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 G has two composition series: 1 = M_0 \vartriangleleft M_1 \vartriangleleft M_2 = G and 1 = N_0 \vartriangleleft \cdots \vartriangleleft N_k = G. Show that k = 2 and that the composition factors are isomorphic up to a permutation.

Note that M_1 is simple and that N_{k-1} \cap M_1 is normal in G. Because N_{k-1} \cap M_1 \leq M_1 is normal, we have either N_{k-1} \cap M_1 = M_1 or N_{k-1} \cap M_1 = 1.

If N_{k-1} \cap M_1 = M_1, then M_1 \leq N_{k-1}. Now N_{k-1}/M_1 \leq G/M_1 is normal, and because G/M_1 is simple either N_{k-1} = G (a contradiction since the inclusion N_{k-1} \leq G is strict) or N_{k-1} = M_1. Because M_1 is simple, we have N_{k-2} = 1, so that k = 2, and clearly the composition factors are isomorphic up to the identity permutation.

Suppose N_{k-1} \cap M_1 = 1. Now M_1 N_{k-1} \leq G is a subgroup, so that M_1 N_{k-1} / M_1 \leq G/M_1. Because G/M_1 is simple, either M_1 N_{k-1} = M_1 or M_1 N_{k-1} = G. If M_1 N_{k-1} = M_1, then N_{k-1} \leq M_1, so that N_{k-1} = 1. But then G is simple, a contradiction because M_1 \leq G is normal. If M_1 N_{k-1} = G, then by the Second Isomorphism Theorem (twice) we have G/M_1 \cong N_{k-1} and G/N_{k-1} \cong M_1. In particular, N_{k-1} is simple so that N_{k-2} = 1 and we have k = 2. Moreover, N_2/N_1 \cong M_1 and N_1/N_0 \cong N_1 while M_2/M_1 \cong N_1 and M_1/M_0 \cong M_1; thus the composition factors are isomorphic up to the permutation (1\ 2).

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  • kavo1  On September 19, 2010 at 12:21 pm

    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?

    • nbloomf  On September 19, 2010 at 12:35 pm

      Because 1 = N_0 \vartriangleleft \cdots \vartriangleleft N_k = G is a composition series, the intermediate quotients N_{t+1}/N_t are simple. Since we don’t consider the trivial group to be simple, the inclusions N_t \leq N_{t+1} are strict, and in particular N_{k-1} < G.

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

  • neal  On July 12, 2011 at 1:28 am

    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?

    • nbloomf  On July 18, 2011 at 1:01 pm

      This follows because N_{k-1} is normal in G.

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