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 '14 at 08:43
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@Martín-BlasPérezPinilla I do mean $\mathbb{Q}$, the field of rational numbers. – Singhal Feb 20 '14 at 08:45
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6Yes. ${\rm PSL}(n,K)$ is simple for any $n \ge 2$ and any field $K$, except when $n=2$ and $|K|=2$ or $3$. – Derek Holt Feb 20 '14 at 08:48
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@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 '14 at 08:51
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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.
Derek Holt
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Do you know what happens for $G = {\rm PSL}(n, R)$ where $R$ is a commutative ring? I've seen that for $R=\Bbb Z$ the commutator subgroup $[G,G]$ has index $12$ in $G$, so $G$ is not simple. What about other rings? For instance, are there some ring of integers $\mathcal O_K$ (where $K$ is a number field) such that ${\rm PSL}(n, \mathcal O_K)$ is simple? – Watson May 30 '17 at 15:53
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