Two subgroup $A, B \le G$ are said to permute if $AB = BA$. A subgroup $H \le G$ is called $S$-semipermutable in $G$ if $H$ permutes with every Sylow $q$-subgroup of $G$ for primes $q$ not dividing $|H|$.
Theorem: Let $H$ be an $S$-semipermutable $\pi$-subgroup of $G$. Then $H^G$ contains a nilpotent $\pi$-complement, and all $\pi$-complements in $H^G$ are conjugate. Also, if $\pi$ consists of a single prime, then $H^G$ is solvable.
Corollary: If a Sylow $p$-subgroup of $G$ is $S$-semipermutable, then $G$ is $p$-solvable.
Proof: Let $P$ be an $S$-semipermutable Sylow $p$-subgroup of $G$. Then $P^G$ is solvable by the above Theorem. Since $P^G$ is normal and has $p'$-index in $G$, it follows that $G$ is p-solvable. $\square$
Why does $P^G$ normal and $p'$-index implies it is $p$-solvable?