Prove that $G$ is a solvable group if and only if when $G$ has a series of subgroups $G = P_n \supset P_{n-1} \supset ... \supset P_1 \supset P_0=\{e\}$ where $P_i \triangleleft P_{i+1}$ and $|P_{i+1}/P_i|$ is prime.
Here's what I'm thinking though I may be reaching:
If I can reason that $P_{i-1}/P_i$ is Abelian, since $G$ has a normal series, $P_n \triangleleft P_{n-1} \triangleleft ... \triangleleft P_1 \triangleleft P_0 = \{e\}$, then this is a simple problem and $G$ is solvable. I can't recall or find in the notes that $|P_{i+1}/P_i|$ being prime implies that it is Abelian though, so I ask.
Any group of prime order is cyclic. So $|P_{i-1}/P_{i}|$ prime implies that it is cyclic and hence abelian.