Degrees of irreducible representations need not determine derived length

Statement
It is possible to have two finite solvable groups $$G_1$$ and $$G_2$$ of the same order and with the same fact about::degrees of irreducible representations over a splitting field but with different derived length values.

Similar facts

 * Conjugacy class size statistics need not determine nilpotency class
 * Conjugacy class size statistics need not determine derived length
 * Degrees of irreducible representations need not determine nilpotency class
 * Degrees of irreducible representations need not determine conjugacy class size statistics
 * Conjugacy class size statistics need not determine degrees of irreducible representations

Opposite facts

 * Sum of squares of degrees of irreducible representations equals order of group, so the degrees of irreducible representations determine the order of the group.
 * Number of one-dimensional representations equals order of abelianization, hence the degrees of irreducible representations do determine the order of the abelianization, and hence the index of the derived subgroup. By Lagrange's theorem and the fact that they already determine the order of the whole group, they determine the order of the derived subgroup.
 * The number of conjugacy classes of size 1 equals the order of the center, so the conjugacy class size statistics do determine the order of the center and hence also the order of the inner automorphism group

Proof
See linear representation theory of groups of order 128, and note that there do exist collections of degrees of irreducible representations that occur in groups of derived length 2 as well as 3.