Let $G$ be a group, a few definitions:
1) If $H \le G$, then a complement of $H$ in $G$ is a subgroup $K \le G$ such that $$ G = HK \quad \mbox{and} \quad H \cap K = \{ 1 \}. $$
2) Let $p$ a prime number, then a $p$-complement is a subgroup $K \le G$ such that its index $|G : K|$ is the highest $p$-power dividing $|G|$.
3) Let $\pi$ be a set of primes, a subgroup $K \le G$ is called a $\pi$-complement if $|K|$ is not divisible by any prime in $\pi$, and every prime divisor of its index $|G : K|$ lies in $\pi$.
Definition 1) seems quite natural to me, here complement is defined with respect to $G$ and a subgroup $H$, in the other cases I am wondering if there are also subgroups such that the complement relates to them like in 1).
For example for 2) if I have a $p$-complement, then there exist a Sylow-$p$-subgroup $P$ such that $P \cap K = \{ 1 \}$ (by order arguments) and also we must have $KP = G$ by considering the size of $KP$.
But if I have a Sylow-$p$-subgroup, then there does not need to exists a subgroup $K \le G$ such that $P \cap K = \{1\}$ and $PK = G$. So I think the notion of $p$-complement generalises the notion of complement with respect to a subgroup, right? Because every $p$-complement is also a complement in the sense of 1) with respect to some Sylow-$p$-subgroup, which is guaranteed by the Sylow-Theorems.
For 3), this obviously generalises 2). But under what circumstances can we find to a $\pi$-complement $K$ in $G$ a subgroup $H \le G$ such that 1) is fullfilled, namely such that $$ G = HK \quad \mbox{and} \quad H \cap K = \{ 1 \}? $$
I know one case, if $K$ is normal then by Schur-Zassenhaus we can find such an $H$, but guess there could be anything more said about that.