Let $M$ Denote the set of 2x2-matrices of the form $$A=\pmatrix{1&m\\0&1}$$ where the entries are integers. Show that $M$, with respect to matrix multiplicaation, is an infinite cyclic group.
To prove this, i should just prove that the set $M$ and $C_{\infty}$ are isomorphic right? Since if they are isomorphic then they are same, which means that $M$ is a infinite cyclic group.
So, i contruct a bijection $f$ as $$f: M->C_{\infty}$$ such that $f(m_1*m_2)=f(m_1)*f(m_2)$. If i can show this, then $M$ and $C_{\infty}$ are isomorphic and therefore $M≈C_{\infty}$. This bijection $f$ takes $A$ to 1 $A^2$ to 2 etc or?
I have now construct an bijection. Next thing I have to do is to find the element $x$ such that every member of $C_{\infty}$**(or shall i take $M$?)** is a power of $x$. This element $x$ is said to generate $C_{\infty}$; $C_{\infty}$ = < x>.
After I have done that, i define $f$ by the rule $$f(n)=x^n$$ there $n$ belongs to $Z$. This generator $x$ is the matrix $A$ above: $$f(n)=x^n=\pmatrix{1&m\\0&1}^n$$ I get stuck here and i don't know if i have done right so far :/
The short solution in the book
The function taking matrix $A$ to m is an isomorphic from $M$ to $Z$(why the integer?). The significnt is that $$\pmatrix{1&m\\0&1}*\pmatrix{1&n\\0&1}=\pmatrix{1&m+n\\0&1}$$
Simply define a group homomorphism $\phi:M \rightarrow \mathbb{Z}$ as follows:
$$\begin{bmatrix} 1 & m \\[0.3em] 0 & 1 \\[0.3em] \end{bmatrix} \mapsto m$$
Of course, you'll want to prove that this is actually a homomorphism. From there, it's not too difficult to confirm that $\phi$ is both surjective and injective $\Longrightarrow$ $\phi$ is bijective and thus an isomorphism $\Longrightarrow$ $M \cong \mathbb{Z} \Longrightarrow M$ is an infinite cyclic group since $\mathbb{Z}$ is an infinite cyclic group. (Note that $\mathbb{Z}$ is a cyclic group under addition generated by $1$).