Question about normal subgroup and relatively prime index Suppose $G$ is finite, $K$ is a normal subgroup in $G$, $H$ is a subgroup of $G$, and $|K|$ is relatively prime to $[G:H]$. Show that $K$ is a subgroup of $H$.
I don't know where to begin...
 A: I'll prove a more general result:

If $K\triangleleft G$, $|K|$ finite, $H\leq G$, $[G:H]$ finite and $|K|$, $[G:H]$  are relatively prime then $H\leq K$

We need to prove the next Lemma for a very elemental proof.

Lemma: If $H$ and $K$ are subgroups of a group $G$, then $[H:H\cap K]\leq [G:K]$. If $[G:K]$ is finite then $[H:H\cap K]=[G:K]$ if and only if $G=KH$.

Proof: If $A$ is collection of all left cosets of $H\cap K$ in $H$ and $B$ is the collection of all left cosets of $K$ in $G$. The map $\phi:A\to B$ given by $\phi(h(H\cap K))=hK$ (remember that $h$ must be in $H$) is well defined since $h(H\cap K)=h'(H\cap K)$ implies $h'h^{-1}\in (H\cap K)$ in particular $h'h^{-1}\in K$, then $h'K=hK$. Map is inyective since $h'K=hK$ with $h\in H$ implies $h'h^{-1}\in K$ and $h'h^{-1}\in H$ therefore $h'h^{-1}\in H\cap K$ and $h(H\cap K)=h'(H\cap K)$. Then $[H:H\cap K]=|A|\leq|B|=[G:K]$. Let $[G:K]=n<\infty$ then $[H:H\cap K]=[G:K]$ if and only if $\phi$ is surjective if and only if $G=KH$ $\square$.
Since $K$ is normal, $KH$ is a subgroup. By Lemma we have $[K:H\cap K]=[HK:K]$ but $|K|$ is finite then $|H\cap K|$ so is and we know that $[K:H\cap K]|H\cap K|=|K|$ then we have that $[K:H\cap K]$ is finite and $[G:H]=[G:HK][HK:H]$.
Note that $[HK:H]=[K: H\cap K]$ (second isomorphism theorem) and both are finite. Clearly $[HK:H]$ divides $[G:H]$ and $[K: H\cap K]$ divides $|K|$ but are relatively primes then $[K: H\cap K]=1$ implies $K\leq H$.
A: You don't need $K$ to be normal, you just need $HK$ to be a subgroup (which it is when $K$ is normal). Note that $K\subseteq H$ if and only if $H\cap K=K$. 
In general, we know that $|HK||H\cap K| = |H||K|$. Therefore, if all quantities are finite, we have
$$\frac{|HK|}{|H|} = \frac{|K|}{|H\cap K|}.$$
If $HK$ is a subgroup and $G$ is finite, then $|HK|$ divides $|G|$, so
$$\frac{|K|}{|H\cap K|} = \frac{|HK|}{|H|} \text{ divides }\frac{|G|}{|H|} = [G:H].$$
From this you should be able to finish.
A: 1) Suppose $\,H\triangleleft G\,$ , and let us look at $\,kH\in G/H\,\,,\,\,k\in K\,$ , so one hand we have $\,\left(kH\right)^{|K|}=k^{|K|}H=H\,$, and OTOH we also have $\,\left(kH\right)^{[G:H]}=k^{[G:H]}H=H\, $ , so both numbers $\,|K|\,,\,[G:H]\,$ have to be multiples of this element's order, which can be only if its order is 1, and we're done.
2) If H is not normal in G, just take $\,\displaystyle{\cap_{g\in G}H^g=}\,$ the core or H in G, instead of H.
