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

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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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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.