Show that $\varphi:KN\rightarrow K$ is a homomorphism I am having trouble with a particular mapping, determining whether or not it is a homomorphism.  
I know that $K,N$ are normal subgroups of $G$, $K\cap N=\{e\},$ and $KN=NK$ where
$$KN=\{kn|k\in K, n\in N\}$$
I know the implication above show that $kn$ does not necessarily equal $nk$, but that
$$kn=n'k'$$
for some $n'\in N, k'\in K$.
The mapping suggested was 
$$\varphi:KN\rightarrow K$$
$$kn\mapsto k$$
But when I test whether or not it is a homomorphism, i get stuck here;
let $k_1n_1, k_2n_2\in KN$.  Then
$$\varphi(k_1n_1k_2n_2)=\varphi(k_1k_2'n_1'n_2)=k_1k_2'=\varphi(k_1n_1)\varphi(k_2'n_2')$$
but I need $\varphi(k_2n_2),$ not $\varphi(k_2'n_2')$...any help here would be great...
 A: In fact, $kn$ does equal $nk$ for all $k\in K$ and $n \in N$. This is not implied by $KN = NK$ alone, but with the additional information that $K \cap N = 1$ and $K \lhd G$ and $N \lhd G$, it is true.
To see this, examine the element $x = knk^{-1}n^{-1}$. Notice that $knk^{-1} \in N$ since $N \lhd G$. Therefore, $x \in N$. Also, notice that $nk^{-1}n^{-1} \in K$ since $K \lhd G$. Therefore, $x \in K$. This means that $x \in N \cap K = 1$, so $x = 1$, which means that $knk^{-1}n^{-1} = 1$, so $kn = nk$.

Another way to think about it, which you will probably learn soon if you continue studying group theory: when you have $G = KN$ and $K$ and $N$ are both normal subgroups with $K \cap N = 1$, we call this a direct product. In this case, it turns out that $G$ is isomorphic to the group $K \times N$, which is the set of ordered pairs of the form $(k,n)$ where $k\in K$ and $n \in N$, with multiplication defined "coordinatewise": $(k_1,n_1)(k_2,n_2) = (k_1 k_2, n_1 n_2)$.

In this form it is obvious that elements of $K$ and $N$ commute, because $K$ is isomorphic to the set of elements of the form $(k,1)$ and $N$ is isomorphic to the set of elements of the form $(1,n)$. Then $(k,1)(1,n) = (k,n) = (1,n)(k,1)$.
