quotient group of matrix group

Solution 1:

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 $H/K \cong (\mathbb{R}^2,+,0)$