Prove part 2 of the Jordan-Holder Theorem by induction on the shortest length of a composition series. [Hint: apply the inductive hypothesis to and use the preceding exercises.]

First we prove some lemmas.

Lemma 1: Let be a group. Let be subgroups such that , , and is simple. Then either or . Proof: Note that is normal in , so that is normal in . Now is normal, so by the Lattice Isomorphism Theorem we have . Because is simple, there are two possibilities.

- If , then since we have

= = by the Second Isomorphism Theorem.

- If , then . Now if , we have , so that .

Lemma 2: Let be a group, subgroups such that , , and is simple, and . Then either or . Proof: We have . By the Lattice Isomorphism Theorem, . The conclusion follows because is simple.

Lemma 3: Let be a group, with simple, and a family of subgroups of such that for , , and for all . Then for all . Proof: We proceed by induction. For the base case, by hypothesis. For the inductive step, suppose such that . Now . By the Lattice Isomorphism Theorem, . Since , . Thus by induction we have for all .

Now we move to the main result.

Note that every composition sequence of a finite group is necessarily finite. For a finite group , we let the *length* of be the shortest length of any composition series of .

Let be a finite group.

If the length of is 1, then is simple and thus has only one composition series; hence all composition series of have the same length and isomorphic composition factors up to a permutation.

If the length of is 2, then we saw in the previous exercise that all composition series of have length 2 and isomorphic composition factors up to a permutation.

Suppose now that for some , if is a finite group of length or less then all composition series of have the same length and isomorphic composition factors up to a permutation. Suppose has length , and let and be composition series of . Let be any other composition series of ; clearly .

If , note that and are composition series for of length and , respectively. By the induction hypothesis, we have and a permutation exists of the set such that . Moreover, trivially. Thus the composition series and have the same length and are equivalent via the permuation .

Suppose now that . If , then by the Lattice Isomorphism Theorem we have . Because is simple, either or ; either case yields a contradiction. Hence ; similarly, .

Note that for all , we have , so that . Because , by the Lattice Isomorphism Theorem we have . Again because is simple, either or . In the first case we have , a contradiction. Thus .

Note that . Let the integer be maximal such that ; note that . By Lemma 3, we have for . Consider now the sequence . Note that we have the starred equality by Lemma 2, as since is maximal.

- If , then is simple.
- If , then . Suppose by way of contradiction that . Now , by the Second Isomorphism Theorem. Since is simple and , we have . Then , a contradiction since . By Lemma 1, then, we have .

In particular, because all intermediate quotients are simple, is a composition series for of length .

Similarly, if the integer is maximal such that , then is a composition series for of length since for the quotient is simple and for the group is simple.

By the inductive hypothesis, we have , so , and there exists a bijection such that .

Finally, note that and by the Second Isomorphism Theorem. Thus, via the permutation , the composition series and are equivalent.

## Comments

Here is a much simpler proof

http://www.math.lsa.umich.edu/~speyer/594/JordanHolder.pdf