Finite solvable groups can have only finitely many distinct subnormal series

Question

My lecture notes say:

If $G$ is a finite solvable group then, $G$ has a chain of subgroups $$G = G_0 > G_1 > G_2 > \cdots > G_n = 1$$ such that $G_i \unlhd G_{(i-1)}$ and $G_{i-1}/G_{i}$ is abelian for all $i$. Note that $G$ can have only finitely many such series.

Could someone point me to a proof of the part in bold?

Answer 1

Suppose a set $A$ has cardinality $k$. Then there are only finitely many strictly nested sequences of subsets $A=A_0 > A_1 > A_2 > ... > A_n$:

1. First choose the proper subset $A_1$; there are at most $2^k$ choices.
2. Then choose the proper subset $A_2$; there are at most $2^{k-1}$ choices.

Continue by induction.

Altogether there are at most $2^k \cdot 2^{k-1} \cdot ... \cdot 2 = 2^{(k^2+k)/2} $ choices.