Groups involving matrices Let $$H=
  \left\{\left( {\begin{array}{ccc}
   1 & b \\
    0 & 1 \\
  \end{array} } \right)\Bigg\vert b\in \mathbb{R}\right\}$$
How do I show that H is a group and under which operation? I have not done so much with matrices in algebra.
I believe there is a pattern/guide of which I have to follow like when showing subgroups.
 A: For $H$ to be a group, $H$ must be closed, satisfy associativity, contain an identity element, and contain an inverse element for each matrix belonging to $H$.
Also, the operation here appears to be multiplication. 
Let $b,c \in \mathbb{R}$. The set $H$ satisfies closure because $$\begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & c \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & b+c \\ 0 & 1 \end{pmatrix},$$ where $b+c \in \mathbb{R}$. The product matrix belongs to $H$; hence, $H$ is closed under multiplication. 
Also, $H$ satisfies the associative property. Matrix multiplication is associative.
There is an identity element, or in this case, an identity matrix $$\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.$$
There exists an inverse element (in this case, an inverse matrix) in $H$. The inverse matrix is $$\begin{pmatrix} 1 & -b \\ 0 & 1 \end{pmatrix}$$ and belongs to $H$ since $-b \in \mathbb{R}$.
A: I assume the candidate operations are matrix addition and multiplication.  We can rule out matrix addition since
$\begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix} + \begin{pmatrix} 1 & c \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 2 & b  + c \\ 0 & 2 \end{pmatrix} \notin H; \tag{1}$
$H$ is not closed under matrix addition.  However, we have
$\begin{pmatrix} 1 & b \\ 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & c \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} 1 & b + c \\ 0 & 1 \end{pmatrix} \in H, \tag{2}$
which not only shows $H$ is closed under matrix multiplication, which is in fact associative (as pointed out by dragon), but that
$I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \tag{3}$
is the identity element for $H$; also, we clearly have
$\begin{pmatrix} 1 & c \\ 0 & 1 \end{pmatrix}^{-1} = \begin{pmatrix} 1 & -c \\ 0 & 1 \end{pmatrix} \in H; \tag{4}$
$H$ contains the multiplicative inverses of all its elements.  So it is indeed an (abelian) group.  Interestingly enough, all these group properties may be read off from (2), without referring them to the space of all matrices, e.g. it is clear from (2) that multiplication must be associative on $H$ etc. etc. etc.
