# Fitting subgroup 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., Fitting subgroup) 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 Fitting subgroup of a group need not be an isomorph-free subgroup.

## Proof

### Example of the symmetric group

Further information: symmetric group:S4

Let $G$ be the symmetric group on the set $\{ 1, 2, 3, 4 \}$. The Fitting subgroup of this is a Klein-four group described as: $H := \{ (), (1,2)(3,4), (1,3)(2,4), (1,4)(2,3) \}$. $H$ is not isomorph-free in $G$: it is isomorphic to the group: $K := \{ (), (1,2), (3,4), (1,2)(3,4) \}$.

### A more generic example: the von Dyck group

Suppose $H$ is a nilpotent group and $K$ is a Fitting-free group, containing a subgroup $L$ isomorphic to $H$. Let $G = H \times K$. The Fitting subgroup 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.