Pi-separable and pi'-core-free implies pi-core is self-centralizing

Statement
Suppose $$\pi$$ is a set of primes and $$G$$ is a finite group that is separable for the prime set $$\pi$$. Further, suppose the $$\pi'$$-core of $$G$$, namely $$O_{\pi'}(G)$$, is trivial. Then, the $$\pi$$-core of $$G$$, namely $$O_\pi(G)$$, is a self-centralizing subgroup of $$G$$:

$$\! C_G(O_\pi(G)) \le O_\pi(G)$$.

Facts with similar proofs

 * Solvable implies Fitting subgroup is self-centralizing
 * Hall and central factor implies direct factor

Applications

 * p-solvable implies p-constrained

Facts used

 * 1) uses::Pi-separability is subgroup-closed
 * 2) uses::Characteristicity is centralizer-closed
 * 3) uses::Normality satisfies transfer condition
 * 4) uses::Characteristicity is transitive + uses::Characteristic implies normal
 * 5) uses::Normal Hall implies permutably complemented: Note that this only uses the case where the normal Hall subgroup is abelian, which does not require the odd-order theorem.
 * 6) uses::Normality satisfies intermediate subgroup condition
 * 7) uses::Cocentral implies normal
 * 8) uses::Equivalence of definitions of normal Hall subgroup: A normal Hall subgroup is the same thing as a characteristic Hall subgroup.

Proof
Given: A prime set $$\pi$$, a $$\pi$$-separable group $$G$$ such that $$O_{\pi'}(G)$$ is trivial; in other words, $$G$$ has no nontrivial normal $$\pi'$$-subgroup.

To prove: $$C_G(O_{\pi}(G)) \le O_{\pi}(G)$$.

Proof: Let $$H = O_\pi(G)$$ and $$C = C_G(H)$$. Let $$Z = Z(H)$$. By definition $$Z = C \cap H$$.