TITLE: The normalizer of $\mathrm{GL}(n,\mathbf Z)$ in $\mathrm{GL}(n,\mathbf Q)$
QUESTION [15 upvotes]: 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 $G$ (one may guess so by applying the description of automorphisms of groups $\mathrm{GL}(n,\mathbf Z)$ by Hua and Reiner).  Is there, however, a simpler and direct 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 question has been earlier posted at mathunderflow).

REPLY [4 votes]: Let me simplify a bit (if I may) a nice argument by Matthew Emerton, by omitting the part with $p$-reductions. We start as above: let $g \in\mathrm{GL}(n,\mathbf Q)$ normalize $\mathrm{GL}(n,\mathbf Z).$ Then the subgroup
$$
zg(\mathbf Z^n) 
$$
is in $\mathbf Z^n$ for a certain $z \in \mathbf Q$ and is invariant under all elements of $\mathrm{GL}(n,\mathbf Z).$ By the description of the subgroups of free abelian groups, there is a basis $f_1,\ldots,f_n$ of $\mathbf Z^n$ and integers $m_1,\ldots,m_n$ with
$m_k | m_{k+1}$ $(k=1,\ldots,n-1)$ such that
$$
zg(\mathbf Z^n) =\langle m_1 f_1,m_2 f_2,\ldots,m_n f_n \rangle.
$$
Hence
$$
(z/m_1)g(\mathbf Z^n) =\langle f_1,(m_2/m_1) f_2,\ldots,(m_n/m_1) f_n \rangle \leqslant \mathbf Z^n.
$$
Thus the subgroup $(z/m_1)g(\mathbf Z^n)$ contains a unimodular/basis element (namely, $f_1$) and then since $(z/m_1)g(\mathbf Z^n)$ is invariant under $\mathrm{GL}(n,\mathbf Z)$ we have that
$$
(z/m_1)g(\mathbf Z^n) =\mathbf Z^n.
$$
Thus $(z/m_1) g \in \mathrm{GL}(n,\mathbf Z).$