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.
Best Answer
It doesn't generalise -- the dihedral group with $8$ elements is an example, and a more general one is the set of matrices $$\begin{bmatrix} 1 & x & y \\ 0 & 1 & z \\ 0 & 0 & 1 \end{bmatrix}$$ with $x,y,z \in \mathbb{Z_p}$.
For a group of order $p^2$, the most common way to prove that it is abelian is to look at its center, $Z(G)$, the set of terms which commute with every other term. The center has to be nontrivial: if you consider the conjugacy classes of $G$ on itself, each must be of size $p^k$ for some non-negative $k$. But the conjugacy class of $e$ is trivial, and thus there exist at least $p-1$ other such cases.
Suppose the center has size $p$. Then $G/Z(G)$ must be a cyclic group on $p$ elements, and thus $G$ must be abelian. (for any group $H$, if $H/Z(H)$ is cyclic then $Z(H)=H$).