Matrix group isomorphic to $\mathbb Z$. The set  $G=\left\{\begin{pmatrix}1 & n  \\
        0 & 1  \\
        \end{pmatrix}\mid  n\in \Bbb Z\right\}$ with the operation of matrix multiplication is a group. Show that $$\phi:\Bbb Z \to G,$$ $$\phi(n)=\begin{pmatrix}1 & n  \\
        0 & 1  \\
        \end{pmatrix}$$
is a group isomorphism (where the operation on $\Bbb Z$ is ordinary addtion).
TO show it's isomorphism:  I know I must show one-to-one, onto and homomorphism. I've done these examples before but never with matrices.  
How can I show if $\phi(a)=\phi(b)$ then $a=b$? Same question for onto and operation preserving with matrices.
Thank you!
 A: Hint: For your first question, write down $\phi(a)$ and $\phi(b)$ (go ahead, write down the matrices on a sheet of paper). Now, if those two are equal, what does it tell you?
The other parts are obviously different, but the idea is the same: just look at the matrices involved and use what you know about matrix multiplication (for the last part).
A: Hint: 
$\begin{pmatrix}1 & n  \\
        0 & 1  \\
        \end{pmatrix}\begin{pmatrix}1 & m  \\
        0 & 1  \\
        \end{pmatrix} = \begin{pmatrix}1 & n+m  \\
        0 & 1  \\
        \end{pmatrix}$
That should help with proving $\phi$ is a homomorphism.
Once you have proven that a homomorphism exists, you must prove it is bijective to prove the mapping is a isomorphism. You already have that the mapping in injective, you must prove it is surjective. 
A: So let me straight things up, since there are multiple answer which get this incorrectly. First of all you need to show that $\phi$ is a homomorphism (for that see other answers, they are fine). Then you can show that it is an isomorphism by constructing the inverse. And by the inverse we mean the function inverse, i.e. a function
$$g:G\to\mathbb{Z}$$
such that $\phi\circ g=\mbox{id}_{G}$ and $g\circ \phi=\mbox{id}_{\mathbb{Z}}$. By general property if such inverse exists then it is a homomorphism as well and so $\phi$ is an isomorphism.
So you can easily check that
$$g\bigg(\begin{bmatrix}
1 & n\\
0 & 1
\end{bmatrix}\bigg)=n
$$
is the inverse of $\phi$.
A: Just note that
$$
\begin{pmatrix}1 & 1 \\
        0 & 1  \\
        \end{pmatrix}^n
=
\begin{pmatrix}1 & n  \\
        0 & 1  \\
        \end{pmatrix}
$$
This implies that $\phi$ is a surjective homomorphism because exponents add when you multiply powers. It is clear that $\phi$ is injective because $\phi(n)_{12}=n$.
A: Hint:
There are other ways to show that this map is one-to-one. 
Use that the $ker(\phi)=0$. 
Does there exist an $n$ such that the rank of 
$\phi(n)=\begin{pmatrix}1 & n  \\
        0 & 1  \\
        \end{pmatrix}$ is not $2$?
Alternatively, construct $\phi^{-1}(n)=\begin{pmatrix}1 & -n  \\
        0 & 1  \\
        \end{pmatrix}$. The existence of an inverse with the homomorphic property implies it's an isomorphism.
