## If a group has a composition series of length 2, then all composition series have length 2 and the composition factors are isomorphic up to a permutation

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 .

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## Comments

I was reading over your solution and I was unsure why it is a contradiction for N_k-1=G. Could you elaborate on why this cannot happen?

Because is a composition series, the intermediate quotients are simple. Since we don’t consider the trivial group to be simple, the inclusions are strict, and in particular .

I’ve updated the solution to make this clear. Thanks!

beginning of last paragraph has N_k-1 inersect M_1 is 1, so M_1 N_k-1 is a subgroup of G, is this a fact from an earlier excercise?

This follows because is normal in .