# Is $\mathrm{PSL} ( 2, \mathbb{Q} )$ a simple group?

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 8:43
@Martín-BlasPérezPinilla I do mean $\mathbb{Q}$, the field of rational numbers. – Singhal Feb 20 '14 at 8:45
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 '14 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 '14 at 8:51

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 $n=2$.