I am looking pure group-theoretical proof of the statement that:

$G$ be a finite group then $|G|=\sum_{k=1}^r n_k^2$ where $r$ is the number of conjugacy classes of the group and $n_k$ are some positive integers.

This result is from the theory representation of finite groups where each $n_k$ is the dimension of the irreducible $\mathbb C[G]$ in the decomposition of the $\mathbb C[G]$ but since the result is independent of the chosen algebraicaly closed field.

I am suprised not to found this result in Google, which is very powerful tool to compute or to be sure you got the full list of conjugacy classes.

  • 3
    $\begingroup$ What would count as "pure group-theoretical" if representations don't make the cut? $\endgroup$
    – Bruno B
    May 11, 2023 at 2:11
  • $\begingroup$ that's tha interesting point to me. Since it can be also seen independently from representation theory, the connection would be interesting. Since someone says to you order of the group is sum of squares, and these squares must be number of conjugacy classes $\endgroup$ May 11, 2023 at 2:13

1 Answer 1


I think this statement is not so strong as you think. Suppose that $k\ge 4$. Since every integer is the sum of four squares, we can write $|G|-k+4=a^2+b^2+c^2+d^2$. Now just add some $1^2$. Of course one needs to be careful if some of $a,b,c,d$ are zero, but this shouldn't be a big problem. It would be more interesting to require that all summands divide $|G|$.


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