Finding a homomorphism between groups with a given kernel What is a homomorphism defined on the group of invertible upper-triangular $3\times 3$ matrices whose kernel consists of matrices $\begin{bmatrix} 1 & 0 & a \\ 0 & 1 & 0 \\ 0 &0 & 1\end{bmatrix}$?
I want to use this to study the quotient group, also is there always a way to find a group homomorphism, once the kernel is given and one the group is known in general ?
What changes when the diagonal entries in the group of upper triangular matrices are all equal to one .
I prefer a correct answer as much as easy to follow explanation, because my backgroud in algebra does not go beyond "first abstract algebra course"
Thank you.
 A: 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
A: I have also not done much beyond a first abstract algebra course, but this should be correct:
I am assuming you ask about homomorphisms to study quotient groups.
To answer "is there always a way to find a homomorphism given a kernel group and a known group"...when the given subgroup (kernel) N is a normal subgroup of G, the map π:G→G/N defined by π(x)=xN is a homomorphism with Kernel N.
When the given kernel is not normal then there is no factor group, as the operation of this group is not well-defined, as there will exist different representatives of at least one coset that, when "multiplied" with a representative of another coset, will end in different cosets!
For the next question: "what happens when all the diagonal entries are all equal to one"
one way to see what happens in the factor group is to see the details in proving the normality of the kernel.
let $$G = \begin{pmatrix}a&b&c\\0&d&e\\0&0&f\end{pmatrix} in \ the \ set \ of \ invertible \ upper \ trianglular \ matrices$$ 
Then by elementary methods we calculate its inverse: $$G^-1 = \begin{pmatrix}1/a&-b/ad&bc/af\\0&1/d&-e/fd\\0&0&1/f\end{pmatrix}$$
using matrix multiplication we check the normality of our kernel by checking if...
$$\begin{pmatrix}a&b&c\\0&d&e\\0&0&f\end{pmatrix}\begin{pmatrix}1&0&x\\0&1&0\\0&0&1\end{pmatrix} \begin{pmatrix}1/a&-b/ad&bc/af\\0&1/d&-e/fd\\0&0&1/f\end{pmatrix}$$
is also in the kernel. we get
$$\begin{pmatrix}1&0&(bcd-be+axd+dc)/fd\\0&1&0\\0&0&1\end{pmatrix}$$
which works in any field F and we are still in the kernel.
IF the entries of the diagonal are all one then the matrix multiplication above becomes 
$$\begin{pmatrix}1&0&bc-be+x+c\\0&1&0\\0&0&1\end{pmatrix}$$
let me point out that this modified version of the group makes it so that the entries of the elements of the cosets never have to use the "division" operation of the field as the initial matrices above.
The cosets using the original group are the cosets that contain the multiplication of an arbitraray representative of the kernel and of G or 
$$\begin{pmatrix}a&b&c\\0&d&e\\0&0&f\end{pmatrix}\begin{pmatrix}1&0&x\\0&1&0\\0&0&1\end{pmatrix} = \begin{pmatrix}a&b&ax+c\\0&d&e\\0&0&f\end{pmatrix}$$
and of course if the entries of the diagonals are all one then it turns into
$$\begin{pmatrix}1&b&c\\0&1&e\\0&0&1\end{pmatrix}\begin{pmatrix}1&0&x\\0&1&0\\0&0&1\end{pmatrix} = \begin{pmatrix}1&b&x+c\\0&1&e\\0&0&1\end{pmatrix}$$
so when you restrict the group of invertible 3x3 matrices to the diagonals being linear, all cosets are of size 3|F| (if F is finite) as we can manipulate the matrixes so that the x+c can be ANY element in the field F. In the original group, we have situations where this cannot happen, such as when a=2 and c=2 and ax+c can never be odd. 
AS TO THE ORIGINAL QUOTIENT GROUP: it is defined as the answer above.
when all the diagonals are 1 then for coset multiplication we get
$$\begin{pmatrix}1&y&?\\0&1&u\\0&0&1\end{pmatrix}\begin{pmatrix}1&y'&?\\0&1&u'\\0&0&1\end{pmatrix}=\begin{pmatrix}1&y'+y'&?\\0&1&u'+u\\0&0&1\end{pmatrix}$$
assuming a finite field F: 
in our modified group (all diagonal entries are one)
we get bigger cosets as a result of making Our group a bit more restricted, and the above shows how.
