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Hello

i wonder if there is a simple argument to show that no group of order 48 can have an irreducible character of degree larger than 4?

Thanks, Karim

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  • $\begingroup$ Edited to change 6 to 4 per karim's clarification. $\endgroup$ – Will Sawin May 21 '12 at 21:26
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    $\begingroup$ I wonder whether it's an interesting question to ask for groups of order $n$ with an irreducible character of degree $d$ with $(d+1)^2\ge n$? The group $A_4$, of order 12, has an irreducible character of degree 3; there's a group of order 20 with an irrep of degree 4, and a group of order 42 with an irrep of degree 6. $\endgroup$ – Gerry Myerson May 21 '12 at 23:25
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    $\begingroup$ @Mark is right. Why bother asking questions about any group of order less than 2000? $\endgroup$ – Steve D May 21 '12 at 23:34
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    $\begingroup$ @Gerry: They have been known for a while. See arxiv.org/abs/math/0603239 $\endgroup$ – S. Carnahan May 22 '12 at 0:05
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    $\begingroup$ @Steve D: I am not sure you understand what you are saying. There are 4725 groups of order 864, for example, and you need a CAS to treat them. There are only 52 groups of order 48, and you can treat them by hand. Each one of them has a normal subgroup of order 16 or 8. In the first case, it is an extension of a group of order 16 by a group of order 3, in the second case it is an extension of a group of order 8 by $S_3$. Also not every question about a small group is trivial. For example, construct a hyperbolic 6-manifold with isometry group $S_3$, if you can (hint: it can be done). $\endgroup$ – user6976 May 22 '12 at 0:18