This is often called the Jordan Holder theorem because it was developed by Jordan (biography) and Holder (biography).
This theorem is often compared to unique factorization in the integers. You always get the same primes, possibly in a different order.
Make sure you are familiar with the previous theorem before you proceed.
Let U and V be two subnormal series for the group G. Build a matrix M with the series U running down the left hand side, and the series V running across the top. Using the notation of the previous section, let Mi,j = Ui+1×(Ui∩Vj).
Watch what happens as we move across the ith row of M. As we move from column j to column j+1, we find the groups Ui+1×(Ui∩Vj) and Ui+1×(Ui∩Vj+1). If we want to use the variables of the previous section, we have S×(A∩B) and S×(A∩T). By Zassenhaus, the latter is normal in the former. The ith row of M builds a subnormal series. The first entry in the ith row is Ui+1×(Ui∩G), or Ui. The last entry is Ui+1×(Ui∩e), or Ui+1. Hence the ith row of M is a subnormal series from Ui to Ui+1.
Run the entire matrix from top to bottom, left to right. This produces a subnormal series from U1 to U2, from U2 to U3, from U3 to U4, and so on, until we have a subnormal series for all of G. This may not fit the technical definition, since many of the groups in the series are the same. But if we don't mind the duplicates, the groups form a descending chain from G all the way down to e. The factor groups are Ui+1×(Ui∩Vj) mod Ui+1×(Ui∩Vj+1), for all i and j.
Next, build a parallel matrix N, with U down the left hand side and V across the top. This time let Ni,j = Vj+1×(Vj∩Ui). This is a form of role reversal. Walk down the jth column of N and find a subnormal series from Vj to Vj+1. Traverse the columns of N, from left to right, and build a subnormal series for all of G.
The factor groups in this series are Vj+1×(Vj∩Ui) mod Vj+1×(Vj∩Ui+1), for all i and j.
Now for the magic. Let i = 3 and j = 7, just to illustrate. The series from the matrix M produces the factor group U4×(U3∩V7) mod U4×(U3∩V8). The series from the matrix N produces the factor group V8×(V7∩U3) mod V8×(V7∩U4). By Zassenhaus, these quotient groups are the same. Both matrices produce the same factor groups, in a different order.
If U and V are composition series, there are no proper refinements. the refinements produced by M and N merely duplicate some of the subgroups in U, or the subgroups of V, respectively. Both M and N produce the same factor groups, many of them trivial. The nontrivial factors must agree, and these are the factor groups produced by U and by V.
If G has a composition series, the simple factor groups of G are unique up to order.
Remember that every finite group G has a composition series, hence G also has a well defined set of factor groups.