Translations in two dimensions - Group theory I have just started learning Lie Groups and Algebra. Considering a flat 2-d plane if we want to translate a point from $(x,y)$ to $(x+a,y+b)$ then can we write it as :
$$   \left( \begin{array}{ccc}
x+a  \\
y+a  \end{array} \right) =
  \left( \begin{array}{ccc}
x  \\
y  \end{array} \right) + 
  \left( \begin{array}{ccc}
a  \\
b  \end{array} \right)$$
Now the set of all translations  $ T =   \left( \begin{array}{ccc}
a  \\
b  \end{array} \right) $ form a two parameter lie group (I presume) with addition of column as the composition rule.
If that is so, how do I go about finding the generators of this transformation.
I know the generators of translation are linear momenta in the corresponding directions. But I am not able to see this here.
PS:  In my course I have been taught that the generators are found by calculating the Taylor expansion of the group element about the Identity of the group. For instance, $\operatorname{SO}(2)$ group 
$$ M =   \left( \begin{array}{cc}
\cos \:\phi & -\sin \:\phi \\
\sin \:\phi &  \cos \:\phi \end{array} \right) $$
I obtain the generator by taking 
$$ \frac{\partial M}{\partial \phi}\Bigg|_{\phi=0} = 
 \left( \begin{array}{cc}
0  & -1 \\
1 &  0 \end{array} \right) $$
Now if I exponentiate this, I can obtain back the group element. My question how do I do this for Translation group. 
EDIT :This edit is to summarise and get a view of the answers obtained.
Firstly, the vector representation of the translation group (for 2D) would in general have the form : 
$$ \begin{pmatrix} 1 & 0 & a_x\\ 0 & 1 & a_y \\ 0 & 0 & 1 \end{pmatrix}\ $$
with generators (elements of Lie algebra) 
$$  T_x =\begin{pmatrix} 0 & 0 & i\\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix}\ , \;\; 
T_y = \begin{pmatrix} 0 & 0 & 0\\ 0 & 0 & i \\ 0 & 0 & 0 \end{pmatrix}\  $$
Secondly, the scalar-field representation of the same is given by the differential operators
$$ exp^{ i(a_x\frac{\partial}{\partial x}+ a_y\frac{\partial}{\partial y} )} $$
with generators 
$$ T_x^s = i\frac{\partial}{\partial x},\;\;T_y^s = i\frac{\partial}{\partial y} $$
The Lie algebra is two-dimensional and abelian : $ [T_x,T_y] = 0$ 

As suresh mentioned if the vector is just a two component object then you can't translate it without expanding the vector. However, if you consider the vector to be variable (which are essentially infinite vectors) then it can be translated. 
To find the differential form of a translation, start with the translation of a 1D dimensional vector, $x$:
\begin{align} 
e ^{ i\epsilon {\cal P} } x & = x + \epsilon \\ 
\left( 1 + i \epsilon {\cal P} \right) x &= x + \epsilon \\ 
{\cal P} x & = - i  
\end{align} 
Thus we must have $ {\cal P} =  - i \frac{ \partial }{ \partial x } $. 
Now it is easy to extend this to two dimensions:
\begin{align} 
e ^{ i\epsilon _x  {\cal P} _x + i \epsilon _y {\cal P} _y  } \left( \begin{array}{c} 
x \\  
y 
\end{array} \right)
& = \left( \begin{array}{c} 
x + \epsilon _x  \\  
y + \epsilon _y  
\end{array} \right) \\
i \left( \epsilon _x {\cal P} _x + \epsilon _y {\cal P} _y \right) \left( \begin{array}{c} 
x \\  
y 
\end{array} \right) &= \left( \begin{array}{c} 
x + \epsilon _x  \\  
y + \epsilon _y  
\end{array} \right) 
\end{align} 
where we have two different generators since you have two degrees of freedom in the transformation you gave in your question. This expression requires,
\begin{align} 
& {\cal P} _x = \left( \begin{array}{cc} 
- i \frac{ \partial }{ \partial x } & 0 \\  
0 & 0
\end{array} \right) \\ 
& {\cal P} _y = \left( \begin{array}{cc} 
0 & 0 \\  
0 & - i \frac{ \partial }{ \partial y } 
\end{array} \right)
\end{align}
