Centre of the special linear group $SL_2(\mathbb R)$ or $SL(2,\mathbb R)$ Algebra by Michael Artin Def 2.5.12



Obviously $I$ and $-I$ are in the centre: $AI=IA,A(-I)=(-I)A$.
How exactly do I go about doing this?
I was thinking to solve for $j,f,g,h$ below
$$\begin{bmatrix}
a & b\\ 
c & d
\end{bmatrix} \begin{bmatrix}
j & f\\ 
g & h
\end{bmatrix} = \begin{bmatrix}
j & f\\ 
g & h
\end{bmatrix} \begin{bmatrix}
a & b\\ 
c & d
\end{bmatrix}, ad-bc=1=jh-fg.$$
So, I plug in b*g=c*f, a*f+b*h=b*j+d*f, c*j+d*g=a*g+c*h, j*h-f*g=1, a*d-b*c=1 in Wolfram Alpha or here or here (or here) to get a bunch of complicated solutions sets, some of which include the desired $\pm I$.
Ugh, where can I find a proof online?
Or if this is still folklore, how do I begin?


*

*Do I for example take cases $c=0, c \ne 0$ and then solve for the centre in each case('s subcases)?

*Another thing I thought was to suppose on the contrary that $f \ne 0$ and then arrive at a contradiction and then assume $f=0$ when supposing on the contrary that $g \ne 0$ and then assuming $f=g=0$ when supposing on the contrary that $j \ne \pm 1$ and then assuming $f=g=0, j = \pm 1$ when supposing on the contrary that $h \ne \pm 1$.
Not looking for a full solution, just a little nudge in the right direction. I've been lost in subcases of subcases of cases (like here) that I think I'm missing something elegant or simple. I guess I've been doing maths for quite awhile that I've forgotten how to do arithmetic.
To clarify, I am looking for a way to do this by systems or at least nothing high level like using facts like these About the Center of the Special Linear Group $SL(n,F)$ Note that this chapter is on homomorphisms. The reader just finished cyclic groups. The reader didn't reach Lagrange's Theorem, fields, rings or even isomorphisms.
 A: A matrix $M$ is in the centre iff $MA=AM$ for all elements $A$ of $\text{SL}_2(\Bbb R)$. This means that $MA'=A'M$ for all $A'$ in the linear span of elements
of $\text{SL}_2(\Bbb R)$. If you can show that all $2\times 2$ matrices can
be such an $A'$ then $M$ will commute will all matrices, and so be a scalar matrix.
ADDED IN EDIT
For a slick proof, let $T=\pmatrix{1&1\\0&1}$ and $U=\pmatrix{1&0\\1&1}$.
Then $T$, $U\in\text{SL}_2(\Bbb R)$ and the only matrices $M$
with $MT=TM$ and $MU=UM$ are the scalar matrices $M=aI$.
A follow-up exercise is to extend this to $\text{SL}_n(\Bbb R)$.
A: We must show $centre(SL_2(\mathbb R)) = \{ \pm I \}$.
Pf: $\supseteq$ is easy. For $\subseteq$, we must show that for any $N=\begin{bmatrix}
j & f\\ 
g & h
\end{bmatrix}$ in $centre(SL_2(\mathbb R))$, we will have $N=\pm I$.
Let $N \in centre(SL_2(\mathbb R))$. By definition of $centre(SL_2(\mathbb R))$, we have that for any $M = \begin{bmatrix}
a & b\\ 
c & d
\end{bmatrix}$ in $SL_2(\mathbb R)$, $MN=NM$.
Observe that $MN=NM$ gives the following equations
$$bg=fc, c(j-h)=(a-d)g, (a-d)f=b(j-h) \tag{1}$$
For the fixed $N$, we must have that $MN=NM$ holds for any $M$ in $SL_2(\mathbb R)$, i.e. for any real numbers $a,b,c,d$ s.t. $ad-bc=1$. Therefore, by careful selection of matrix $M$, i.e. by careful selection of real numbers $a,b,c,d$, we can deduce $e,f,g,h$.
First careful selection: Let us choose $M$ s.t. $c=0$ and $b \ne 0$. For $c=0$, since $N \in centre(SL_2(\mathbb R))$, we must have that $(1)$ becomes:
$$bg=0, 0=(a-d)g, (a-d)f=b(j-h)$$
From the first two inequalities, we have that ($g=0$ or $b=0$) and ($g=0$ or $a=d$). Therefore, $g=0$ or ($a=d$ and $b=0$). Now $b \ne 0$ implies that we must have $g=0$.
Second careful selection: Similarly, let us choose $M$ s.t. $b=0$ and $c \ne 0$. Then $f=0$.
Third careful selection: Let us choose $M$ s.t. $b \ne 0$. From our deductions that $f=0=g$, we have that $(1)$ becomes:
$$0=0, c(j-h)=0, 0=b(j-h)$$
From the third equation, we have, $j=h$.
Hence, we have come up with three deductions as follows:
$$1. f=0$$
$$2. g=0$$
$$3. j=h$$
Hence, $N=\begin{bmatrix}
j & 0\\ 
0 & j
\end{bmatrix}=jI$. Now, $1=\det(N) \implies j = \pm 1$.
Therefore, we have shown that if $N$ is in $centre(SL_2(\mathbb R))$, $N=\pm I$. QED
