Let $G = N_1 \times N_2 \times \ldots \times N_n$ be a finite group and the direct product of the normal subgroups $N_i$. Let $\pi$ be a set of primes, then a $\pi$-complement is a subgroup $K$ such that $|K|$ is only divisible by primes not in $\pi$ and the index of $K$ in $G$ is only divisible by primes from $\pi$.
Now if $K$ is a $\pi$-complement of $G$, then $$ K = K_1 \times K_2 \times \cdots \times K_n $$ with $\pi$-complement $K_i$ of $N_i$, and vice versa if $K_i$ are $\pi$-complements of $N_i$, then $K = K_1 \times K_2 \times \cdots \times K_n$ is a $\pi$-complement of $G$.
Does the above result holds? I nowhere found this result, so I am unsure, also how would you going to prove it?
I have a proof of the following related result: If $G = N_1 \times N_2 \times \ldots \times N_n$ is a finite group and the direct product of the normal subgroups $N_i$, then every Sylow $p$-subgroup of $G$ has the form $$ S_1 \times S_2 \times \ldots \times S_n $$ for $S_i \in \mbox{Syl}_p(N_i)$ and vice versa for $S_i \in \mbox{Syl}_p(N_i)$ we have $S := S_1 S_2 \cdots S_n = S_1 \times S_2 \times \ldots \times S_n$ and $S \in \mbox{Syl}_p(G)$.