Solvable implies Fitting subgroup is self-centralizing

This article gives the statement, and possibly proof, of the fact that in any solvable group, the subgroup obtained by applying a given subgroup-defining function (i.e., Fitting subgroup) always satisfies a particular subgroup property (i.e., self-centralizing subgroup)
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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 $C_{G}(F(G))$ 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:

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, $B\leq C$, 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)