Solvable implies Fitting subgroup is self-centralizing

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This article gives the statement, and possibly proof, of a particular subgroup of kind of subgroup in a group being self-centralizing. In other words, the centralizer of the subgroup in the group is contained in the subgroup
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Statement

In a solvable group, the Fitting subgroup is self-centralizing: it contains its centralizer in the whole group.

Facts used

  1. Characteristicity is centralizer-closed
  2. Characteristicity is intersection-closed
  3. Members of the derived series of a group are characteristic subgroups (follows from the fact that characteristicity is commutator-closed)
  4. Characteristicity is quotient-transitive
  5. Characteristicity is transitive

Proof

Given: A solvable group G. F(G) denotes the Fitting subgroup of G, and CG(F(G)) denotes its centralizer in G

To prove: CG(F(G))F(G)

Proof: Let C=CG(F(G)) and H=CF(G). Note that every element of H commutes with every element of C, so HZ(C), and in particular, H is normal in C.

Consider the derived series of C/H. Since G is solvable, so is C/H, so its derived series terminates at the identity in finitely many steps. Let B be the inverse image of the term just before the trivial subgroup in this derived series. Then, BG is a subgroup with the property that [B,B]H. But since H commutes with every element of C, it also commutes with every element of B, so [[B,B],B] is trivial. Hence B is nilpotent of class two.

We now show that B is normal, through a series of observations:

  1. F(G) is a characteristic subgroup
  2. Since characteristicity is closed under taking centralizers, C=CG(F(G)) is also characteristic in G
  3. Since characteristicity is closed under intersections, H is characteristic in G
  4. The quotient B/H is a characteristic subgroup of C/H, being a member of the derived series
  5. Hence, using the fact that characteristicity is quotient-transitive, B is a characteristic subgroup of C
  6. Since C is already characteristic in G, and using the fact that characteristicity is transitive, we see that B is characteristic in G

Thus, B is a characteristic subgroup, hence a normal subgroup. So, B is a nilpotent normal subgroup. Moreover, BC, but B is not contained in H, so B cannot be contained in F(G), contradicting the defining feature of F(G) as the subgroup generated by all nilpotent normal subgroups.

References

Textbook references