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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?

edited Apr 5 2010 at 13:02

Ben Webster♦
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asked Apr 5 2010 at 9:05

Tommaso Centeleghe
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By the way, you're strongly encouraged to write your title in the form of a question so that people on the front page know what it is, rather than just that it has something to do with finite groups. I've changed it, but if you don't like the title I changed it to, go ahead and edit it to something else. – Ben Webster♦ Apr 5 2010 at 13:05

I thought about it and I could not come up with a compact form for a more precise title. Thanks for doing it! – Tommaso Centeleghe Apr 5 2010 at 13:16

Kevin Buzzard · Accepted Answer · 2010-04-05 10:18:52Z UTC

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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.

edited Apr 5 2010 at 10:18

answered Apr 5 2010 at 10:10

Kevin Buzzard
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Dear Kevin Buzzard, many thanks for your answer. My "what can we say about G?" was just a preamble to the actual mathematically (and uninteresting) well-defined question "Is G forced to be abelian?". – Tommaso Centeleghe Apr 5 2010 at 12:06

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At the opposite extreme, a finite simple group of Lie type can't exhibit this odd numerical behavior due to the existence of Steinberg characters (and the fact that class sizes divide the group order in general). I have no idea about other non-nilpotent cases, but I suspect Marty Isaacs (Wisconsin) would know how to answer the original question even without the help of Magma if he was motivatedto do so. In any case, Kevin is correct to start with nonabelian 2-groups since these are the prime suspects for counterexamples and there are a lot of them. – Jim Humphreys Apr 5 2010 at 12:27

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Generally speaking, the complex group algebra does seldom impose very strict conditions for G. Just look at the theorems of Maschke and Wedderburn and you will see that there are many different groups with isomorphic complex group algebra. By the way, your question is equivalent to asking whether $h=|G|$. I'm guessing that there is a simple reason why Kevin found nilpotent groups. Most groups of small order are 2-groups (ok almost every group is a 2-group). As the condition is very arithmetical it is no wonder that he found a 2-group. I don't believe that $G$ must be nilpotent. – Tilemachos Vassias Apr 5 2010 at 12:31

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@Jim: let me underline my stupidity. I didn't start with non-abelian 2-groups: I started with groups of order 1. I (by which I mean my computer) then moved on to groups of order 2 and so on. Ever since I realised that I had at my desk a machine that was capable of iterating over all finite groups of order at most 2000 I've been easy bait for this kind of question. – Kevin Buzzard Apr 6 2010 at 6:02

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@Prof Isaacs: if you make a comment on an answer to a question that's over a year old then probably the only person that will notice the comment is the person who wrote the answer, which is me in this case. I was notified of the comment by the software but probably no-one else well be. If you want anyone other than me to read what you wrote, you'd be better off asking a new question, or writing a new answer to this question (which will bump the question back to the front page). I think the former option would be the most logical... – Kevin Buzzard Feb 8 2012 at 21:34

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Glasby · Answer 2 · 2012-05-25 00:18:52Z UTC

@Marty Isaacs: There exist non-nilpotent groups whose conjugacy class sizes are all squares. For example, let $G$ be Magma's 93rd group order 540. It has class sizes 1,4,9. Indeed, |G|=15*1+30*4+45*9. Also $|Z(G)|=15$ and $G/Z(G)$ is centerless. Thus G is non-nilpotent, and each conjugacy class size is a square.

When do the sizes of conjugacy classes and squares of degrees of irreps give the same partition for a finite group?

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