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Jan 10, 2023 at 6:50 vote accept kindasorta
Jan 9, 2023 at 15:08 answer added Arno Fehm timeline score: 3
Jan 9, 2023 at 11:24 comment added kindasorta Yeah, I am precisely curious about this fact.
Jan 9, 2023 at 10:02 comment added Arno Fehm I think it is known that the (closed) subgroup generated by finitely many of these is the (profinite) free product of these. If this is the kind of result you are interested in, I can look it up.
Jan 9, 2023 at 9:35 comment added YCor The embedding $G_{\mathbf{Q}_p}\to $G_{\mathbf{Q}}$, if I'm correct, is only defined up to conjugation, and in particular its image is only well-defined up to conjugation.
Jan 9, 2023 at 9:04 history asked kindasorta CC BY-SA 4.0