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1d
comment Radical and Centric not Essential P-group
Yes, I am the author of that paper. Some parts of it could be proved much more efficiently!
1d
comment Formula for getting a value that doubles the amount of the previous value?
You might google "geometric progression".
1d
comment Is it possible to have a research career while checking the proof of every theorem that you cite?
When I was at Chicago, I remember Raghavan Narasimhan telling me that he would not use a result in his work which he could not prove himself.
1d
comment Great Mathematicians Without a PhD
Apart from the fact that it is off-topic for the site, one might ask "If a Mathematician is "great", and seen as such in historical context, why does it matter whether or not they have a Ph.D.?".
May
2
comment For $P$ $\mathbb{Z}G$-projective, $\mathbb{Q}\otimes P$ is $\mathbb{Q}G$-free
@EhudMeir : Here $\mathbb{Z}_{p}$ denotes the $p$-adics, (not the field with $p$ elements ) and characters afforded by projective $\mathbb{Z}_{p}G$-modules do genuinely vanish on all $p$-singular elements of $G$ ( see Curtis and Reiner (1962 or later two part edition, for example)). I retyped this comment.
May
2
comment Radical and Centric not Essential P-group
Suzuki, or Aschbacher- well-known texts on finite group theory ( there are many others- Carter, for finite groups of Lie type).
May
2
comment Radical and Centric not Essential P-group
I think you have to do some reading. But if you look at the case of $G/U \cong {\rm GL}(3,2)$ this quotient is a group of order 168, and it has two conjugacy classes of maximal parabolic subgroups each isomorphic to $S_{4}$. These have index $7$ of course, so they are not strongly ($2$)-embedded ( as when $Y$ is a strongly $p$-embedded subgroup of $Y$, we have that $[Y:X] i \equiv 1$ (mod $p^{a}$) where $p^{a}$ is the order of a Sylow $p$-subgroup of $Y$.
May
1
comment For $P$ $\mathbb{Z}G$-projective, $\mathbb{Q}\otimes P$ is $\mathbb{Q}G$-free
Tensor the $\mathbb{Z}G$-module with $\mathbb{Z}_{p}$ for any prime $p$. Then it is a projective $\mathbb{Z}_{p}G$-module, so its character vanishes on each element of order divisible by $p$. Since $p$ is arbitrary, the character afforded by the $\mathbb{Q}G$-module vanishes on all non-identity elements of $G$, hence its character is an integer multiple of the regular character. By Maschke, it is free.
Apr
29
revised Character Values for Alternating Groups of degree $\geq 7$
clarification
Apr
28
revised Character Values for Alternating Groups of degree $\geq 7$
Updated in view of the answer of Jeremy Rickard.
Apr
28
comment Radical and Centric not Essential P-group
My name is G.R. Robinson, but there is nothing new in what I am saying.
Apr
28
comment Radical and Centric not Essential P-group
From no reference, just from understanding the definitions.
Apr
28
comment Radical and Centric not Essential P-group
I meant "strongly $p$-embedded" in the last result.
Apr
28
comment Radical and Centric not Essential P-group
Well, I suppose the parabolic $P$ itself is a smaller example. In general, the smallest example, whatever it is, will be the fusion system of genuine group $G$ with $U = O_{p}(G)$ elementary Abelian and $C_{G}(U) = U$, such that $G/U$ has no strongly embedded subgroup. A candidate might be the case that $p = 2$, $U$ elementary of order $8$, and $G/U \cong {\rm GL}(3,2)$, but I have not checked.
Apr
28
comment Constructing the largest finite group with a fixed number of conjugacy classes
E.Landau proved in around 1895 that for a fixed $k$, there are only finitely many solutions to $\sum_{j=1}^{k} \frac{1}{n_{j}} = 1$ in positive integers. Apply this to the class equation of a finite group.
Apr
27
answered Radical and Centric not Essential P-group
Apr
27
comment Character Values for Alternating Groups of degree $\geq 7$
@PeterMueller : I believe that the answer to your stronger question should be found (one way or other) in the proof of the result of James and Kerber I mention.
Apr
27
revised Character Values for Alternating Groups of degree $\geq 7$
Amended answer
Apr
27
comment Character Values for Alternating Groups of degree $\geq 7$
@PeterMueller : The easiest way would be show that for such a suitably chosen $g$, there might be more than two irreducible characters $\chi$ with $\chi(g)$ non-real. But it is conceivable that there is always just one complex conjugate pair of irreducible characters with this property. Unfortunately, I don't have access to that book at the moment, but I'm sure someone on here will.
Apr
27
comment Character Values for Alternating Groups of degree $\geq 7$
@PeterMueller :Mmm, thanks, you are right that it doesn't explicitly answer the question as it stands ( I slightly misread the question). I need to check out the key lemma from James and Kerber used in the referenced paper with Thompson to see whether that covers things. The Lemma of James and Kerber states that if $g \in A_{n}$ has all its disjoint cycles of distinct odd length, then the field generated by the character values at $g$ is a genuinely quadratic extension of $\mathbb{Q}$. In any case, the answer to the question lies in that result of James and Kerber.