Perhaps this is a simple question, but I'm curious about it and have no idea how to even approach answering it.
I know that for a finite group $G$, one has the formula
$$ n_G=\sum_\mu n_\mu^2, \tag{$\ast$} $$
where $n_G$ is the order of the group, the sum is over all (finite-dimensional) irreducible, inequivalent representations $\mu$ of $G$, and $n_\mu$ is the dimension of the representation $\mu$. Moreover, the number of $\mu$ is given by the number of conjugacy classes of $G$.
Question: Is it possible for two groups $G$ and $G'$ (say $n_G,n_{G'}>4$ in order to allow for nonabelian-ness, as what I'm going to ask would be trivially true for abelian groups) to be nonisomorphic, but have precisely the same realization of $(\ast)$? Meaning: can they have precisely the same number of irreps in every dimension?
Best Answer
Yes : the quaternion group and the dihedral group of order $8$ both have $4$ representations of degree $1$ and one irreducible representation of degree $2$.
Note that there are not many way of obtaining $8$ as a sum of squares : every dimension must be less than or equal to $2$, and there must always be a representation of dimension $1$ (the trivial one), so for a non-abelian group of order $8$ the only possibility is to write it as $8=1+1+1+1+2^2$.