Matrix group isomorphic to $\mathbb Z$. The set  $G=\left\{\begin{pmatrix}1 &amp; n  \\
        0 &amp; 1  \\
        \end{pmatrix}\mid  n\in \Bbb Z\right\}$ with the operation of matrix multiplication is a group. Show that $$\phi:\Bbb Z \to G,$$ $$\phi(n)=\begin{pmatrix}1 &amp; n  \\
        0 &amp; 1  \\
        \end{pmatrix}$$
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!
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).
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 
$\phi(n)=\begin{pmatrix}1 &amp; n  \\
        0 &amp; 1  \\
        \end{pmatrix}$ is not $2$?

Alternatively, construct $\phi^{-1}(n)=\begin{pmatrix}1 &amp; -n  \\
        0 &amp; 1  \\
        \end{pmatrix}$. The existence of an inverse with the homomorphic property implies it's an isomorphism.