If $G$ is a group whose order is $p^n$($p$ is prime), then $G$ is solvable.
How am I going to show this? Any help is appreciated. Thank you.
Try by induction on the power of $p$. If $n=1$, $G$ is solvable by definition as a cyclic group of prime order.
Suppose that statement is true for all $k\leq n-1$. Suppose $|G|=p^n$. By the class equation, the center $Z(G)$ is nontrivial. So $Z(G)$ is normal in $G$ and abelian, hence solvable.
So either $G/Z(G)$ is a $p$-group of smaller order, or it is trivial.