I should admit the question below does not have a serious motivation. But still I found it somehow natural.


Let $G$ be a finite group of order $n$ with $h$ conjugacy classes. If $c\_1,\ldots,c\_h$ are the orders of the conjugacy classes of $G$, then clearly


$n=c\_1+c\_2+\ldots+c\_h$.


Let now $\pi\_1,\ldots,\pi\_h$ be the pairwise non-isomorphic, irreducible complex representations of $G$. It is well known that another partition of $n$ of length $h$ is given by the squares of the degrees $d\_i$'s of the $\pi\_i$'s:


$n=d\_1^2+d\_2^2+\ldots+d\_h^2$.


Question: Assume that, up to reordering, the two partitions of $n$ described above are the same. Then what can we say about $G$? Is $G$ forced to be abelian?



When do the sizes of conjugacy classes and squares of degrees of irreps give the same partition for a finite group?
My standard rant about "what can we say about $G$": what we can say about $G$ is that the two partitions are the same. If the questioner doesn't find that a helpful answer then they might want to consider the possibility that they asked the wrong question ;-)


But as to the actual question: "is $G$ forced to be abelian?", the answer is no, and I discovered this by simply looping through magma's database of finite groups. Assuming I didn't make a computational slip, the smallest counterexample has order 64, is the 73rd group of order 64 in magma's database, which has 8 representations of degree 1, 14 representations of degree 2, 8 elements in the centre and 14 more conj classes each of order 4.


Letting the loop go further, I see counterexamples of size 64, 128, 192 (I guess these are just the counterexamples of size 64 multiplied by Z/3Z) and then ones of order 243 (a power of 3). So I guess all examples I know are nilpotent. Are they all nilpotent? That's a question I don't know the answer to.


