Two subgroups with coprime orders 
Let $G$ be a group of order $mn$, where $m$ and $n$ have no common factor. If $G$ contains exactly one group of order $m$ and exactly one subgroup of order $n$, prove that $G$ is the direct product of $M$ and $N$.

In order to prove $G$ is the direct product of two subgroups, we need prove these two subgroups are both normal，Intersection is trivial and every element of $G$ is the product of two elements in $M$ and $N$.
If we know $M$ and $N$ are both normal, I can use lagrange theorem to prove their intersection is trivial. I don't know how to prove they are normal and $G=MN$.
 A: Hint: To show that they are normal, recall that conjugation preserves order of subgroups:
$$|gMg^{-1}| = |M|\quad\text{and}\quad |gNg^{-1}|=|N|.$$
To see that $G=MN$ it suffices to show that $|G|=|MN|$ (so in this case, $MN$ is a subgroup of $G$ of the same order). After showing that the intersection $M\cap N=\{e\}$, you can apply the following equality:
$$|MN| = \frac{|M| |N|}{|M\cap N|}.$$
A: Let $x\in M, y\in N$ Consider $f_x:G\rightarrow G$, defined by $f_x(g)=xgx^{-1}$, $f_x$ is an automorphism, this implies that the order of $f_x(N)$ is $n$ since $N$ is the unique subgroup of $G$ of cardinal $n$, we deduce that $f_x(N)=N$. Similarly, $f_y(M)=M$. Consider  $[x,y]=xyx^{-1}y^{-1}=(xyx^{-1})y^{-1}\in N=x(yx^{-1}y^{-1})\in M$, this implies that $xy=yx$ since $N\cap M=\{e\}$ since the order of $N\cap M$ divides $n$ and $m$.
This implies that $f:N\times M\rightarrow G$ defined by $f(x,y)=xy$ is well-defined as a homomorphism of group and is an isomorphism since its kernel is trivial: $f(x,y)=xy=e$ implies that $x=y^{-1}$ and $x\in N\cap M$ this implies that $x=y=e$.
