Solvable Groups

Derived series, abelian-quotient chains, and which groups are solvable.

The idea

Abelian groups are the ones we understand completely. The next best thing to being abelian is being assembled out of abelian pieces, and solvability is what makes that precise.

The tool is the commutator subgroup $G' = [G, G]$, generated by all the elements $xyx^{-1}y^{-1}$. Each of those records the failure of one particular pair to commute, so $G'$ is trivial exactly when $G$ is abelian. More than that, $G'$ is the smallest normal subgroup with abelian quotient: $G/N$ is abelian if and only if $N$ contains $G'$. Passing to $G/G'$ is thus the cheapest way to force commutativity — it discards the non-commutativity and nothing else.

So do it again. Form $G'' = (G')'$, then $(G'')'$, and keep going. This derived series has two possible fates: it reaches $1$ after finitely many steps, in which case the successive quotients are abelian and $G$ stands displayed as a stack of abelian layers, or it jams at some nontrivial subgroup and stays there forever. The first case is what solvable means, and the equivalent formulation is a chain $1 = G_{0} \trianglelefteq G_{1} \trianglelefteq \cdots \trianglelefteq G_{n} = G$ with every quotient $G_{i+1}/G_{i}$ abelian.

Jamming is not exotic. A non-abelian group with no proper nontrivial normal subgroup has nowhere to descend to, so it defeats the process on the very first step: $G' = G$, and the series never moves. The name comes from polynomials: a polynomial turns out to be solvable by radicals exactly when the group attached to it is solvable in this sense.

Ways to work on it

• Walkthrough. The abelian-quotient chain, the derived series, and why A_5 fails.
• Practice. Decide whether a named group is solvable.
• Hardest. Compute the derived series of S_4 and read off solvability.

Not sure where to start? Take the ten-question placement test.