Let $g \in GL(n,\mathbb Q)$ normalize $GL(n,\mathbb Z)$. Consider the lattice $g(\mathbb Z^n) \subset \mathbb Q^n$; it is preserved by $GL(n,\mathbb Z)$. Replacing $g$ by $gz$ for some appropriate scalar matrix $z$, we may assume that $g(\mathbb Z^n) \subset \mathbb Z${\mathbb Z}^n$, but that$g(\mathbb Z^n)\not\subset p \mathbb Z^n$for any prime$p$. Suppose now that$p$divides the index$[\mathbb Z^n:g(\mathbb Z^n)]$. Then the image of$g(\mathbb Z^n)$is a proper subspace of$\mathbb F_p^n$(by the assumption that the$p$divides the index) which is non-zero (by the assumption that$g(\mathbb Z^n)$is not contained in$p\mathbb Z^n$). It is preserved by$GL(n,\mathbb F_p)$. [Added: As tomasz points out in a comment below,$GL(n,\mathbb Z)$does not surject onto$GL(n,\mathbb F_p)$. However, its image does contain$SL(n,\mathbb F_p)$, so the argument below goes through, if we replace$GL(n,\mathbb F_p)$by$SL(n,\mathbb F_p)$.] But this is a contradiction, since$\mathbb F_p^n$is an irreducible$GL(n,\mathbb F_p)$-representation. Consequently, no such$p$exists, and so$g(\mathbb Z^n) = \mathbb Z^n$. Thus$g \in GL(n,\mathbb Z)$, and so we have shown that the normalizer of$GL(n,\mathbb Z)$is equal to$Z(G) \cdot GL(n,\mathbb Z),$as required. This argument (assuming that it's correct!) extends at least to the case when$R$is a PID. 2 added 262 characters in body Let$g \in GL(n,\mathbb Q)$normalize$GL(n,\mathbb Z)$. Consider the lattice$g(\mathbb Z^n) \subset \mathbb Q^n$; it is preserved by$GL(n,\mathbb Z)$. Replacing$g$by$gz$for some appropriate scalar matrix$z$, we may assume that$g(\mathbb Z^n) \subset \mathbb Z$, but that$g(\mathbb Z^n)\not\subset p \mathbb Z^n$for any prime$p$. Suppose now that$p$divides the index$[\mathbb Z^n:g(\mathbb Z^n)]$. Then the image of$g(\mathbb Z^n)$is a proper subspace of$\mathbb F_p^n$(by the assumption that the$p$divides the index) which is non-zero (by the assumption that$g(\mathbb Z^n)$is not contained in$p\mathbb Z^n$). It is preserved by$GL(n,\mathbb F_p)$. [Added: As tomasz points out in a comment below,$GL(n,\mathbb Z)$does not surject onto$GL(n,\mathbb F_p)$. However, its image does contain$SL(n,\mathbb F_p)$, so the argument below goes through, if we replace$GL(n,\mathbb F_p)$by$SL(n,\mathbb F_p)$.] But this is a contradiction, since$\mathbb F_p^n$is an irreducible$GL(n,\mathbb F_p)$-representation. Consequently, no such$p$exists, and so$g(\mathbb Z^n) = \mathbb Z^n$. Thus$g \in GL(n,\mathbb Z)$, and so we have shown that the normalizer of$GL(n,\mathbb Z)$is equal to$Z(G) \cdot GL(n,\mathbb Z),$as required. This argument (assuming that it's correct!) extends at least to the case when$R$is a PID. 1 Let$g \in GL(n,\mathbb Q)$normalize$GL(n,\mathbb Z)$. Consider the lattice$g(\mathbb Z^n) \subset \mathbb Q^n$; it is preserved by$GL(n,\mathbb Z)$. Replacing$g$by$gz$for some appropriate scalar matrix$z$, we may assume that$g(\mathbb Z^n) \subset \mathbb Z$, but that$g(\mathbb Z^n)\not\subset p \mathbb Z^n$for any prime$p$. Suppose now that$p$divides the index$[\mathbb Z^n:g(\mathbb Z^n)]$. Then the image of$g(\mathbb Z^n)$is a proper subspace of$\mathbb F_p^n$(by the assumption that the$p$divides the index) which is non-zero (by the assumption that$g(\mathbb Z^n)$is not contained in$p\mathbb Z^n$). It is preserved by$GL(n,\mathbb F_p)$. But this is a contradiction, since$\mathbb F_p^n$is an irreducible$GL(n,\mathbb F_p)$-representation. Consequently, no such$p$exists, and so$g(\mathbb Z^n) = \mathbb Z^n$. Thus$g \in GL(n,\mathbb Z)$, and so we have shown that the normalizer of$GL(n,\mathbb Z)$is equal to$Z(G) \cdot GL(n,\mathbb Z),$as required. This argument (assuming that it's correct!) extends at least to the case when$R\$ is a PID.