Omega subgroups are isomorph-free

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This article gives the statement, and possibly proof, of the fact that for any group, the subgroup obtained by applying a given subgroup-defining function (i.e., omega subgroups of a group of prime power order) always satisfies a particular subgroup property (i.e., isomorph-free subgroup)}
View subgroup property satisfactions for subgroup-defining functions | View subgroup property dissatisfactions for subgroup-defining functions


Statement with symbols

Let P be a group of prime power order. Then, for any natural number j, the subgroup:

\Omega_j(P) = \langle g \in P \mid g^{p^j} = e \rangle

is an isomorph-free subgroup.

Facts used

  1. Omega subgroups are homomorph-containing
  2. Homomorph-containing implies isomorph-free for co-Hopfian subgroup


The proof follows directly from facts (1) and (2), along with the fact that any finite group is co-Hopfian.