Problem of finding subgroup without Sylow's Thm. Let $G$ is a group with order $p^n$  where $p$ is prime and $n \geq 3$.
By Sylow's Thm, we know that $G$ has a subgroup with order $p^2$.
But, I wonder to proof without Sylow's Thm.

Let's try a direct approach assuming the weaker condition $G$ finite with $p^2\mid |G|\quad$ (i.e., $G$ not necessarily a $p$-group):
By Cauchy, there exists an element $x \in G$ of order $p$.
Case 1: $p$ divides the index of the normalizer $\mathop{N}_G(\langle x \rangle)$ in $G$
$\langle x \rangle$ acts by right multiplication on the set of cosets $\mathop{N}_G(x)\backslash G = \{\mathop{N}_G(\langle x \rangle)g : g\in G\}\quad$. All orbits of $\langle x \rangle$ have order $1$ or $p$, implying that the number of fixed points of $\langle x \rangle$ equals the index of $\mathop{N}_G(\langle x \rangle)$ in $G$ modulo $p$, which is $0$ modulo $p$ in this case. As $\mathop{N}_G(\langle x \rangle)$ is a fixed point of $\langle x \rangle$, there has to be another fixed point, say $\mathop{N}_G(\langle x \rangle)g = \mathop{N}_G(\langle x \rangle)gx\quad$, which is equivalent to $gxg^{-1} = x^{g^{-1}} \in \mathop{N}_G(\langle x \rangle)\quad$.
As $g\not\in\mathop{N}_G(\langle x \rangle)\quad$, $x^{g^{-1}}$ is an element of order $p$ not contained in $\langle x \rangle$ that normalizes $\langle x \rangle$. Hence $\langle x, x^{g^{-1}} \rangle$ is a subgroup of order $p^2$ of $G$.
Case 2: The index of the normalizer $\mathop{N}_G(\langle x \rangle)$ in $G$ is coprime to $p$
Then the quotient group $\mathop{N}_G(\langle x \rangle)/\langle x \rangle\quad$ has order divisible by $p$. By Cauchy, it contains a subgroup of order $p$, whose preimage has order $p^2$.
