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18h
revised Number of elements of “$\mathrm{SL}_n(\mathbb{F}_p^\times)$” mod $p$
minor typo
21h
revised Number of elements of “$\mathrm{SL}_n(\mathbb{F}_p^\times)$” mod $p$
gave more precise statement.
1d
comment Over which fields is a $G$-module reducible?
It's your question!
1d
comment Over which fields is a $G$-module reducible?
Another way to say is that you want to find those $L$ such that ${\rm End}_{LG}(V_{L})$ is not a division algebra.
1d
revised Number of elements of “$\mathrm{SL}_n(\mathbb{F}_p^\times)$” mod $p$
added comments about answer (mod $p$).
1d
revised Number of elements of “$\mathrm{SL}_n(\mathbb{F}_p^\times)$” mod $p$
added 186 characters in body
1d
answered Number of elements of “$\mathrm{SL}_n(\mathbb{F}_p^\times)$” mod $p$
1d
revised Generating subgroups of large index by a large chunk of a conjugacy class
minlor typo
1d
revised Generating subgroups of large index by a large chunk of a conjugacy class
added extra example
1d
comment Number of elements of “$\mathrm{SL}_n(\mathbb{F}_p^\times)$” mod $p$
A little remark is that the number of such matrices is an integer multiple of $(p-1)^{n-1}.$ The group $T$ of invertible diagonal matrices acts on such matrices by conjugation and only scalar matrices fix anything in the action.
1d
answered Generating subgroups of large index by a large chunk of a conjugacy class
Dec
18
revised automorphism of prime order for group of Lie type in
added extra example
Dec
17
answered automorphism of prime order for group of Lie type in
Dec
17
comment A generalisation of the theorem of Maschke
@JimHumphreys : I missed the question at the time, and just noticed it. I was motivated to give the reference to clarify what Burnside had done about subgroups of ${\rm GL}(n,\mathbb{C})$ (periodic such subgroups of bounded period are finite) and Schur had done ( finitely generated periodic such subgroups are finite and, in general, such periodic subgroups are completely reducible).
Dec
17
comment ULU Decomposition of a matrix
If it could be done, it could be done with $u_{1}$ and $u_{2}$ unipotent.
Dec
17
answered A generalisation of the theorem of Maschke
Dec
16
awarded  Necromancer
Dec
16
answered Does a referee have to check carefully the proof ?
Dec
15
comment Why do sporadic simple groups have so few conjugacy classes?
As a matter of interest, It is proved in a 2006-ish paper of Bob Guralnick and myself that in general, $\frac{k(G)}{|G|} \to 0$ as $[G:F(G)] \to \infty$ ( for finite $G$).
Dec
15
comment Why do sporadic simple groups have so few conjugacy classes?
@S.Carnahan : Well, it's maybe not quite equivalent. Having few conjugacy classes is equivalent to the mean irreducible character degree being large. For example, $M_{12}$ has an irreducible complex representation of degree $11$, Suzuki's sporadic group of order 448,345,497,600 has a $12$-dimensional irreducible complex representation.