I would like to find the center of this group 
Show that the center of $SL(2, \mathbb{C})$ is $\pm Id.$ Hint: Use $
    \begin{pmatrix}
    1 & 1  \\
    0 & 1  \\
        \end{pmatrix}
$, $
    \begin{pmatrix}
    1 & 0  \\
    1 & 1  \\
        \end{pmatrix}
$.

The center of a group is $C(G) = \{ g \in G : g\ast h = h \ast g \quad \forall h \in G  \}.$ And it is true that $
    \begin{pmatrix}
    1 & 0  \\
    0 & 1  \\
        \end{pmatrix}
$, $
    \begin{pmatrix}
    -1 & 0  \\
    0 & -1  \\
        \end{pmatrix}
$, $
    \begin{pmatrix}
    1 & 1  \\
    0 & 1  \\
        \end{pmatrix}
$, $
    \begin{pmatrix}
    1 & 0  \\
    1 & 1  \\
        \end{pmatrix}
$, have determinant $1$, hence, they are in the group, and particularily, the suggested matrices commute with both $\pm Id.$ But this doesn't prove that only two matrices, $Id, -Id$, are in the center. And It doesn't show that,  $Id, -Id$ commutes with all the matrices in the group. So why would I pick those matrices in the first place? What am I missing? 

Let $g = \pmatrix{a & b \\ c & d} \in C(G)$. One then has $$\pmatrix{a & b \\ c & d}\pmatrix{1 & 1 \\ 0 & 1} = \pmatrix{1 & 1 \\ 0 & 1}\pmatrix{a & b \\ c & d}$$ $$\pmatrix{a & a+b \\ c & c+d} = \pmatrix{a + c & b+d \\ c & d}$$
One then has $c=0$ and $a=d$.
With matrix $\pmatrix{1 & 0 \\ 1 & 1}$, you will get $b=0$.
So now you have $g = \pmatrix{a & 0 \\ 0 & a}$.
