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

## 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

This proof uses a tabular format for presentation. Provide feedback on tabular proof formats in a survey (opens in new window/tab) | Learn more about tabular proof formats|View all pages on facts with proofs in tabular format

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]
2 $Z \le O_\pi(C)$ Step (2) [SHOW MORE]
3 $O_\pi(C)$ is a normal $\pi$-subgroup of $G$ Facts (2), (4) [SHOW MORE]
4 $O_\pi(C) \le C \cap H = Z$ [SHOW MORE]
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]
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]
8 If $C$ is strictly bigger than $Z$, $M$ is nontrivial normal Hall in $K$ Fact (7) Steps (6), (7) [SHOW MORE]
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]
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.