# Solvable radical not is isomorph-free

From Groupprops

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) doesnotalways 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

## Contents

## 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 is a solvable group and is a Fitting-free group (in particular, the solvable core of is trivial), containing a subgroup isomorphic to . Let . The solvable core of is . However, this subgroup is isomorphic to .

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