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.
In the proof of the following corollary I do not understand why we need to suppose that $H$ is a Hall subgroup, further I think $\pi$-complement to a group $H$ means an complement such that this complement is a $\pi$-group? (I nowhere found a definition, but I guess it means this). So now to the corollary:
Corollary: Suppose that $H$ is a nilpotent Hall subgroup of $G$ and that $H$ is $S$-semipermutable. Then $H^G$ is solvable.
Proof: Let $\pi$ be the set of prime divisors of $|H|$. Then $H^G$ has a nilpotent $\pi$-complement $K$, and thus $H^G = HK$. Since both $H$ and $K$ are nilpotent, the solvability of $H^G$ follows by the Wieland-Kegel Theorem. $\square$