Timeline for When do the sizes of conjugacy classes and squares of degrees of irreps give the same partition for a finite group?
Current License: CC BY-SA 2.5
12 events
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| Feb 8, 2012 at 21:34 | comment | added | Kevin Buzzard | @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... | |
| Feb 8, 2012 at 17:04 | comment | added | Marty Isaacs | I wonder if even the weaker condition that all class sizes are squares implies that G is nilpotent. Does anyone know a counterexample to that? A very tiny step in that direction is the observation that if |G| is divisible by some prime p to the first power only, then a Sylow p-subgroup is central. That is because no class size can be divisible by p, and so the centralizers of the Sylow p-subgroups cover G. But in general, conjugates of a proper subgroup of a finite group can never cover the group. | |
| Apr 6, 2010 at 7:22 | comment | added | Tommaso Centeleghe | @Kevin: No prob! :-) | |
| Apr 6, 2010 at 6:02 | comment | added | Kevin Buzzard | @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. | |
| Apr 6, 2010 at 6:02 | comment | added | Kevin Buzzard | @Tommaso: just ignore my rant about "what can we say about G"---I had obviously got out of the wrong side of bed that day. Sorry. | |
| Apr 5, 2010 at 12:31 | comment | added | Tile | 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. | |
| Apr 5, 2010 at 12:27 | comment | added | Jim Humphreys | 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. | |
| Apr 5, 2010 at 12:06 | comment | added | Tommaso Centeleghe | 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?". | |
| Apr 5, 2010 at 11:49 | vote | accept | Tommaso Centeleghe | ||
| Apr 5, 2010 at 11:47 | vote | accept | Tommaso Centeleghe | ||
| Apr 5, 2010 at 11:49 | |||||
| Apr 5, 2010 at 10:18 | history | edited | Kevin Buzzard | CC BY-SA 2.5 |
typo
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| Apr 5, 2010 at 10:10 | history | answered | Kevin Buzzard | CC BY-SA 2.5 |