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I am a new poster but I don't think this question has been asked before. Pardon me if it is.

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$PSL(2,\mathbb{Q})$ or $PSL(2,q)$? –  Martín-Blas Pérez Pinilla Feb 20 at 8:43
    
@Martín-BlasPérezPinilla I do mean $\mathbb{Q}$, the field of rational numbers. –  Singhal Feb 20 at 8:45
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Yes. ${\rm PSL}(n,K)$ is simple for any $n \ge 2$ and any field $K$, except when $n=2$ and $|K|=2$ or $3$. –  Derek Holt Feb 20 at 8:48
    
@DerekHolt Thank you. Is the proof easy? If not, can you provide a reference or a sketch of it (may be just for $\mathbb{Q}$ only)? –  Singhal Feb 20 at 8:51

1 Answer 1

up vote 5 down vote accepted

As I said in my comment, ${\rm PSL}(n.K)$ is simple for all $n \ge 2$ and all fields $K$, except for $n=2$, $|K| \le 3$.

The proof is not exactly easy, but it is not impossibly difficult either. The field $K$ plays virtually no role, except in one place where we need at least $4$ distinct elements in $K$ when $n=2$.

It is in various books, such as Huppert's "Endliche Gruppen I", which is unfortunately in German. I have extracted the proof from some lecture notes of mine, which I took from Huppert's book. The proof assumes some results in group theory, such as a normal subgroup of a primitive permutation group being transitive.

Anyway, I hope this helps. You can find it here.

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Thank you so much! –  Singhal Feb 20 at 9:22

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