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

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## Statement

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

## Proof

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

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

Proof:

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.