Solvable implies Fitting subgroup is self-centralizing

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

Facts used

 * 1) uses::Characteristicity is centralizer-closed
 * 2) uses::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 $$C_G(F(G))$$ denotes its centralizer in $$G$$

To prove: $$C_G(F(G)) \le F(G)$$

Proof: Let $$C = C_G(F(G))$$ and $$H = C \cap F(G)$$. Note that every element of $$H$$ commutes with every element of $$C$$, so $$H \le Z(C)$$, 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, $$B \le G$$ is a subgroup with the property that $$[B,B] \le H$$. 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, $$C = C_G(F(G))$$ 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, $$B \le C$$, 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.

Textbook references

 * , Page 218, Theorem 1.3 (Section 6.1)