Constrained for a prime divisor implies not simple non-abelian

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.

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:

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

Facts used

 * 1) uses::Prime power order implies not centerless

Proof
We prove the contrapositive, which is somewhat easier.

Given: A finite simple non-abelian group $$G$$

To prove: $$G$$ is not $$p$$-constrained for any prime divisor $$p$$ of the order of $$G$$.

Proof: Since $$p$$ divides the order of $$G$$, we obtain that $$O_{p',p}(G)$$ is trivial, and hence $$P \cap O_{p',p}(G)$$ is trivial for any $$p$$-Sylow subgroup $$P$$ of $$G$$. We thus get:

$$C_G(P \cap O_{p',p}(G)) = G$$

Since $$O_{p',p}(G)$$ is trivial, we see that the $$p$$-constraint condition is violated.