Let $P$ and $P'$ be Sylow $p$-subgroups of a finite group $G$. Prove that if $P'\subset N_G(P)$ then $P'=P$.
Proof: Since $P\vartriangleleft N_G(P)\leq G$ and it's easy to check that $P$ is also Sylow $p$-subgroup in $N_G(P)$. Because it's normal then the number of Sylow $p$-subgroups of $N_G(P)$ is equal to one. By condition $P'\subset N_G(P)$ then $P'$ is also Sylow $p$-subgroup of $N_G(P)$. Hence $P'=P$.
I have spent a whole day in order to solve this easy problem. However, it became easy to me after solution. The key point is that the $P$ is also Sylow $p$-subgroup in $N_G(P)$.
Is my solution correct?
Thanks in advance for evaluation.