Showing the Heisenberg group is the central extension of the additive group Context and some work so far:
I found out about the Heisenberg group on Youtube. I'm a Physics student. I wanted to know more about it, and I realized there was more to learn.
Here is what I found out.
1.) The Heisenberg group is a group of upper triangular matrices of the form:
\begin{pmatrix}
   1 & a & b\\
   0 & 1 & c\\
   0 & 0 & 1
\end{pmatrix}
I found this out by looking through the Wikipedia page
2.) I think I can now construct the corresponding Lie algebra.
Generally,

*

*A matrix exponential $e^{At} = I + At + \frac{1}{2} A^2t^2+ . . . $


*The derivative  $\frac{d}{dt} e^{At} = $ A  +  A^2 t. . .  = A.e^{At}$
Let
$e^{At} =   \begin{pmatrix}
   1 & a(t) & b(t)\\
   0 & 1 & c(t)\\
   0 & 0 & 1
\end{pmatrix}$
then
$\frac{d}{dt} e^{At}|_0= \begin{pmatrix}
   0 & a'(0) & b'(0)\\
   0 & 0 & c'(0)\\
   0 & 0 & 0   
   \end{pmatrix}$
with basis:
$ A= \begin{pmatrix}
   0 & 1 & 0\\
   0 & 0 & 0\\
   0 & 0 & 0   
   \end{pmatrix}$, $B =\begin{pmatrix}
   0 & 0 & 0\\
   0 & 0 & 1\\
   0 & 0 & 0   
   \end{pmatrix}$
and
$C =\begin{pmatrix}
   0 & 0 & 1\\
   0 & 0 & 0\\
   0 & 0 & 0   
   \end{pmatrix}$
Then $[A,B], [A, C]$ and $[B, C]$ can be computed
The central element can then be seen from the commutation relations
Question:
I have read that the Heisenberg group is the central extension of the additive group V. How can one show this?
More context:
I have found this question that tackles the problem without invoking cocycles. It looks at the commutator of two elements of the group and finds the conditions under which the commutator must vanish to find the center of the group. The argument then terminates with the statement that the element that satisfies that argument is isomorphic to the additive group.
More context:
In this Wikipedia link , under  "On symplectic vector spaces" the abstract group law of the Heisenberg group is given which will be reproduced here : $(v,t).(v',t') = (v+v',t+t', \frac{1}{2} \omega(v,v'))$
 A: I've read that the additive group where the operation is addition.
Let $C(G)$ be the center of the Heisenberg group.
To find the center, $C(G)$ it suffices to find the condition necessary for two heisenberg matrices to form a vanishing Lie bracket.
The below was done in Mathematica not by hand
In practice $\begin{pmatrix}
             1 & a_1 & b_1\\
             0 & 1 & c_1\\
             0 & 0 & 1 
             \end{pmatrix}$ $\begin{pmatrix}
             1 & a_2 & b_2\\
             0 & 1 & c_2\\
             0 & 0 & 1 
             \end{pmatrix}$ -  $\begin{pmatrix}
             1 & a_2 & b_2\\
             0 & 1 & c_2\\
             0 & 0 & 1 
             \end{pmatrix}$$\begin{pmatrix}
             1 & a_1 & b_1\\
             0 & 1 & c_1\\
             0 & 0 & 1 
             \end{pmatrix}$ = 0
Hence $\begin{pmatrix}
             1 & a_1 + a_2 & b_1 + b_2 + a_1 c_2\\
             0 & 1 & c_1 +c_2\\
             0 & 0 & 1 
             \end{pmatrix}$ - $\begin{pmatrix}
             1 & a_1 + a_2 & b_1 + b_2 + a_2 c_1\\
             0 & 1 & c_1 +c_2\\
             0 & 0 & 1 
             \end{pmatrix}$ = $\begin{pmatrix}
             0 & 0 & - a_2 c_1 + a_1 c_2\\
             0 & 0 & 0\\
             0 & 0 & 0 
             \end{pmatrix}$
So $a_2 c_1 = a_1c_2$  to obtain vanishing commutator
Let center then be matrices of the form
C = $\begin{pmatrix}
         1 & 0 & b_3\\
         0 & 1 & 0\\
         0 & 0 & 1 
         \end{pmatrix}$
It is easy to check that the commutator with any upper diagonal matrix is vanishing.
So matrices of the form $C$ form the center of the group  $C(G)$
The Heisenberg group is the central extension of the additive group , and the center of the group is C(G)
Let the  quoteient group be $G/C(G)$
According to Wikipedia's article on group extensions
$1 \rightarrow N \rightarrow G \rightarrow Q \rightarrow 1$
$C(G)$ of Heisenberg group is the matrices with $(a,b,c)$ as defined above to be $(0,b_3,0)$ as shown by computation.
$C(G)$ is  $N$ the normal subgroup of $G$
so $1 \rightarrow C(G) \rightarrow G \rightarrow R^2 \rightarrow 1$
$ G/C(G)$ ~  $\mathbb{R^2}$
It would seem that it suffices to show this map exists. See page 625 of this
