Solvable radical not is isomorph-free

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)
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Statement

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

Related facts

• Fitting subgroup not is isomorph-free
• Fitting subgroup is normal-isomorph-free in finite
• Solvable core is normal-isomorph-free in finite
• Perfect core is homomorph-containing

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.