Finite group of order $mn$ with $\gcd(m,n) = 1$ . Let $G$ be a finite group of order $mn$ with $\gcd(m,n) = 1$ and $H$ be a normal subgroup of order $m$. Prove that it is the only subgroup of order $m$.
This is somewhat different from this question: Finite group of order $mn$ with $m,n$ coprime and I don't think we are allowed to use homomorphisms(it is in the next section)
I let $K$ be a subgroup of order $m$. If $K = H$, I am done.
Otherwise, I supposed $K \neq H$. Based on previous results, I know that $HK = \{hk | h\in H, k \in H\}$ would be another subgroup of $G$ if $H$ is normal. We would have that $\exists k \in K$ such that $k \notin H$ or $\exists h \in H$ such that $h \notin K$. In either case, $|HK| > m$ which may cause problems with some theorems of number theory and Lagrange's theorem but I am unsure of the specifics.
I don't know if I can say more about $|HK|$ or whether I am even on the right track. Thoughts?
 A: Let $G$ be a group and $1_G$ denotes the identity of this group.
Fact 1. Is $H$ is a normal subgroup of $G$ and $[G:H]=n$, then we have $$x^m\in H$$ for every $x\in G$.
Fact 2. If $K$ is a subgroup such that $|K|=n$, then we have $$x^n=1_G$$ for each $x\in K$.

Now let us assume that $K$ is any subgroup of order $m$ in the situation described in the question. From $\gcd(m,n)=1$ we get existence of integers $u$ and $v$ such that $$mu+nv=1$$ and thus, for any $x\in K$, we get $$x=x^{mu+nv}=(x^m)^u\cdot (x^n)^v = (x^n)^v \in H$$
using $x^m=1$ and $x^n\in H$.
So we see that $K\subseteq H$ and together with $|K|=|H|$ this implies $K=H$.

Fact 2 is consequence of Lagrange's theorem applied to the group $K$. (We are basically saying that if $x\in K$, then order of $x$ divides $|K|$.|
The proof of Fact 1 can be found, for example, here: Let $H$ be a normal subgroup of index $n$ in a group $G$. Show that for all $g \in G, g^n \in H$ and Prove that if a normal subgroup $H$ of $ G$ has index $n$, then $g^n \in H$ for all $g \in G$.
