Wikipedia:Reference desk/Archives/Mathematics/2016 April 16
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April 16
Isomorphic to all nontrivial normal subgroups
Is there a group G that is neither simple nor isomorphic to Z, but that is isomorphic to all of its nontrivial normal subgroups?
talk) 15:20, 16 April 2016 (UTC)
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- Take an arbitrary non-identity element. That element generates a cyclic subgroup. --talk) 15:51, 16 April 2016 (UTC)]
- This proof would be correct if it were all subgroups instead of just normal ones. The cyclic subgroup is not necessarily normal. talk) 17:33, 16 April 2016 (UTC)]
- My apologies, I read too quickly. --talk) 17:35, 16 April 2016 (UTC)]
- My apologies, I read too quickly. --
- This proof would be correct if it were all subgroups instead of just normal ones. The cyclic subgroup is not necessarily normal.
Apparently free groups of infinite rank also have this property (but I don't have a proof or a reference at hand). —Kusma (t·c) 12:14, 19 April 2016 (UTC)
- Since normal subgroups of such a group can't be generated by a set of lower cardinality than the big group (- e.g. finite sets) - (conjugate by a generator of the big group that isn't in the smaller set of letters used to form generators of the subgroup) - this follows from the Nielsen–Schreier theorem.John Z (talk) 01:20, 22 April 2016 (UTC)