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Solvable groups

A chain of normal subgroups whose every quotient is abelian: which groups have one, why the property survives subgroups, quotients and extensions, and where it stops.

Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.

1. What you will learn

By the end of this lesson you will be able to define a solvable group, exhibit a chain with abelian factors for a small group, state the equivalent condition on composition factors, decide whether a named group is solvable, use the closure of solvability under subgroups, quotients and extensions, and prove that an extension of solvable groups is solvable.

2. Factors, and now a condition on them

The last two lessons built composition series and showed that the multiset of factors belongs to the group rather than to the chain. That is what makes this lesson's definition legitimate: solvable is a condition on the factors, and if different chains could give different factors the word would depend on which chain somebody happened to write down.

3. Solvable, abelian factor, extension, perfect

$G$ is solvable when it has a chain $1 = G_0 \trianglelefteq \cdots \trianglelefteq G_n = G$ with every factor $G_{i+1}/G_i$ abelian. $G$ is an extension of $Q$ by $N$ when $N \trianglelefteq G$ and $G/N \cong Q$. A group equal to its own commutator subgroup is perfect, and a perfect non-trivial group is never solvable.

4. Every factor abelian, not just one

Definition. $G$ is solvable when there is a chain

$$1 = G_0 \trianglelefteq G_1 \trianglelefteq \cdots \trianglelefteq G_n = G$$

in which every factor $G_{i+1}/G_i$ is abelian.

An equivalent test for finite groups. Refine such a chain to a composition series. The factors are then simple and abelian, and an abelian simple group is cyclic of prime order. So a finite group is solvable exactly when every composition factor has prime order. Jordan-Hölder makes this independent of the chain, which is what makes the definition sound.

Which groups are solvable.

• Every abelian group, trivially.
• Every $p$-group: the centre is non-trivial, so quotient and induct.
• Every group of order $pq$, $p^{2}q$, or more generally $p^{a}q^{b}$ — that last is Burnside's theorem, proved with character theory.
• Every group of order less than $60$.
• $S_3$ and $S_4$, which is why cubics and quartics have radical formulas.

Which are not. $A_n$ and $S_n$ for $n \ge 5$. $A_5$ is simple and not abelian, so it is its own only composition factor and no refinement exists. Every non-abelian simple group is an obstruction in the same way.

Closure properties. Solvability passes to subgroups (intersect the chain), to quotients (push the chain forward), and to extensions (splice a chain of $N$ below the pull-back of a chain of $G/N$). The last is the useful one: it means a group can be proved solvable layer by layer.

The name. It comes from equations, and it will not be justified until the last unit. Briefly: a radical formula for a polynomial corresponds to a tower of field extensions each obtained by adjoining a root, and each such step contributes an abelian layer to a group. So a formula exists exactly when the group has a chain with abelian factors — which is this condition, arrived at from a completely different direction.

5. Where solvability stops, and why sixty

Every group of order less than $60$ is solvable, and this is not a coincidence — $60$ is the smallest order of a non-abelian simple group, and $A_5$ realises it.

The proof that smaller orders are safe is the Sylow counting of the previous lessons, applied order by order. For most orders a Sylow count is forced to be $1$; for the rest, element counting or the action on cosets finishes the job. Every such argument ends with a proper normal subgroup, and induction on order does the rest — because both the subgroup and the quotient are smaller and therefore solvable, and an extension of solvable by solvable is solvable.

The general results worth knowing.

• Burnside's $p^{a}q^{b}$ theorem. A group whose order has at most two distinct prime factors is solvable. Only three primes are needed for a counterexample, and $60 = 2^{2} \cdot 3 \cdot 5$ has exactly three.
• The Feit-Thompson theorem. Every group of odd order is solvable. This is a much deeper result — the original paper runs to 255 pages — and it says every non-abelian finite simple group has even order.
• Solvable equals derived series terminates, which is the next lesson's reformulation and the one that is actually computed with.

A warning about the word. Solvable and simple are nearly opposite conditions for non-abelian groups: a solvable group has plenty of normal subgroups to climb through, a simple group has none. The one overlap is a cyclic group of prime order, which is both, and that is exactly the case where the two words agree.

6. Where solvability is misjudged

Reading solvable as abelian. $S_4$ is solvable and thoroughly non-commutative. The condition is on the factors, not on the group.

Accepting one abelian factor. $S_5$ has the chain $1 \lhd A_5 \lhd S_5$ with an abelian top factor of order $2$, and is not solvable. Every factor has to be abelian.

Requiring the terms to be normal in $G$. As with any series, normality in the next term up is all that is asked.

Thinking a group with a normal subgroup is solvable. Having somewhere to climb is necessary, not sufficient.

Expecting simple and solvable to be exclusive. A cyclic group of prime order is both. For non-abelian groups they are exclusive, and that is the content.

Assuming solvable means easy. Solvable groups can be enormous and complicated; what is guaranteed is a layered structure, not simplicity of any other kind.

7. Every group of order $p^{3}$ is solvable

1. The centre is non-trivial by the class equation, so there is a normal subgroup $Z$ with $|Z| \in \{p, p^{2}, p^{3}\}$.

   A step is always available.

2. $Z$ is abelian, and $G/Z$ has order $p^{2}$ or less, so it is abelian too by the earlier result.

   Both ends abelian.

3. So $1 \lhd Z \lhd G$ has abelian factors and $G$ is solvable — and the same induction handles $p^{n}$ for every $n$.

   Solvable, in two steps.

8. $S_5$ is not solvable

1. Its composition series is $1 \lhd A_5 \lhd S_5$, with factors $A_5$ and a group of order $2$.

   Two factors.

2. $A_5$ is simple, so no refinement can break that factor up, and it is not abelian.

   One bad factor.

3. By Jordan-Hölder no other chain can do better, so $S_5$ is not solvable — and neither is any group containing $A_5$ as a composition factor.

   The verdict is chain-independent.

9. Your turn: is every group of order $100$ solvable?

1. $100 = 2^{2} \cdot 5^{2}$, which has only two distinct prime factors.

   Factor the order.

2. Burnside's theorem says every group of order $p^{a}q^{b}$ is solvable.

   A general theorem applies.

3. Your turn: work this step out. Its working is at the end of the packet.

   So yes, without looking at any particular group of order $100$. A direct argument also works: $n_5$ divides $4$ and is $1$ modulo $5$, so it is $1$, giving a normal Sylow $5$-subgroup of order $25$ — which is abelian, with an abelian quotient of order $4$.

10. Guided practice

The chain $1 \lhd V_4 \lhd A_4 \lhd S_4$ shows that $S_4$ is solvable. The orders of the terms are given. Fill in the order of each factor, and the product of them all.

11. Guided practice

Is this solvable: every group of order $p^{k}$ for a prime $p$?

• Solvable
• Not solvable

12. Practice

Select every statement that is true of solvable groups.

This task has no paper form; do it on a device.

13. Practice

Match each construction to what it does to solvability.

14. Practice

Put in order the steps of deciding whether a finite group is solvable.

Number the steps in order (write the number in the box):

• Find a proper normal subgroup whose quotient is abelian.
• Check that the quotient really is abelian.
• Repeat inside that normal subgroup, which is smaller.
• Stop at the trivial group, and declare the group solvable.

15. Somewhere new

Build the proof that if $N$ is normal in $G$ and both $N$ and $G/N$ are solvable, then $G$ is solvable.

This task has no paper form; do it on a device.

16. Lesson test

Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.

17. Test question

The chain $1 \lhd V_4 \lhd A_4 \lhd S_4$ shows that $S_4$ is solvable. The orders of the terms are given. Fill in the order of each factor, and the product of them all.

18. What you can do now

You can decide whether a finite group is solvable and exhibit the chain when it is. Say in your own words why one abelian factor is not enough, and name the smallest group that is not solvable. Next: the derived series, which finds the chain for you.

Working for the steps left to you

9. Your turn: is every group of order $100$ solvable?, step 3