# 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.

## 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.