Fitting subgroup 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., 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.

Related facts

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

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Another generic example: the product of a Fitting-free group with a nilpotent 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.