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Jul 19, 2019 at 13:51 comment added Andrea Ferretti Many good answers here, but I have chosen this one, as it actually address how one would go on about proving the theorem I presented as an example, even though for a particular choice of definition of finite sets
Jul 19, 2019 at 13:46 vote accept Andrea Ferretti
Jul 18, 2019 at 17:10 comment added François G. Dorais With this notion of finite, you also need that the range of $H$ is a decidable subgroup, otherwise finding "the one with lowest index" doesn't work. Equivalently, you can assume that $H$ is finite in the same sense ($H$'s appropriate $L$ is nonempty).
Jul 18, 2019 at 16:29 history answered Neil Strickland CC BY-SA 4.0