## 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 $(1\ 2)$.

• 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 $G$.