Isomorphism between Matrix and Field Let U be $\begin{pmatrix}1 & c \\ 0 & 1\end{pmatrix} c\in K$. Show that U is isomorphic to the group (K,+).
I know that I have to show that there is a bijective Grouphomomorphism $U \rightarrow K$. But I don't understand what we are mapping here exactly. Are we just summing up the entrys of the matrix?
 A: The point is that
$$ \pmatrix{1 & a\cr 0 & 1\cr} \pmatrix{1 & b\cr 0 & 1\cr} = \pmatrix{1 & a+b\cr 0 & 1\cr}$$
The mapping takes $\pmatrix{1 & c\cr 0 & 1\cr}$ to $c$.
A: I would like to add something to the answer given by @Robert Israel.
The set $U$ of matrices :
$$ U_c=\begin{pmatrix}1 & c \\ 0 & 1\end{pmatrix}$$
is a subgroup of $GL(2,\mathbb{K})$ (for multiplication, of course).
Mapping:
$$f: U \rightarrow \mathbb{K}, \ \  U_c \mapsto c$$
is a group isomorphism with the additive group $(\mathbb{K},+)$ because it is bijective and 
$$\tag{1}U_{c}U_{c'}=U_{c+c'}.$$
You may have noticed a similarity with the correspondence between addtion and multiplication that is usually found in exponentiation. This shouldn't come as a surprise: there is a linear algebra concept called matrix exponentiation. 
If $V_c=\begin{pmatrix}0 & c \\ 0 & 0\end{pmatrix}$ then $exp(V_c)=\begin{pmatrix}1 & c \\ 0 & 1\end{pmatrix}$. In this way, property $(1)$ can be "trivially" written:
$$\tag{2}exp(V_c)exp(V_{c'})=exp(V_{c+c'})$$
