Number of conjugacy classes in group of prime power order is congruent to order of group modulo prime-square minus one

From Groupprops
Jump to: navigation, search


Suppose p is a prime number and P is a group of prime power order with underlying prime p. Then the number of conjugacy classes of P (which is the same as the number of irreducible representations) is congruent to the order of P modulo (p^2 - 1).

Facts used

  1. Number of irreducible representations equals number of conjugacy classes
  2. Degree of irreducible representation divides order of group: For a p-group, in particular, this means that it is a power of p.
  3. Sum of squares of degrees of irreducible representations equals order of group


Given: A prime number p. A finite p-group P of order p^r with n conjugacy classes.

To prove: n \equiv p^r \pmod{(p^2 - 1)}


Step no. Assertion/construction Facts used Given data used Previous steps used Explanation
1 n equals the number of irreducible representations of P. Fact (1) n is the number of conjugacy classes.
2 All irreducible representations of P have degree p^i for some nonnegative integer i, with 0 \le i \le r. Fact (2) P has order p^r [SHOW MORE]
3 Let a_i be the number of irreducible representations of degree p^i for i = 0,1,2,\dots,r. Then, a_0 + a_1 + a_2 + \dots + a_r = n. Steps (1), (2) Step-combination direct.
4 a_0(p^0)^2 + a_1(p^1)^2 + a_2(p^2)^2 + \dots + a_r(p^r)^2 = p^r Fact (3) P has order p^r Step (3) Step-fact direct.
5 a_i \equiv a_i(p^i)^2 \pmod{(p^2 - 1)} [SHOW MORE]
6 a_0 + a_1 + a_2 + \dots + a_r \equiv a_0(p^0)^2 + a_1(p^1)^2 + a_2(p^2)^2 + \dots + a_r(p^r)^2 \pmod{(p^2 - 1)} Step (5) Sum up Step (5) for 0 \le i \le r.
7 n \equiv p^r \pmod{(p^2 - 1)} Steps (3), (4), (6) Step-combination direct.