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 $R?$
The normalizer of $\mathrm{GL}(n,\mathbf Z)$ in $\mathrm{GL}(n,\mathbf Q)$
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linear-algebra
group-theory
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1You need to add the scalar matrices, at least. – 2011-11-10
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0@Plop: I will, thank you. – 2011-11-10
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0Perhaps you should try asking this on MathOverflow. – 2011-11-11
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0Right. $\phantom{abc}$ – 2011-11-11
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3The interested readers may find Emerton's answer to the question at http://mathoverflow.net/questions/80667/the-normalizer-of-mathrmgln-mathbf-z-in-mathrmgln-mathbf-q – 2011-11-11
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0@Emerton Maybe you want to copy your answer on mathoverflow here, so that this question gets removed from the unanswered tab. – 2013-08-18
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0@Julian. Emerton = Matt E here, I think? Anyway, you can't ping him, as he hasn't participated in this thread. – 2013-08-23