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

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

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

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

Related facts

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. Pi-separability is subgroup-closed
2. Characteristicity is centralizer-closed
3. Normality satisfies transfer condition
4. Characteristicity is transitive + Characteristic implies normal
5. 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. Normality satisfies intermediate subgroup condition
7. Cocentral implies normal
8. Equivalence of definitions of normal Hall subgroup: A normal Hall subgroup is the same thing as a characteristic Hall subgroup.

Proof

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Given: A prime set $\pi$, a $\pi$-separable group $G$ such that $O_{{\pi }^{\prime }}(G)$ is trivial; in other words, $G$ has no nontrivial normal ${\pi }^{\prime }$-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 $Z=C\cap H$.

Step no.	Assertion/construction	Facts used	Given data used	Previous steps used	Explanation
1	$Z$ is normal in $C$	Fact (3)			[SHOW MORE] $H=O_{\pi }(G)$ is normal in $G$ so $Z=C\cap H$ is normal in $C$.
2	$Z\le O_{\pi }(C)$			Step (2)	[SHOW MORE] Since $Z\le H$, $Z$ is a $\pi$-subgroup of $C$. The previous step tells us that $Z$ is a normal $\pi$-subgroup of $C$, so $Z\le O_{\pi }(C)$.
3	$O_{\pi }(C)$ is a normal $\pi$-subgroup of $G$	Facts (2), (4)			[SHOW MORE] Since $H$ is characteristic, so is $C$ (by Fact (2)). Further, $O_{\pi }(C)$ is characteristic in $C$. Fact (4) tells us that $O_{\pi }(C)$ is normal in $G$.
4	$O_{\pi }(C)\le C\cap H=Z$				[SHOW MORE] Since $O_{\pi }(C)$ is a normal $\pi$-subgroup of $G$, $O_{\pi }(C)\le O_{\pi }(G)=H$. Also, $O_{\pi }(C)\le C$ by definition, so $O_{\pi }(C)\le C\cap H=Z$.
5	$Z=O_{\pi }(C)$:			Steps (2), (4)	Step-combination direct.
6	If $C$ is strictly bigger than $Z$, then $K=O_{\pi ,{\pi }^{\prime }}(C)$ is strictly bigger than $Z$	Fact (1)	$G$ is $\pi$-separable.	Step (5)	[SHOW MORE] Since $G$ is $\pi$-separable, so is $C$. Thus, the upper $\pi$-series of $C$ terminates at $C$. If $Z=K$, then we get $O_{\pi }(C)=O_{\pi ,{\pi }^{\prime }}(C)$, and so the series terminates at $Z=O_{{\pi }^{\prime }}(C)$, a contradiction to $C$ being strictly bigger than $Z$.
7	If $C$ is strictly bigger than $Z$, there exists a nontrivial complement, say $M$, to $Z$ in $K$	Facts (5), (6)		Steps (1), (6)	[SHOW MORE] $Z$ is a normal subgroup of $C$, and hence, of $K$ (fact (6)). Moreover, since $K/Z$ is a ${\pi }^{\prime }$-group, $Z$ is normal Hall in $K$. Thus, fact (5) yields that there exists a permutable complement $M$ to $Z$ in $K$.
8	If $C$ is strictly bigger than $Z$, $M$ is nontrivial normal Hall in $K$	Fact (7)		Steps (6), (7)	[SHOW MORE] To see this, note that $Z$ by definition centralizes $C$, so $Z$ is in the center of $K$. Hence, $M$ is cocentral, so fact (7) yields that $M$ is normal in $K$. By construction, $M$ is a ${\pi }^{\prime }$-Hall subgroup of $K$, so $M$ is normal in $K$.
9	If $C$ is strictly bigger than $Z$, $M$ is a nontrivial normal ${\pi }^{\prime }$-subgroup of $G$, i.e., $M\le O_{{\pi }^{\prime }}(G)$	Facts (2), (4), (8)		Steps (6), (7), (8)	[SHOW MORE] By fact (8), $M$ is characteristic in $K$, and $K$ in turn is characteristic in $C$ (since $K=O_{\pi ,{\pi }^{\prime }}(C)$) which in turn is characteristic in $G$, because it is the centralizer of the characteristic subgroup $H$ (fact (2)). By fact (4), a characteristic subgroup of a characteristic subgroup of a characteristic subgroup is characteristic, so $M$ is characteristic, and in particular, normal.
10	If $C$ is strictly bigger than $Z$, we obtain the required contradiction to the assumption that $O_{{\pi }^{\prime }}(G)$ is trivial. Thus, $C=Z$ and in particular we get $C\le H$, as desired.		$G$ is ${\pi }^{\prime }$-core-free.	Steps (1), (9)	Step (9) yields a nontrivial normal ${\pi }^{\prime }$-subgroup of $G$, so $O_{{\pi }^{\prime }}(G)$ is trivial.

References

Textbook references

• Finite Groups by Daniel Gorenstein, ISBN 0821843427, Page 228, Theorem 3.2, Section 6.3 (pi-separable and pi-solvable groups), ^{More info}