Mapping to Group that is multiplying two groups The context, I was trying to show 
$$H\cong \frac{HN}{N}$$
when $H\le G, N\lhd G,$ $H\cap N=\{e\}$
But i got stuck, so I had to check the argument and I was given at the first line
that
when setting $HN=\{hn|h\in H, n\in N\}$
There is surjective homomorphism $\varphi : H \to HN, \varphi (h)=hN$
Which seems doesn't make sense to me about 2 things.


*

*$\varphi : H \to HN$ However,  $\varphi (h)=hN$ which is -I think- form of function that $H \to\ H/N$, the quotient group mapping function.
HN has element that is multiplication of each element from H and N.
However $\varphi (h)=hN$ maps h to hN which is form of coset. 
For me this set of function $\varphi$ doesnt fit for given situation.

*Given argument state that $\varphi$is surjective. But Unless N is singleton group, How does the mapping 
$\varphi : H \to HN$ can be surjective? Afterall $|H|\lt |HN|$
I think the problem here is that My knowledge about quotient group mapping notation is wrong such as $\phi (g) = gN$ Doesnt always have to indicate mapping to quotient group maybe?...
Im very beginning at the learing groups pls help me LV.100 mathematicians super society...
 A: Note: $hN$ is not an element of $HN$, which is equal to
$$HN = \{hn\mid h\in H, n\in N\}.$$
So your proposed map does not even make sense (and is certainly not a surjective homomorphism).
Better: let’s define a map $f\colon HN\to H$, and then prove that it is a homomorphism, is onto, and the kernel is $N$.
What map? Well, let $f(hn) = h$.
First, we show it is well-defined: if $h_1n_1 = h_2n_2$, then $h_2^{-1}h_1 = n_2n_1^{-1}$. As the left hand side is in $H$ and the right hand side is in $N$, and $H\cap N = \{e\}$, then $h_2^{-1}h_1 = n_2n_1^{-1} = e$, so $h_1=h_2$, $n_1=n_2$; that is, the expression of elements of $HN$ as a product of something in $H$ and something in $N$ is unique, so $f$ is well-defined.
Second, we show that $f$ is onto: this is easy, since given $h\in H$, we know that $he \in HN$, and $f(he) = h$ by definition. So $f$ is onto.
Thirdly, we show that $f$ is a homomorphism. To show this, we let $h_1,h_2\in H$, $n_1,n_2\in N$. We need to figure out how to write $(h_1n_1)(h_2n_2)$ as an element of $HN$, so we know what is the image of the product. We have
$$\begin{align*}
(h_1n_1)(h_2n_2) &= h_1(h_2h_2^{-1})n_1h_2n_2\\
&= h_1h_2 (h_2^{-1}n_1h_2)n_2.
\end{align*}$$
Now, since $N$ is normal, $h_2^{-1}n_1h_2=n_3$ for some $n_3\in N$, so $(h_1n_1)(h_2n_2) = (h_1h_2)(n_3n_2)$.
Can you take it from here?
