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The subgroup generated by the commutators, why it is the smallest with abelian quotient, and the canonical chain it builds — which reaches the trivial group exactly when the group is solvable.
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.
By the end of this lesson you will be able to compute the derived subgroup of a small group, recognise it as the kernel of the largest abelian quotient, build a derived series and read its length, say why a group is solvable exactly when its derived series reaches the trivial group, identify a perfect group, and prove the minimality of the derived subgroup.
The last lesson defined solvability as the existence of a chain with abelian factors, and left the finding of one to ingenuity. That is unsatisfactory in two ways: a failed search proves nothing, and there is no canonical answer to point at. The derived series fixes both — it is built with no choices, and if it fails no chain can succeed.
The commutator of $a$ and $b$ is $[a, b] = aba^{-1}b^{-1}$, trivial exactly when they commute. The derived subgroup $G' = [G, G]$ is generated by all of them. The quotient $G^{\mathrm{ab}} = G/G'$ is the abelianisation. $G$ is perfect when $G' = G$. The derived series is $G \ge G' \ge G'' \ge \cdots$, and its length is how many steps it takes to reach the trivial group.
The commutator. $[a, b] = aba^{-1}b^{-1}$ measures the failure of $a$ and $b$ to commute: it is the identity exactly when $ab = ba$. The derived subgroup $G' = [G, G]$ is the subgroup generated by all of them.
Three facts, in order of usefulness.
The third is the one that does the work. It says every homomorphism from $G$ to an abelian group factors through $G/G'$, which is why $G^{\mathrm{ab}}$ is called the abelianisation.
The derived series. Iterate:
$$G = G^{(0)} \ \trianglerighteq\ G^{(1)} = G' \ \trianglerighteq\ G^{(2)} = (G')' \ \trianglerighteq\ \cdots$$
Every factor is abelian by fact 2. So if the series reaches $1$, the group is solvable with that chain as witness.
And the converse. If $1 = G_0 \trianglelefteq \cdots \trianglelefteq G_n = G$ has abelian factors, then $G/G_{n-1}$ abelian forces $G' \subseteq G_{n-1}$ by fact 3, and inductively $G^{(i)} \subseteq G_{n-i}$, so $G^{(n)} = 1$. A group is solvable exactly when its derived series reaches the trivial group, and no searching is required.
Perfect groups. If $G' = G$ the series never moves. Any non-abelian simple group is perfect — $G'$ is normal and non-trivial, so it is everything — so $A_5$ has derived series $A_5, A_5, A_5, \ldots$ and is not solvable. For $S_5$: $S_5' = A_5$ and then it stalls.
Computing it. Rarely by listing commutators. The quick route is fact 3 backwards: find the largest abelian quotient, and its kernel is $G'$. For $S_n$ the sign map has abelian image of order $2$, and no larger abelian quotient exists, so $S_n' = A_n$.
Another way: picture
Think of $G'$ as the sediment left when a group is asked to commute. Everything that stops $G$ being abelian is pushed down into $G'$, and what floats above is the cleanest abelian group $G$ can be turned into. Repeating asks the sediment to commute, and so on down. A solvable group settles out completely; a perfect group is all sediment and nothing ever settles.
Another way: steps
To test solvability with the derived series: 1. Find the largest abelian quotient of $G$; its kernel is $G'$. 2. If $G' = G$, stop: $G$ is perfect and not solvable (unless $G$ is trivial). 3. If $G' = 1$, stop: $G$ is abelian, hence solvable. 4. Otherwise repeat inside $G'$. 5. The derived length is the number of steps to reach $1$; abelian groups have length $1$ and $S_4$ has length $3$.
This is the standard trap, and it deserves a paragraph of its own.
$G'$ is defined as the subgroup generated by the commutators. In most familiar groups every element of $G'$ happens to be a single commutator, which makes the distinction invisible. But it is not a theorem. The smallest group in which some element of $G'$ is a product of two commutators and not a commutator itself has order $96$, and examples become common among larger groups.
So the safe statements are: every commutator lies in $G'$, and $G'$ consists of products of commutators and their inverses. The inverse of a commutator is a commutator — $[a,b]^{-1} = [b,a]$ — so $G'$ is exactly the set of finite products of commutators.
Derived length is not composition length. $S_4$ has composition length $4$ and derived length $3$; $\mathbb{Z}_{12}$ has composition length $3$ and derived length $1$. Derived length counts how many abelian layers are needed, which is much coarser and is the quantity solvability is about.
A table worth carrying.
| Group | $G'$ | Derived length |
|---|---|---|
| abelian | $1$ | $1$ |
| $S_3$ | $A_3$ | $2$ |
| $S_4$ | $A_4$ | $3$ |
| the symmetries of a square | order $2$ | $2$ |
| $A_5$ | $A_5$ | never reaches $1$ |
| $S_5$ | $A_5$ | never reaches $1$ |
The last two rows are the whole reason the quintic has no formula, arriving three lessons before the Galois theory that will say so.
Taking $G'$ to be the set of commutators. It is the subgroup they generate; the difference is real, if rare in small groups.
Expecting $G'$ to be abelian. Only the quotient is. $[S_5, S_5] = A_5$.
Computing $[a,b]$ as $ab - ba$. That is the Lie bracket of matrices, a different object in an additive setting. In a group the commutator is $aba^{-1}b^{-1}$.
Thinking a non-trivial $G'$ means not solvable. $S_4$ has $G' = A_4$ and is solvable; what matters is whether the series eventually reaches $1$.
Confusing derived length with the length of a composition series. They count different things and are rarely equal.
Forgetting to check normality of the intermediate terms. They are automatically normal in the term above — each $G^{(i+1)}$ is the derived subgroup of $G^{(i)}$ — and in fact each is normal in $G$, which is a bonus not shared by general solvable chains.
The sign map is the largest abelian quotient, of order $2$, so $S_4' = A_4$.
Step one.
$A_4$ has $V_4$ as a normal subgroup with quotient of order $3$, and no larger abelian quotient, so $A_4' = V_4$.
Step two.
$V_4$ is abelian, so $V_4' = 1$. The series is $S_4, A_4, V_4, 1$ — three steps, so $S_4$ is solvable with derived length $3$.
Reaches the bottom.
$A_5$ is simple, so $A_5'$ is normal and therefore $1$ or $A_5$.
Only two options.
It is not $1$, since that would make $A_5$ abelian.
So $A_5' = A_5$: perfect.
The derived series is constant, so it never reaches $1$ and $A_5$ is not solvable. For $S_5$ the first step gives $A_5$ and then the same stall occurs.
Not solvable, with no search required.
The group has order $8$ and is not abelian, so $G'$ is not trivial.
Something is there.
The largest abelian quotient: the group has three subgroups of index $2$, and the quotient by the centre — the identity with the half turn — has order $4$ and is abelian.
A quotient of order $4$.
So $G'$ is contained in that centre, and being non-trivial it equals it: $G'$ has order $2$. Then $G'$ is abelian, so $G'' = 1$ and the derived length is $2$ — the group is solvable, as every group of order $8$ must be.
Match each group to its derived subgroup $[G, G]$.
| The trivial subgroup | $A_3$, of order $3$ | $A_4$, of order $12$ | The whole group | $V_4$, of order $4$ | |
|---|---|---|---|---|---|
| Any abelian group | |||||
| $S_3$ | |||||
| $S_4$ | |||||
| $A_5$ |
How many elements does the derived subgroup of $S_4$ have?
Answer:
The derived series of $S_4$ runs $S_4$, then $A_4$, then $V_4$, then the trivial subgroup. Fill in the order of each term and the order of the quotient by the term below it.
| Order of this term | Order of the quotient by the next term | |
|---|---|---|
| $S_4$ itself | ||
| Its derived subgroup | ||
| The derived subgroup of that | ||
| One step further down | — |
Select every statement that is true of the derived subgroup $[G, G]$ of a group $G$.
This task has no paper form; do it on a device.
Put in order the steps of computing the derived series of a finite group.
Number the steps in order (write the number in the box):
Build the proof that $[G, G]$ is the smallest normal subgroup of $G$ with abelian quotient.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
Match each group to its derived subgroup $[G, G]$.
| The trivial subgroup | $A_3$, of order $3$ | $A_4$, of order $12$ | The whole group | $V_4$, of order $4$ | |
|---|---|---|---|---|---|
| Any abelian group | |||||
| $S_3$ | |||||
| $S_4$ | |||||
| $A_5$ |
You can compute a derived series and read solvability from where it stops. Say in your own words why the derived subgroup is the smallest normal subgroup with abelian quotient. Next: the groups at the far end of the scale, with no normal subgroups at all.
9. Your turn: the derived subgroup of the symmetries of a square, step 3