Proving $x\in\text{SL}(n,\mathbb Q)$ given finite indices of $x^{-1}Gx$ in $G$ and $x^{-1}Gx$

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Denote $G=\text{SL}(n,\mathbb Z)$ and let $x\in \text{SL}(n,\mathbb R)$ such that $$[G:x^{-1}Gx\cap G],[x^{-1}Gx:x^{-1}Gx\cap G]<\infty.$$ Prove that $x\in\text{SL}(n,\mathbb Q)$.

I know that $\text{SL}(n,\mathbb Z)$ is finitely generated but somehow I can't find how to use that to show that the entries of $x$ are all rational.

The best I got is that for $n=2$, by some calculation, $x=\sqrt q M$ where $q\in\mathbb Q,M\in\text{GL}(2,\mathbb Q)$. From there I couldn't prove that $\sqrt q\in\mathbb Q$.

Can I get a hint how to proceed? even for the case of $SL(2,\mathbb Z$)?

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This is a special case of a theorem by A.Borel (in the context of arithmetic subgroups of semisimple algebraic groups):

A. Borel, Density and maximality of arithmetic subgroups. J. Reine Angew. Math. 224 (1966), 78–89.

Maybe there is a simpler proof in the SL(n) case. Venkataramana or D. Witte Morris at Mathoverflow will surely know one if there is such.