# Constrained for a prime divisor implies not simple non-abelian

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## Statement

Suppose $G$ is a finite group and $p$ is prime divisor of the order of $G$. Then, if $G$ is a p-constrained group, $G$ cannot be a simple non-abelian group.

## Definitions used

### p-constrained group

Further information: P-constrained group (?)

Suppose $G$ is a finite group and $P$ is a $p$-Sylow subgroup. We say that $G$ is $p$-constrained if we have:

$C_G(P) \le O_{p',p}(G)$.

## Facts used

1. Prime power order implies not centerless

## Proof

Given: A finite group $G$, a prime divisor $p$ of the order of $G$. $G$ is $p$-constrained.

To prove: $G$ is not a simple non-abelian group.

Proof: Since $p$ divides the order of $G$, there exists a nontrivial $p$-Sylow subgroup $P$ of $G$. By fact (1), $Z(P)$ is nontrivial, so the centralizer $C_G(P)$, which contains $Z(P)$ is nontrivial. Thus, by the definition of $p$-constrained, we have that $O_{p',p}(G)$ is nontrivial. Thus, it is a nontrivial normal subgroup of $G$, so we have $O_{p',p}(G) = G$.

Consider $N = O_{p'}(G)$. If $N$ is trivial, then $G = O_{p',p}(G)/O_{p'}(G)$ is a $p$-group, and hence by fact (1), $Z(G)$ is a nontrivial normal subgroup, forcing $Z(G) = G$, forcing $G$ to be abelian.

On the other hand, if $O_{p'}(G) = G$, then $p$ does not divide the order of $G$, a contradiction.