Solvable radical not is isomorph-free

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

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.