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

Characteristicity is centralizer-closed
Characteristicity is intersection-closed
Members of the derived series of a group are characteristic subgroups (follows from the fact that characteristicity is commutator-closed)
Characteristicity is quotient-transitive
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: $C_{G} (F (G)) \leq F (G)$

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 $H \leq Z (C)$ , 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, $B \leq G$ is a subgroup with the property that $[B, B] \leq H$ . 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:

$F (G)$ is a characteristic subgroup
Since characteristicity is closed under taking centralizers, $C = C_{G} (F (G))$ is also characteristic in $G$
Since characteristicity is closed under intersections, $H$ is characteristic in $G$
The quotient $B / H$ is a characteristic subgroup of $C / H$ , being a member of the derived series
Hence, using the fact that characteristicity is quotient-transitive, $B$ is a characteristic subgroup of $C$
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, $B \leq C$ , 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.

References

Textbook references

Finite Groups by Daniel Gorenstein, ISBN 0821843427, ^{More info}, Page 218, Theorem 1.3 (Section 6.1)