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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 $(p1)^{n1}.$ 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 2006ish 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. 