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.

Well, we could apply the fact that the center of such a group $G$ is nontrivial (proof). Since the center is nontrivial, it either has order $p$ or $p^m$ for $1&ltm\leq n$. In the former case, $G/Z(G)$ is of order $p^{n-1}$ so $Z(G/Z(G))$ is nontrivial, hence has order a power of $p$. Since $Z(G/Z(G))$ is abelian, it is easy to find such a subgroup (i.e. using the Fundamental Theorem of Finite Abelian Groups). The latter case is similarly easy, as $Z(G/Z(G))$ is an abelian group of $p^m$ for $m>1$.
