Problem

Prove the following special case of part (2) of the Jordan-Hölder Theorem: assume the finite group \(G\) has two composition series

$$ 1 = N_0 \trianglelefteq N_1 \trianglelefteq … \trianglelefteq N_r = G $$

and

$$ 1 = M_0 \trianglelefteq M_1 \trianglelefteq M_2 = G. $$

Show that \(r = 2\) and that the list of composition factors is the same. [Use the Second Isomorphism Theorem.]

Solution

We first show that \(N_{r-1} = M_1\).

Note \(N_G(N_{r-1}) = G\) because \(N_{r-1} \trianglelefteq G\). Therefore, \(M_1 \le G = N_G(N_{r-1})\). By the Second Isomorphism Theorem, \(N_{r-1} \trianglelefteq M_1N_{r-1}\).

In general, if \(B \trianglelefteq AB\), \(xBx^{-1} = B\) for all \(x \in \{x = ab \mid a \in A, b \in B\}\), which leads to \(aBa^{-1} = B\) for all \(a \in A\). Therefore, \(B \trianglelefteq AB \implies B \trianglelefteq A\). Therefore, \(N_{r-1} \trianglelefteq M_1\).

Since \(M_1 / M_0 = M_1 / 1 = M_1\) is simple, \(N_{r-1}\) is either \(M_1\) or \(1\). If \(N_{r-1} = 1\), then \(N_1 = 1\), and thus \(N_1 / N_0 = 1 / 1 = 1\) is not simple, contradicting the fact that the first series is a composition series. Therefore, \(N_{r-1} = M_1\).

Now that we know \(N_{r-1} = M_1\), the first composition series can be written as

$$ 1 = M_0 \trianglelefteq N_1 \trianglelefteq … \trianglelefteq N_{r-2} \trianglelefteq M_1 \trianglelefteq M_2 = G. $$

Because \(M_1\) is simple, \(N_{r-2}\) is either \(M_1\) or \(1\). However, \(M_1 / N_{r-2}\) is simple, so it cannot be \(M_1\). Also, \(N_1 / M_0\) is simple, so it cannot be \(1\) either. Thus, no such \(N_{r-2}\) exists. By the same logic, in fact, none of those \(N\)’s can exist between \(M_0\) and \(M_1\) in the series.

Therefore, \(r = 2\).

$$\tag*{$\blacksquare$}$$