Proving the Heisenberg Group is a group I have to prove that the Heisenberg Group, 
\begin{pmatrix}
   1&a&b  \\ 
   0&1&c\\    
   0&0&1
\end{pmatrix}
where $a,b,c\in\mathbb{R}$ is a group. 
I am proving a group under matrix addition. 
For the inverse part of the proof, I have
$$ \det H = 1 (1-c) - a (0) + b (0) = -c.$$ 
But couldn't $c$ be equal to zero so then this is not a group.
Or am I correct in proving that $\det H$ is not equal to zero and therefore an inverse exists?
 A: I like the Heisenberg group; I like it a lot.  I like it so much I'm going to pitch my hat into the ring and answer the question.
It is, course, not true that the set $\mathscr H$ of matrices of the form
$\begin{bmatrix}
   1&a&b  \\ 
   0&1&c\\    
   0&0&1
\end{bmatrix}, \tag{1}$
form a group under the operation of matrix addition, but it does form a group under matrix multiplication, and that group is what's usually known as the Heisenberg group.  See this Widipedia entry for more information--lot's more.
I think the easiest and conceptually cleanest way to deal with $\mathscr H$ is to observe that any $H \in \mathscr H$ is of the form $H = I + N$, where $N$ is a strictly upper triangular matrix of the form
$N = \begin{bmatrix} 0 & a & b \\ 0 & 0 & c \\ 0 & 0 & 0 \end{bmatrix}, \tag{2}$
and that if
$N_1 = \begin{bmatrix} 0 & a_1 & b_1 \\ 0 & 0 & c_1 \\ 0 & 0 & 0 \end{bmatrix} \tag{2}$
and
$N_2 = \begin{bmatrix} 0 & a_2 & b_2 \\ 0 & 0 & c_2 \\ 0 & 0 & 0 \end{bmatrix}, \tag{3}$
then
$N_1 N_2 = \begin{bmatrix} 0 & 0 & a_1 c_2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}. \tag{4}$
Thus if $H_1 = I + N_1, H_2 = I + N_2 \in \mathscr H$ we have
$H_1 H_2 = (I + N_1)(I + N_2) = I + N_1 + N_2 + N_1 N_2$
$= \begin{bmatrix} 1 & a_1 + a_2 & b_1 + b_2 + a_1 c_2 \\ 0 & 1 & c_1 + c_2 \\ 0 & 0 & 1 \end{bmatrix} \in \mathscr H, \tag{5}$
showing that $\mathscr H$ is closed under matrix multiplication.  Clearly, $I \in \mathscr H$.  To show $H \in \mathscr H$ implies the existence of $H^{-1} \in \mathscr H$, consider the formula
$1 - r^n = (1 - r) \sum_0^{n -1} r^k, \tag{6}$
which holds for any $1 \le n  \in \Bbb Z$, the integers, and $r \in R$, a (not necessarily commutative) ring with unit.  If $n = 1$, (6) reads $1 - r = 1 - r$, and if $n = 2$ it becomes $1 - r^2 = (1 - r)(1 + 2)$; taking these as base cases and the $n = m$ case of (6) as the inductive hypothesis, we observe that
$1 - r^{m + 1} = 1 - r^m + r^m - r{m + 1} = (1 - r)\sum_0^{m - 1} r^k + (1 - r)r^m = (1 -r)\sum_0 ^m r^k, \tag{7}$
inductively demonstrating the validity of (6).
We apply (6) to matrices $N$ of the form (2), noting that, by (4),
$N^2 = \begin{bmatrix} 0 &0 & ac \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}, \tag{8}$
from which it is easily seen that $N^3 = 0$.  Thus,  taking $r = -N$ and $n = 3$ in (6) we have
$I = (I + N)(I - N + N^2), \tag{9}$
demonstrating that
$H^{-1} = I - N + N^2 \in \mathscr H \tag{10}$
for $H = I + N \in \mathscr H$.  Indeed, from (2) and (8) we calculate
$H^{-1} = \begin{bmatrix} 1 & -a & ac - b \\ 0 & 1 & -c \\ 0  &0 & 1 \end{bmatrix}, \tag{11}$
in accord with the expression given by our colleague B.S. in his answer.  We therefore see that $\mathscr H$, containing as it does both $I$ and $H^{-1}$ for any $H \in \mathscr H$, and being closed under matrix multiplication, is in fact a group.  QED.
Remarks:  The notion of Heisenberg group may in fact be generalized in several directions, and I thought it might be worthwhile to mention a few here.  First of all, matrices of the form $H = I + N$, with $N$ as in (2), form a group whenever  $a, b, c \in R$, where $R$ is any unital ring, commutative or not.  The proof and all the formulas are indeed exactly the same as in the above treatment, except that all matrix elements, $1, a, b , c \in R$, instead of $\Bbb R$.  A second direction of generalization is the size of the matrices $H \in \mathscr H$.  We may, in fact, consider the set $\mathscr H_n(R)$ of $n \times n$ upper triangular matrices over an arbitrary unital ring with every diagonal entry equal to $1$; it is easy to see these are of the form $I + N$, with $N$ strictly upper triangular and all matrix entries taken from some ring with unit $R$.  Then $N^n = 0$ and the formulas given above appropriately generalize, provided the sum $\sum N^k$ is truncated after the term $N^n$; $\mathscr H_n(R)$ is indeed a group.
Hope this helps.  Cheerio,
and as always,
Fiat Lux!!!
A: The set of that matrices is group under 
multiplication operation. We have 
$$det(H)=\begin{vmatrix} 1 & a & b\\ 0 & 1 & c\\ 0 & 0 &1 \end{vmatrix}=1$$ instead and so by any method you know, we can find:
$$H^{-1}=\begin{vmatrix} 1 & -a & ac-b\\ 0 & 1 & -c\\ 0 & 0 &1 \end{vmatrix}$$ It is a good point that this group has a nice presentation as follos:
$$\mathcal{H}=\langle a,b,c\mid[a,b]=c,[c,b]=[c,a]=1\rangle$$
