I would like to find the center of this group

Question

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?

Answer 1

Suposse that for all $B\in\text{PSL}(2,\mathbb{C})$ holds $AB=BA$. We want to prove $A=\pm\text{Id}$. Let $A=\left(\begin{array}{cc} a & b\\ c & d\end{array}\right)$. Because holds for all matrix, then, in particular $B=\left(\begin{array}{cc} 1 & 0\\ 1 & 1\end{array}\right)$. thus, we have that $$\left(\begin{array}{cc} a+b & b\\ c+d & d\end{array}\right)=\left(\begin{array}{cc} a & b\\ c & d\end{array}\right)\left(\begin{array}{cc} 1 & 0\\ 1 & 1\end{array}\right)=\left(\begin{array}{cc} 1 & 0\\ 1 & 1\end{array}\right)\left(\begin{array}{cc} a & b\\ c & d\end{array}\right)= \left(\begin{array}{cc} a & b\\ a+c & b+d\end{array}\right)$$So we have that $a+b=a$, $b=b$, $c+d=a+c$, $b+d=d$. Then, we conclude that $b=0$. In other way, also the above is $B=\left(\begin{array}{cc} 1 & 1\\ 0 & 1\end{array}\right)$. So, we have that $$\left(\begin{array}{cc} a & b+a\\ c & d+c\end{array}\right)=\left(\begin{array}{cc} a & b\\ c & d\end{array}\right)\left(\begin{array}{cc} 1 & 1\\ 0 & 1\end{array}\right)=\left(\begin{array}{cc} 1 & 1\\ 0 & 1\end{array}\right)\left(\begin{array}{cc} a & b\\ c & d\end{array}\right)=\left(\begin{array}{cc} a+c & b+d\\ c & d\end{array}\right)$$Thus, $a+c=a$, $b+d=b+a$, $c=c$ y $d+c=d$. We conclude that $c=0$ and $d=a$. Well then, we have that $b=c=0$ and that $d=a$. Also, $ad-bc=ad=1$. Thus, $a=d=1$ or $a=d=-1$. Therefore, $A=\left( \begin{array}{cc} 1 & 0\\ 0 & 1 \end{array}\right)$ or $A=\left(\begin{array}{cc} -1 & 0\\ 0 & -1\end{array}\right)$

Answer 2

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 $g = \pmatrix{a & 0 \\ 0 & a}$.