quotient group of matrix group 
In the general linear group $GL_3(\mathbb R)$, consider the subsets
$$H = \begin{bmatrix} 1 & * &*  \\ 0 & 1 & * \\ 0 & 0 & 1\end{bmatrix}$$
$$K = \begin{bmatrix} 1 & 0 &*  \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{bmatrix}$$
where ∗ represents an arbitrary real number.  Identify the quotient group $H/K$.

The solution should be this $$
\begin{bmatrix}
1 & b & 0 \\
0 & 1 & c \\
0 & 0 & 1 \\
\end{bmatrix},
$$
but I don't know how to arrive at this. I know that quotient group is the set of all cosets of $H$, so is should consists of all products of matrices $HK$ but in the solution I've found in order to find $H/K$ the book consider when 2 matrices $A,B$ $\in H$  such that $AK=BK$ with $K \in K$. Why? Is there any other simple method ?
 A: First of all you have to ask yourself if it is possibile that $H/K$ is a group, i.e if $K$ is normal in $H$.
In your case (please verify this fact) you have that $K$ is an abelian group (in particular is isomorphic to $(\mathbb{R},+,0)$ ) and is contained in the center of $H$ (I.e $AK=KA$ for each $A \in H$, and $K\in K$), so that $K$ is clearly normal in $H$.
Thus The quotient set $H/K=\{AK: A \in H\}$ inherits also a group  structure.
Now we have to understand what is the multiplication in $H$ to compute easily the quotient set $H/K$.
Taking an element $A\in H$, I will denote $A$ as $A(a,b,c)$, where $A_{12}=a, A_{13}=b$ and $A_{23}=c$.
Then taking two matrices $A=A(a,b,c)$ and $B=B(a’,b’,c’)$, you have
$AB=AB(a+a’, b+ac’ +b’, c+c’)=BA$
and so if you have a general element $A=A(a,b,c)$, you can observe that $A’(a,0,c) $ is an element in its class on $H/K$, in fact
$A(a,b,c)=A’(a, 0, c)K(0, b+ac’+b’,  0) $
So that you can define the section map $s: H/K \to G$ Sending each class $AK$ to $A’(a,0,c)$, where $(G:=\{ A(a,0,c) : a,c \in \mathbb{R}\},+,0)$ is an additive group.
You can prove that this map is an isomorphism of groups and observing that $A(a,0,c)+B(a’,0,c’)=(A+B)(a+a’, 0, c+c’)$
then you can say that $H/K \cong (\mathbb{R}^2,+,0)$
