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 $H\leq N$. 

It is clear that $H\le HN\le G$ and $[G:H]=[G:HN][HN:H]$so 
$$[HN:H] \mid [G:H]\quad(*)$$
Define $\alpha:\{n(N∩H) | n\in N\} → \{aH | a\in NH\}$ by 
$\alpha(n(H∩N))=nH$. Then $\alpha$ is an isomorphism and so $[HN:H]=[N:N∩H]$, Therefore ($|N|$ is finite)
$$[HN:H]=\frac{|N|}{|N∩H|} \mid |N|\quad(**)$$
$(*),(**)\implies [HN:H] \mid ([G:N],|N|)=1 \implies [HN:H]=1 ⇒ H\le N$.
