Normal subgroup Suppose $N$ is a normal subgroup of $G$ and $H$ is a subgroup of $G$ that $|N|,[G:H]$ are finite and $(|N| , [G : H] ) =1$. Prove that $H\leq N$. 

Hint: $$|HN|=\frac{|H|\cdot|N|}{|H \cap N|}    $$ and $[HN:N]$ divides $[G:N]$. But $[HN:N]=[H:H \cap N]$, which in its turn divides $|H|$.
