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.

You can construct a nonabelian group of order $p^n$, for $p$ an odd prime, $n\gt2$,  by selecting a nontrivial homomorphism $\varphi:\Bbb Z_p\to\rm{Aut}(\Bbb Z_{p^{n-1}})\cong\Bbb Z_{p^{n-1}-p^{n-2}}$.  Let $G=\Bbb Z_{p^{n-1}}\rtimes_\varphi\Bbb Z_p$.
If, on the other hand, $p=2$, consider dihedral groups.
