Let $G$ be a group of order $56$. (We do NOT assume the Sylow-$7$ subgroup to be normal.) Then either the Sylow-$2$ subgroup is normal or the Sylow-$7$ subgroup is normal. How to prove?
My idea: consider the case $n_2=7,n_7=8$. Then the number of elements in all Sylow $2$ -subgroups and Sylow $7$-subgroups is at most exactly $(7-1)*8+(2-1)*7+1=56$. This cannot give a contradiction. We need to prove that $G$ is simple. Then $|G||(8!/2)$. To prove $G$ simple, we need to prove there does not exist normal subgroups of order $2$ and order $4$. But I do not know how to prove.