Normal subgroup

Question

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 $H\leq N$.

Answer 1

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 $|H|$.

Answer 2

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)

$$[HN:H]=\frac{|N|}{|N∩H|} \mid |N|\quad(**)$$

$(*),(**)\implies [HN:H] \mid ([G:N],|N|)=1 \implies [HN:H]=1 ⇒ H\le N$.