$ 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.

Answer 2

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, $H = <h>$ 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 $h$.

Next part is obvious from here to conclude G is cyclic & abelian.