Timeline for Is every $GL_2(\mathbb{Z}/n\mathbb{Z})$-extension contained in some elliptic curve's torsion field?
Current License: CC BY-SA 3.0
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| Jul 14, 2017 at 0:47 | comment | added | Noam D. Elkies | The somewhat remarkable thing is that ${\rm PGL}_2({\bf Z}/4)$ has a central element of determinant $-1$. (There has to be, once we know that ${\rm PGL}_2({\bf Z}/4) \cong S_4$, because $S_4$ has no outer automorphism.) | |
| Jul 13, 2017 at 19:00 | history | edited | David E Speyer | CC BY-SA 3.0 |
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| Jul 13, 2017 at 17:56 | comment | added | Noam D. Elkies | Sorry, it is indeed PGL_2. The GL_2 statement is not just little-known but wrong ;-) :-( | |
| Jul 13, 2017 at 17:40 | comment | added | David E Speyer | @NoamD.Elkies Also, the order is wrong. $|GL_2(\mathbb{Z}/4)|$ is $96$, not $48$. Perhaps you meant $PGL_2(\mathbb{Z}/4)$? | |
| Jul 13, 2017 at 17:38 | comment | added | David E Speyer | @NoamD.Elkies I did not know that! But, wait, I am confused. Both groups have a unique non-identity central element -- namely $- \mathrm{Id}$ in $GL_2(\mathbb{Z}/4)$ and $(-1, \mathrm{Id})$ in $\{ \pm 1 \} \times S_4$. But $-\mathrm{Id}$ has a square root, $\left( \begin{smallmatrix} 0&1 \\ -1&0 \end{smallmatrix} \right)$ and $(-1,\mathrm{Id})$ doesn't. So how can this be right? | |
| Jul 13, 2017 at 17:27 | comment | added | Noam D. Elkies | It's a neat and surprisingly little-known fact that ${\rm GL}_2({\bf Z}/4)$ is isomorphic with $\{\pm1\} \times S_4$. | |
| Jul 13, 2017 at 16:38 | history | edited | David E Speyer | CC BY-SA 3.0 |
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| Jul 13, 2017 at 16:32 | history | answered | David E Speyer | CC BY-SA 3.0 |