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Taking a group apart by a chain of normal subgroups and keeping the quotients: refinement, why a composition series is one whose factors are simple, and why every finite group has one.
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 build a subnormal series for a small group, compute its factors and check that their orders multiply to the order of the group, say why normality is required only in the next term up, refine a series until every factor is simple, recognise a composition series, and prove that every finite group has one.
The first course built $G/N$ for a normal subgroup $N$ and showed that it is a smaller group carrying part of the information in $G$. One quotient simplifies a group once. This lesson takes the quotient over and over, recording the pieces that fall out — and the question of whether the pieces depend on the route is what the next lesson settles.
A subnormal series is a chain $1 = G_0 \trianglelefteq G_1 \trianglelefteq \cdots \trianglelefteq G_n = G$, each term normal in the next. Its factors are the quotients $G_{i+1}/G_i$, and $n$ is its length. A refinement inserts extra terms. A composition series is one with no proper refinement, equivalently one whose factors are all simple.
A subnormal series of $G$ is a chain
$$1 = G_0 \trianglelefteq G_1 \trianglelefteq \cdots \trianglelefteq G_n = G,$$
where each term is normal in the next one up. Its factors are the quotient groups $G_{i+1}/G_i$. For a finite group the orders multiply:
$$|G| = \prod_i |G_{i+1}/G_i|.$$
Normal in the next, not normal in $G$. This is the point most often misread. In the chain $1 \trianglelefteq \{e, (1\,2)(3\,4)\} \trianglelefteq V_4 \trianglelefteq A_4 \trianglelefteq S_4$ the second term is not normal in $S_4$ — it is not even normal in $A_4$ — and the chain is perfectly good. A series in which every term is normal in $G$ is called a normal series, and it is a stricter and less useful notion.
Refinement. Inserting extra terms refines a series. A series that cannot be refined at all is a composition series, and the condition is exactly that every factor is simple: a factor with a proper non-trivial normal subgroup would, by the correspondence theorem, supply a term to insert.
Existence. Every finite group has a composition series. Induct on the order: take a maximal proper normal subgroup $N$, so $G/N$ is simple, and put $G$ on top of a composition series of $N$. Infinite groups may have none — $\mathbb{Z}$ does not, because $n\mathbb{Z}$ always contains $2n\mathbb{Z}$ and the chain never reaches the bottom.
What the factors are for. They are the pieces $G$ is built from, in the sense that they are what remains after every possible simplification. For $\mathbb{Z}_{12}$ they are groups of orders $2, 2, 3$ — the prime factorisation of $12$. For $S_4$ they are $2, 2, 3, 2$. For $A_5$ there is one factor and it is $A_5$ itself.
What they are not. The factors do not determine the group. $\mathbb{Z}_4$ and the Klein four-group have the same factors — two of order $2$ — and are not isomorphic. Knowing the pieces is not knowing how they were assembled, and that gap is the extension problem.
Another way: picture
Think of the group as a tower and the series as floors, each sitting normally on the one below. What is recorded is not the floors but the gaps between them. Refining means adding a floor; a composition series is a tower with no room for another floor anywhere, and then each gap is a simple group — the algebraic equivalent of a prime.
Another way: steps
To build a composition series of a finite group: 1. Find a maximal proper normal subgroup $N$ — often a Sylow subgroup, a kernel, or a subgroup of index $2$. 2. Record the factor $G/N$, which is then simple. 3. Repeat inside $N$, which is smaller. 4. Stop at the trivial group. 5. Check: the orders of the factors should multiply back to $|G|$.
Building a series is mostly a hunt for normal subgroups, and the earlier lessons of this unit supply most of the ammunition.
Index two. A subgroup of index $2$ is always normal, because the two left cosets and the two right cosets are both the subgroup and everything else. That is how $A_n \trianglelefteq S_n$ is got, and it is the first step of nearly every series involving a symmetric group.
A unique Sylow subgroup. If some $n_p = 1$, that subgroup is normal and gives a step. The last two lessons were largely about forcing this to happen.
A union of conjugacy classes. $V_4$ is normal in $A_4$ because it is the identity's class together with the class of double transpositions. Reading the class equation lists the candidates.
The centre. $Z(G)$ is always normal, and for a $p$-group it is non-trivial — so a $p$-group always has a step available. Iterating gives the upper central series and shows every $p$-group has a composition series whose factors all have order $p$.
A worked example: $S_4$. $A_4$ has index $2$, so it is normal and the top factor has order $2$. Inside $A_4$, the subgroup $V_4$ is a union of classes, so it is normal and the next factor has order $3$. $V_4$ is abelian, so any of its three subgroups of order $2$ is normal in it, giving factors of order $2$ and $2$. The series has length $4$ and factors $2, 3, 2, 2$ reading downwards — every one of prime order, which is what will make $S_4$ solvable and the quartic solvable by radicals.
Requiring every term to be normal in $G$. Only normality in the next term up is asked for. Demanding more usually makes a composition series impossible to build.
Thinking normality is transitive. It is not: $\{e, (1\,2)(3\,4)\}$ is normal in $V_4$ and $V_4$ is normal in $A_4$, but the small subgroup is not normal in $A_4$. This is exactly why the definition says in the next one up.
Recording the subgroups instead of the factors. The factors are what the theory is about, and two very different chains can produce the same list.
Assuming the factors determine the group. $\mathbb{Z}_4$ and the Klein four-group disagree while having identical factors.
Expecting a composition series always to exist. Only for finite groups, and more generally for groups satisfying both chain conditions. $\mathbb{Z}$ has none.
Abelian, so every subgroup is normal and any increasing chain will do. Take $1 \lhd \mathbb{Z}_2 \lhd \mathbb{Z}_{10} \lhd \mathbb{Z}_{30}$.
A chain through the divisors.
The factors have orders $2$, $5$ and $3$, each prime and so each simple.
Divide successive orders.
No refinement is possible, and $2 \times 5 \times 3 = 30$. The factors are the prime factorisation of $30$, which is what a composition series of a cyclic group always produces.
A composition series.
$A_5$ is simple, so its only normal subgroups are $1$ and itself.
No room to climb.
The only series is $1 \lhd A_5$, of length $1$, with the single factor $A_5$.
One step.
So a group of order $60$ can have a series of length $1$ while $\mathbb{Z}_{60}$ has one of length $4$. Length is not determined by the order — it is determined by how far the group can be broken down.
Length measures decomposability, not size.
The group has order $8$, and its centre — the identity with the half turn — is normal of order $2$.
A step at the bottom.
The four rotations form a subgroup of index $2$, so they are normal; and the centre is normal inside them.
A chain $1 \lhd Z \lhd R \lhd G$.
The factors have orders $2, 2, 2$, all prime, so this is a composition series of length $3$ — and $2 \times 2 \times 2 = 8$ as it must be. Choosing a different subgroup of order $4$, such as a pair of reflections with the centre, gives a different chain with the same three factors.
These five subgroups form a composition series of $S_4$. Put them in order, starting from the trivial subgroup.
Number the steps in order (write the number in the box):
How many factors does a composition series of $\mathbb{Z}_{30}$ have?
Answer:
Here is a composition series of $\mathbb{Z}_{12}$, given by the orders of its terms. Fill in the order of each factor, and the product of them all.
| Order below | Order above | Order of the factor | |
|---|---|---|---|
| The first step | 1 | 2 | |
| The second step | 2 | 4 | |
| The third step | 4 | 12 | |
| The three factors multiplied together | — | — |
Select every statement that is true of a composition series of a finite group $G$.
This task has no paper form; do it on a device.
Match each group to the orders of the factors in a composition series of it.
| Orders $2, 2, 3$ | Orders $3, 2$ | Orders $2, 2, 3, 2$ | A single factor of order $60$ | Orders $2, 2, 3, 5$ | |
|---|---|---|---|---|---|
| $\mathbb{Z}_{12}$ | |||||
| $S_3$ | |||||
| $S_4$ | |||||
| $A_5$ |
Build the proof that every finite group has a composition series.
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.
These five subgroups form a composition series of $S_4$. Put them in order, starting from the trivial subgroup.
Number the steps in order (write the number in the box):
You can build a composition series for a small group and list its factors. Say in your own words why a term need not be normal in the whole group, and give the example that shows normality is not transitive. Next: whether a different route up the tower gives different pieces.
9. Your turn: a composition series of the symmetries of a square, step 3