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April 16[edit]

Isomorphic to all nontrivial normal subgroups[edit]

Is there a group G that is neither simple nor isomorphic to Z, but that is isomorphic to all of its nontrivial normal subgroups? GeoffreyT2000 (talk) 15:20, 16 April 2016 (UTC)[reply]

Take an arbitrary non-identity element. That element generates a cyclic subgroup. --JBL (talk) 15:51, 16 April 2016 (UTC)[reply]

This proof would be correct if it were all subgroups instead of just normal ones. The cyclic subgroup is not necessarily normal. GeoffreyT2000 (talk) 17:33, 16 April 2016 (UTC)[reply]

My apologies, I read too quickly. --JBL (talk) 17:35, 16 April 2016 (UTC)[reply]

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)[reply]

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)[reply]