Is a group of prime-power order always abelian? Let $G$ be a group of order $p^n$, with $p$ prime. By Sylow's first theorem, there exists at least one  subgroup of order $p^n$ (the number of subgroups of order $p^i$ is $1$ mod $p$ per $i$). The subgroups with order $p^n$ are all Sylow-$p$ groups.
Now, by Sylow's third theorem, because the group is of order $p^n$, the number $m_{p^{n}}$ of such subgroups must divide $\#G/p^n =1$, and only $1$ divides $1$, so there is only one subgroup of order $p^n$.
By Sylow's second theorem, all Sylow-$p$ groups are conjugated to each other by at least one element $g\in G$, so, for any $S$ and $S'$, we have $S=gS'g^{-1}$. In this case, there is only one Sylow-$p$ group, so it is conjugated to itself.
Of course, that one subgroup is the group itself. We now have $gG=Gg$ for some $g$ in $G$. Can we get to the entire group being abelian, from here?
I ask because my textbook on Abstract Algebra states that any group of order $p^2$ is abelian, and I'm curious whether it generalises.
Edit: As has been pointed out, everything I've proved above is quite trivial. Below it is discussed that the essential question is actually "How does one prove that groups of order $p^2$ are abelian using Sylow theory?", since my textbook explicitly mentions this property as an application of Sylow's theorems.
Edit 2: One of the authors has confirmed that they accidentally mixed some classic classification theorems into the list of applications of Sylow theory, and that this was one of them.

A group of order $p^n$ is always nilpotent. This is a natural generalisation of abelian. The examples of $Q_8$ and $D_4$ of order $8$ are nilpotent but non-abelian. The group of upper-unitriangular matrices over $\Bbb F_p$ is the Heisenberg group, which is $2$-step nilpotent, and also non-abelian.
Reference:
Prove that every finite p-group is nilpotent.
