Degrees of irreducible representations need not determine nilpotency class

Statement

It is possible to have two finite nilpotent groups $G_{1}$ and $G_{2}$ of the same order and with the same Degrees of irreducible representations (?) over a splitting field but with different nilpotency class values.

Related facts

Similar facts

• Degrees of irreducible representations need not determine derived length
• Conjugacy class size statistics 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

Further information: linear representation theory of groups of order 32#Degrees of irreducible representations, linear representation theory of groups of prime-fifth order#Degrees of irreducible representations

There are many examples among groups of order $p^{5}$ for $p$ a prime number. For instance, for order $2^{5}=32$, there are groups of nilpotency class both 2 and 3 with the following degrees of irreducible representations: 8 of degree 1, 6 of degree 2.