Timeline for What does equality modulo $p$ of $p$-adic linear groups imply?
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Nov 23, 2016 at 16:15 | comment | added | Geoff Robinson | Yes, it is indeed. | |
| Nov 23, 2016 at 16:07 | comment | added | kneidell | Hi @GeoffRobinson - is absolutely irreducible the same as absolutely simple? | |
| Nov 23, 2016 at 15:53 | comment | added | Geoff Robinson | I should have said "absolutely irreducible" above. | |
| Nov 23, 2016 at 15:44 | comment | added | Geoff Robinson | So maybe the right question is: is some conjugate of $H$ commensurable with $G$, since the condition is blind to conjugation by elements of $1 + pM_{n}(\mathbb{Z}_{p})$. ( also, I think that if the groups remain irreducible (mod $p$), then the groups upstairs must be conjugate by an element of $1 + pM_{n}(\mathbb{Z}_{p})$ if they are conjugate at all). | |
| Nov 23, 2016 at 10:21 | comment | added | David Loeffler | They will not be commensurable in general. Take e.g. $G$ to be the diagonal matrices in $\mathbf{SL}_2$ and $H$ to be the conjugate of $G$ by $\begin{pmatrix} 1 & p \\ 0 & 1 \end{pmatrix}$. Then $G$ and $H$ have the same image mod $p$ but $G \cap H$ is trivial. | |
| Nov 23, 2016 at 10:09 | history | edited | kneidell | CC BY-SA 3.0 |
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| Nov 23, 2016 at 10:04 | history | asked | kneidell | CC BY-SA 3.0 |