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Any two composition series of a finite group agree in length and in factors: what that makes an invariant, why the subgroups still differ, and why the factors never rebuild the group.
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 state the Jordan-Hölder theorem, compute the factors of two different composition series of the same group and check they agree as multisets, outline the refinement argument that proves it, use the factor list as an invariant to separate groups, say why equal factor lists never prove isomorphism, and derive the uniqueness of prime factorisation from the theorem.
The last lesson built composition series and found that a group usually has several — $\mathbb{Z}_{12}$ has two obvious ones passing through completely different subgroups. Whether the pieces they produce agree was left hanging, and it is not a small question: if the answer were no, the phrase the composition factors of $G$ would mean nothing.
A property is an invariant of a group when isomorphic groups share it, so a difference in it proves non-isomorphism. A multiset is a collection in which repetition counts but order does not — which is the right shape for a list of factors. Schreier refinement intersects the terms of two series to refine both at once, and Zassenhaus's lemma (the butterfly lemma) identifies the factors that come out.
The Jordan-Hölder theorem. Any two composition series of a finite group have the same length, and their factors agree up to isomorphism and reordering.
So the multiset $\{G_{i+1}/G_i\}$ depends on $G$ alone. It is called the multiset of composition factors of $G$, and both its size (the composition length) and its contents are invariants.
What the proof does. Given two series, intersect every term of one with every term of the other. The resulting grid refines both — Schreier's theorem — and Zassenhaus's lemma shows that the factor produced by crossing row $i$ with column $j$ is the same whichever series it is read from. So the two refinements have matching factors. A composition series admits no proper refinement, so each refinement is its own series with trivial factors padded in, and deleting those leaves the two original lists equal as multisets.
The analogy with primes, and its limits. Simple groups are to finite groups what primes are to integers: every finite group breaks into them, and the list is unique. Indeed applying the theorem to $\mathbb{Z}_n$ is the fundamental theorem of arithmetic.
But the analogy stops at one crucial place. An integer is recovered from its prime factorisation by multiplying; a group is not recovered from its composition factors. $\mathbb{Z}_4$ and the Klein four-group have the same factors and are not isomorphic; so do $\mathbb{Z}_6$ and $S_3$. Reassembly is the extension problem, and it is hard — given $N$ and $Q$, the groups $G$ with a normal $N$ and $G/N \cong Q$ can be many and are classified by cohomology rather than by counting.
Why it matters anyway. It makes the classification of finite simple groups worth having. Knowing all the simple groups is knowing all the possible pieces, which bounds what any finite group can be made from — and it makes solvable, the property the next lessons study, well defined: it is a condition on the multiset of factors, so it cannot depend on how the group was taken apart.
Another way: picture
Two people dismantle the same machine by different routes. They remove different assemblies in different orders, and at no point is either following the other. When both finish, the piles of irreducible parts on the floor match exactly, item for item. The theorem says this always happens — and it also warns that neither pile is enough to say how the machine was put together, because different machines leave the same pile.
Another way: steps
To use the theorem: 1. Build any one composition series; the choice does not matter. 2. Record its factors as a multiset. 3. That multiset is an invariant, so it may be quoted as the composition factors of the group. 4. To prove two groups non-isomorphic, compare their multisets. 5. Do not try to prove two groups isomorphic this way — equal multisets prove nothing.
It can prove groups different. $\mathbb{Z}_{12}$ has factors of orders $2, 2, 3$; $A_4$ has factors of orders $2, 2, 3$ as well, so this particular test fails to separate them — and indeed a sharper invariant is needed there. But $S_5$ has factors $A_5$ and $\mathbb{Z}_2$, and no group of order $120$ built from abelian pieces can be isomorphic to it, which is a real conclusion drawn from the list alone.
It defines solvability. A group is solvable exactly when every composition factor is abelian, hence of prime order. Jordan-Hölder is what makes that a property of the group rather than of a chain: if one series had all-abelian factors and another did not, the word would be meaningless.
It cannot rebuild the group. The standard pairs are worth memorising, because they come up whenever somebody hopes the list is enough:
| Same factors | The groups |
|---|---|
| $2, 2$ | $\mathbb{Z}_4$ and the Klein four-group |
| $2, 3$ | $\mathbb{Z}_6$ and $S_3$ |
| $2, 2, 2$ | $\mathbb{Z}_8$, the symmetries of a square, and the quaternion group |
Each row is a family of non-isomorphic groups with an identical list. The failure is not an accident of small numbers — it is the extension problem, and it is where a large part of modern group theory lives.
It does not order the factors. The theorem says up to reordering, and the order genuinely varies: $\mathbb{Z}_{12}$ can produce $2, 2, 3$ or $3, 2, 2$ depending on the route.
Thinking the subgroups agree. Only the factors do. Two series of $\mathbb{Z}_{12}$ pass through subgroups of order $4$ and of order $3$ respectively, and neither chain contains the other.
Thinking the order of the factors is determined. It is not, and the statement says so.
Concluding two groups are isomorphic from equal factor lists. This is the big one. The list is an invariant, so unequal lists prove non-isomorphism; equal lists prove nothing at all.
Expecting it for infinite groups. $\mathbb{Z}$ has no composition series, so there is nothing for the theorem to say.
*Reading simple as small.* $A_5$ is a single composition factor of order $60$, bigger than most of the groups whose factor lists are three or four entries long.
First route: $1 \lhd \mathbb{Z}_2 \lhd \mathbb{Z}_4 \lhd \mathbb{Z}_{12}$, with factors of orders $2, 2, 3$.
Through the subgroup of order $4$.
Second route: $1 \lhd \mathbb{Z}_3 \lhd \mathbb{Z}_6 \lhd \mathbb{Z}_{12}$, with factors of orders $3, 2, 2$.
Through the subgroup of order $3$.
Different subgroups, different order of appearance, same multiset $\{2, 2, 3\}$ — and that is the prime factorisation of $12$.
The theorem, in the smallest interesting case.
$\mathbb{Z}_4$: the chain $1 \lhd \mathbb{Z}_2 \lhd \mathbb{Z}_4$ gives factors of orders $2$ and $2$.
Two factors.
The Klein four-group: the chain through any of its three subgroups of order $2$ gives factors of orders $2$ and $2$ as well.
The same two factors.
The groups are not isomorphic — one has an element of order $4$ and the other does not. So equal factor lists are no evidence of isomorphism whatsoever.
The limit of the analogy with primes.
$A_5$ has index $2$ in $S_5$, so it is normal and gives a factor of order $2$.
The top step.
$A_5$ is simple, so nothing can be inserted below it, and the chain $1 \lhd A_5 \lhd S_5$ is already a composition series.
Two factors.
The factors are $A_5$ and a group of order $2$. By Jordan-Hölder that list is an invariant, so no chain will ever break $S_5$ into abelian pieces — which is precisely why the general quintic has no radical formula.
Here are two composition series of $\mathbb{Z}_{12}$, each given by the orders of its terms from the bottom up. Fill in the order of each factor, and the product along each row.
| First factor | Second factor | Third factor | Product | |
|---|---|---|---|---|
| Climbing through orders $1, 2, 4, 12$ | ||||
| Climbing through orders $1, 3, 6, 12$ |
Put in order the steps of the argument that two composition series of the same group have the same factors.
Number the steps in order (write the number in the box):
How many factors does a composition series of the cyclic group of order $3^{4}$ have?
Answer:
Select every statement that the Jordan-Hölder theorem makes about a finite group.
This task has no paper form; do it on a device.
Match each group to the list of composition factors that Jordan-Hölder attaches to it.
| Three factors of order $2$ | Factors of orders $2$, $3$ and $5$ | Factors of orders $2$, $2$ and $3$ | $A_5$ and a factor of order $2$ | Four factors of order $2$ | |
|---|---|---|---|---|---|
| $\mathbb{Z}_8$ | |||||
| $\mathbb{Z}_{30}$ | |||||
| $A_4$ | |||||
| $S_5$ |
Build the proof that the prime factorisation of a positive integer is unique, using the Jordan-Hölder theorem.
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.
How many factors does a composition series of the cyclic group of order $3^{4}$ have?
Answer:
You can compute the composition factors of a finite group and treat them as an invariant. Say in your own words why two groups with the same factors can still differ, and give an example. Next: the groups whose factors are all abelian, which are the solvable ones.
9. Your turn: the composition factors of $S_5$, step 3