Solvable radical not is isomorph-free

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This article gives the statement, and possibly proof, of the fact that for a group, the subgroup obtained by applying a given subgroup-defining function (i.e., solvable radical) does not always satisfy 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

The solvable radical of a group need not be an isomorph-free subgroup.

Related facts

Proof

A generic example: the product of a Fitting-free group with a nilpotent group

Suppose H is a solvable group and K is a Fitting-free group (in particular, the solvable core of K is trivial), containing a subgroup L isomorphic to H. Let G = H \times K. The solvable core of G is H \times 1. However, this subgroup is isomorphic to 1 \times L.

An example might be to take K as any non-Abelian simple group, and H as isomorphic to an Abelian subgroup L of K. For instance, K is the alternating group on five letters and H is a cyclic group of order two.