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Jan 13, 2015 at 20:35 comment added LSpice Of course, one must be careful not to rely too much on this sort of luck; $\operatorname{PGL}_n(\mathbb Z_p)$ is not a maximal subgroup of $\operatorname{PGL}_n(\mathbb Q_p)$!
Dec 20, 2011 at 22:49 history edited Alain Valette CC BY-SA 3.0
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Dec 18, 2011 at 10:12 comment added Marc Palm Both answers are very useful. I checked the other simply because it was first. Thanks.
Dec 17, 2011 at 22:00 history edited Alain Valette CC BY-SA 3.0
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Dec 17, 2011 at 21:36 history answered Alain Valette CC BY-SA 3.0