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.
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.
The key theorem to remember is that if $H\unlhd G$ and $H$ is solvable and $G/H$ is solvable, then $G$ is also solvable. If $|G/Z(G)|< p^n$, then by induction $G/Z(G)$ is solvable, so $G$ is solvable. Otherwise you just have $G=Z(G)$.