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).
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$\begingroup$ It is not true in general, as $\begin{pmatrix} 2 \sqrt{-15} & 5 \\ 5 & - \sqrt{-15} \end{pmatrix} \in \mathrm{GL}_2(\mathbb{Q}(\sqrt{-15}))$ conjugates $M_2(\mathbb{Z}[\sqrt{-15}])$ but is not a scalar multiple of an element of $\mathrm{GL}_2(\mathbb{Z}[\sqrt{-15}])$ (its determinant, $5$, is not a square times a unit). $\endgroup$– HenriCommented Nov 12, 2011 at 17:02
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$\begingroup$ In Henri's example, ${\mathbf Z}[\sqrt{-15}]$ is not the full ring of integers of ${\mathbf Q}(\sqrt{-15})$. Perhaps that "explains" the example. Emerton's argument goes through for PIDs. Does it go through more generally for any Dedekind domain, perhaps by a localization argument to reduce to the case of a PID? $\endgroup$– KConradCommented Nov 12, 2011 at 17:21
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$\begingroup$ @KConrad No, the very same matrix also conjugates $M_2(\mathbb{Z}[\frac{1+\sqrt{-15}}{2}])$. Homology's answer shows that $A$ conjugates $\mathrm{GL}_n(R)$ iff $I^n =(det(A))$ where $I$ is the (fractional) ideal generated by the coefficients of $A$, so that one can always take a bigger $R' \subset \mathrm{Frac}(R)$. $\endgroup$– HenriCommented Nov 12, 2011 at 17:31
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$\begingroup$ @KConrad However the UFD case follows from the localization (I edited my answer below). $\endgroup$– HomologyCommented Nov 13, 2011 at 11:58
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3 Answers
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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 $g(\mathbb Z^n) \subset {\mathbb Z}^n$, 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.
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$\begingroup$ @Emerton: thank you very much indeed! $\endgroup$– OlodCommented Nov 11, 2011 at 16:12
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$\begingroup$ The automorphism group of $\mathbb{Z}^n$ (fixing the origin) contains the permutations and the diagonal matrices with $\pm 1$ entries. So $Z(G)$ in the statement (and proof) should be replaced by the semidirect product of these 2 groups, I think. $\endgroup$ Commented Nov 11, 2011 at 19:28
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2$\begingroup$ @Emerton: That's a very fine proof. But there is one point that I don't understand properly: Why is the image of $g(\mathbb{Z}^n)$ preserved by $GL(n,\mathbb{F}_p)$ ? It would be obvious to me, if each matrix in $GL(n,\mathbb{F}_p)$ were a mod p reduction from $GL(n,\mathbb{Z})$, what isn't the case, however. In effect, that doesn't matter, of course, since $g$ also normalizes $SL(n,\mathbb{Z})$ what maps surjectively to $SL(n,\mathbb{F}_p)$. Then one can use the rest of the argument verbatim with $SL$ in place of $GL$. $\endgroup$– tomaszCommented Nov 11, 2011 at 21:15
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$\begingroup$ Dear tomasz, Thanks for the correction. Regards, Matthew $\endgroup$– EmertonCommented Nov 12, 2011 at 2:35
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$\begingroup$ @Abhinav: No, it should not. Clearly, the center of $G$ is contained in the normalizer of any subgroup of $G$. But $Z(G)\cap GL(n,Z)=\{I,-I\}$, so you cannot omit it if you want the full normalizer. On the other hand, Emerton's nice proof indeed requires scaling by an arbitrary scalar (matrix). And neither require adding the permutation matrices, as those are already contained in $GL(n,Z)$. $\endgroup$– Max HornCommented Nov 12, 2011 at 11:20
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It is also true for local rings. Let $A$ be a matrix with coefficients in $K = \mathrm{Frac}(R)$, such that $A \mathrm{GL}_n(R) A^{-1} \subset \mathrm{GL}_n(R)$. Then in particular, $A (I_n + E_{i,j})A^{-1} \in M_n(R)$ and $A (I_n+E_{i,i}+E_{i,j}+E_{j,i})A^{-1} \in M_n(R)$ for all $i \neq j$, so that $a_{i,j} b_{k,l} \in R$ for all $i,j,k,l$, where $a$, $b$ denote the coefficients of $A$, $B:=A^{-1}$.
Since $\sum_j a_{1,j}b_{j,1}=1$, there is a $j_0$ such that $a_{1,j_0} b_{j_0,1} \in R^{\times}$. Let $A'=a_{1,j_0}^{-1} A$. Then the coefficients of $A'^{-1} = a_{1,j_0} B$ are in $R$, and those of $A'$ also because $\frac{a_{i,j}}{a_{1,j_0} }= \frac{a_{i,j}b_{j_0,1}}{a_{1,j_0}b_{j_0,1}} \in R$.
For a general $R$, this just shows that $A \in K^{\times} \mathrm{GL}_n (R_m) $ for any maximal ideal $m$ of $R$.
Edit: This implies that the normalizer is indeed $K^{\times} \mathrm{GL}_n(R)$ in the case $R$ is a UFD. Indeed we have that $I^n=(\det (A))$ where $I=\sum_{i,j} a_{i,j} R$ by localization ($i/\det(A) \in \cap_m R_m = R$ for any $i$ product of $n$ coefficients of $A$). Looking at decompositions, we see that $\mathrm{gcd}(a_{i,j})^n = \det(A)$, and so $A/\\mathrm{gcd}(a_{i,j}) \in \mathrm{GL}_n(R)$.
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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 $(z/m_1) g \in \mathrm{GL}(n,\mathbf Z).$