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 $\pi'$ -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 $G$ :

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

Related facts

Facts with similar proofs

Applications

p-solvable implies p-constrained

Facts used

Pi-separability is subgroup-closed
Characteristicity is centralizer-closed
Normality satisfies transfer condition
Characteristicity is transitive + Characteristic implies normal
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.
Normality satisfies intermediate subgroup condition
Cocentral implies normal
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 $Z = C \cap H$ .

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 $K = O_{\pi,\pi'}(C)$ 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 $O_{\pi}(C) = O_{\pi,\pi'}(C)$ , and so the series terminates at $Z = O_{\pi'}(C)$ , 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 $\pi'$ -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 $\pi'$ -Hall subgroup of $K$ , so $M$ is normal in $K$ .
9	If $C$ is strictly bigger than $Z$ , $M$ is a nontrivial normal $\pi'$ -subgroup of $G$ , i.e., $M \le O_{\pi'}(G)$	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 $K = O_{\pi,\pi'}(C)$ ) 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 $O_{\pi'}(G)$ is trivial. Thus, $C = Z$ and in particular we get $C \le H$ , as desired.		$G$ is $\pi'$ -core-free.	Steps (1), (9)	Step (9) yields a nontrivial normal $\pi'$ -subgroup of $G$ , so $O_{\pi'}(G)$ is trivial.