Timeline for The normalizer of $\mathrm{GL}(n,\mathbf Z)$ in $\mathrm{GL}(n,\mathbf Q)$
Current License: CC BY-SA 3.0
9 events
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| Nov 12, 2011 at 11:20 | comment | added | Max Horn | @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 $GL(n,Z)$. | |
| Nov 12, 2011 at 3:05 | history | edited | KConrad | CC BY-SA 3.0 |
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| Nov 12, 2011 at 2:37 | history | edited | Emerton | CC BY-SA 3.0 |
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| Nov 12, 2011 at 2:35 | comment | added | Emerton | Dear tomasz, Thanks for the correction. Regards, Matthew | |
| Nov 11, 2011 at 21:15 | comment | added | tomasz | @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 $GL$. | |
| Nov 11, 2011 at 19:28 | comment | added | Abhinav Kumar | 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. | |
| Nov 11, 2011 at 16:12 | comment | added | Olod | @Emerton: thank you very much indeed! | |
| Nov 11, 2011 at 16:05 | vote | accept | Olod | ||
| Nov 11, 2011 at 13:58 | history | answered | Emerton | CC BY-SA 3.0 |