normal subgroup and index of subgroup I have a question about subgroups and its orders and indices. This is the exercise problem of Hungerford's Algebra.
If $N \unlhd G$, $|N|$ is finite, $H \leq G$, $[G:H]$ is finite, and $[G:H]$ and $|N|$ are relatively prime, then $N \leq H$.
First I think the complete lattice of this subgroups and consider the 2nd Isomorphism theorem $NH/N \cong H/N \cap H$. Then I get the result $|NH| = [NH:N]|N|$, and $|H| = [H:N \cap H]|N \cap H|$. 
But with only these result I cannot use the counting lemma because I cannot gurantee that the indices are finite. So I think I need an another apporach to solve this problem. Please show me the right way to do it.

Let me try.
Since $N$ is normal, $HN$ is subgroup of $G$.
Now, you have $x = [N: H\cap N] \leq [G:H]$, so $x$ is finite.
Moreover, you have $x = [N: H\cap N] = [HN : H]$.
On other hand, $([G:H],|N|) = 1$ or $([G:HN][HN:H],[N: H \cap N]|H \cap N|) = 1$, so you have $[HN:H] = [N : H\cap N] = 1$, then $N = H\cap N$, or $N \leq H$.
