First of all, let me describe the problem that I try to solve:
Let $p$ be a prime, $G$ a finite group, $P$ a $p$-Sylow subgroup of $G$, and $L$ a set of all elements of $G$ which its order is coprime to $p$. Prove that if $K$ is a normal subgroup of $G$ with $\lvert K \rvert$ is coprime to $p$ and $G = PK$ then $K = L$.
I already have proved that $K \subset L$. (For any element $k$ of $K$, consider $\langle k \rangle$ and use Lagrange's theorem and the hypothesis that $\lvert K \rvert$ is coprime to $p$.)
Then I try to show $K \supset L$, but I cannot. Could you help me? The following is what I tried:
Let $z$ be a element of $L$. From $L \subset G = PK$, it can be written of the form $z = xy$ for some $x \in P$ and $y \in K$. Put $q$ be the highest power of $p$ dividing $\lvert G \rvert$. Since $K$ is normal and $P$ is $p$-Sylow subgroup, $$ z^q = (xy)^q = x^q k = k \in K $$ for some $k \in K$. Hence, at least, $z^q \in K$. But from here, how can I show $z \in K$? If my English is bad or mathematical description is unclear then tell me; I'll fix it. Thank you.