Let $H$ be a subgroup of group $G$ and $K$ be a normal subgroup of $G$ such that $\gcd([G:H], |K|) = 1$. Prove that $K \subseteq H$ [closed]

I am trying to solve this:

Let $H$ be a subgroup of group $G$ and $K$ be a normal subgroup of $G$ such that $\gcd([G : H], |K|) = 1$. Prove that $K \subseteq H$

I tried to use Lagrange's theorem. But here $G$ may be an infinite set, I don't kown how to deal with that. How can I solve this problem？

Since $K$ is normal, $HK$ is a subgroup and $H \subseteq HK \subseteq G$, so $|HK:H|=|K:H \cap K|$, divides $|G:H|$ and $|K|$. Since these last numbers are relatively prime, $H \cap K=K$, that is $K \subseteq H$.

If $G$ is infinite, the following proof avoids the Second Isomorphism Theorem. If $X$ is a group and $Y \leq X$ a subgroup, denote by $Y\backslash X$ the set of left cosets of $Y$ in $X$. Define the map $f: H\backslash KH \longrightarrow (H \cap K) \backslash K$ by $f(kH)=k(H \cap K)$ for $k \in K$. Then this map is well defined: if $kH=k'H$, then $(k')^{-1}k \in H$, but obviously $(k')^{-1}k \in K$, whence $(k')^{-1}k \in H \cap K$ and thus $f(kH)=k(H \cap K)=k'(H \cap K)=f(k'H)$. So the map $f$ does not depend on the coset representatives.

The map is obviously surjective. Finally let's prove that it is also injective: assume $f(kH)=f(k'H)$, that is, $k(H \cap K)=k'(H \cap K)$. It follows that $(k')^{-1}k \in H \cap K$, in particular $(k')^{-1}k \in H$. This implies that $kH=k'H$, hence $f$ is injective.

So the map gives a bijection between the left cosets of $H$ in $KH$ and the left cosets of $H \cap K$ in $K$. Hence $|KH:H|=|K:H \cap K|$ and these numbers where assumed to be finite (see the condition on the gcd of the indices). Now $|KH:H| \mid |G:H|$ and $|K:H \cap K| \mid |K|$, so, since gcd$(|G:H|,|K|)=1$, we must have $|KH:H|=1$, that is, $KH=H$, equivalent to $K \subseteq H$, and $K=H \cap K$, which amounts to the same thing.