Jordan–Hölder theorem

Jordan–Hölder theorem. If \[ 1=G_0\triangleleft G_1\triangleleft\cdots\triangleleft G_m=G \] and \[ 1=H_0\triangleleft H_1\triangleleft\cdots\triangleleft H_n=G \] are composition series of a finite group \(G\), then \(m=n\) and, after reordering, the factors \(G_i/G_{i-1}\) are isomorphic to the factors \(H_j/H_{j-1}\).

We prove this by induction on \(|G|\). The result is trivial for \(G=1\). Let \[ M=G_{m-1},\qquad N=H_{n-1}. \] These are maximal proper normal subgroups of \(G\).

If \(M=N\), apply induction to the two composition series of \(M\) obtained by deleting the final term \(G\); adjoining the common simple factor \(G/M\) proves the result.

Assume \(M\ne N\). Since \(MN\) is a normal subgroup properly containing \(M\), maximality gives \(MN=G\); similarly \(M\cap N\) is proper in both \(M\) and \(N\). The second isomorphism theorem gives \[ G/M\cong N/(M\cap N),\qquad G/N\cong M/(M\cap N). \] Choose a composition series of \(M\cap N\). Extending it to \(M\) and to \(N\) using the two displayed simple quotients gives composition series of \(M\) and \(N\). By induction, the original composition series of \(M\) has the same factors as this refined series of \(M\), and likewise for \(N\).

Consequently the factors below \(M\cap N\) agree, while the two remaining factors are exchanged by \[ G/M\cong N/(M\cap N),\qquad G/N\cong M/(M\cap N). \] Thus the two composition series of \(G\) have the same multiset of composition factors, and hence the same length.