Finding a homomorphism between groups with a given kernel
You might imagine it as the group of upper triangular matrices with the top-right entry marked as unknown or irrelevant, $$\begin{pmatrix}x&y&?\\0&z&u\\0&0&v\end{pmatrix} $$ This is fine because when multiplying two such matrices, the top-right entries are needed only for computing the top-right entry: $$\begin{pmatrix}x&y&?\\0&z&u\\0&0&v\end{pmatrix}\begin{pmatrix}x'&y'&?\\0&z'&u'\\0&0&v'\end{pmatrix}=\begin{pmatrix}xx'&xy'+yz'&?\\0&zz'&zu'+uv'\\0&0&vv'\end{pmatrix} $$
By removing all decoration, this becomes a group structure on the set $$G:=\{\,(x,y,z,u,v,D)\mid yzvD-1=0\,\}$$ and with multiplication rule $$(x,y,z,u,v,D)\cdot(x',y',z',u',v',D')=(xx',xy'+yz',zz',zu'+uv',vv',DD').$$ But I suppose that this explicit rule looks a bit unintuitive, compared to the matrix with irrelevant entry.
As a sidenote: Because the set $G$ and the group operation are defined in terms of polynomials, this is an algebraic group