# The normalizer of $\mathrm{GL}(n,\mathbf Z)$ in $\mathrm{GL}(n,\mathbf Q)$

It seems that the normalizer of $H=\mathrm{GL}(n,\mathbf Z)$ in $G=\mathrm{GL}(n,\mathbf Q)$ is "almost" equal to itself, that is, $$N_G(\mathrm{GL}(n,\mathbf Z))=Z(G) \cdot \mathrm{GL}(n,\mathbf Z)$$ where $Z(G)$ is the centre of $G.$ Is there a simple proof/disproof of this fact? More generally, for which integral domains $R$ it is known that $\mathrm{GL}(n,R)$ "almost" coincides with its normalizer in the group $\mathrm{GL}(n,Q(R))$ where $Q(R)$ is the quotient field of $R?$

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You need to add the scalar matrices, at least. – Plop Nov 10 '11 at 0:00
@Plop: I will, thank you. – Olod Nov 10 '11 at 7:34
Perhaps you should try asking this on MathOverflow. – Derek Holt Nov 11 '11 at 8:58
Right. $\phantom{abc}$ – Olod Nov 11 '11 at 9:44
The interested readers may find Emerton's answer to the question at mathoverflow.net/questions/80667/… – Olod Nov 11 '11 at 16:16