Prove the following special case of part 2 of the Jordan-Holder Theorem. Suppose the finite group has two composition series: and . Show that and that the composition factors are isomorphic up to a permutation.
Note that is simple and that is normal in . Because is normal, we have either or .
If , then . Now is normal, and because is simple either (a contradiction since the inclusion is strict) or . Because is simple, we have , so that , and clearly the composition factors are isomorphic up to the identity permutation.
Suppose . Now is a subgroup, so that . Because is simple, either or . If , then , so that . But then is simple, a contradiction because is normal. If , then by the Second Isomorphism Theorem (twice) we have and . In particular, is simple so that and we have . Moreover, and while and ; thus the composition factors are isomorphic up to the permutation .