# SL(2,IR) is the Commutator Subgroup of GL(2,IR)

Here is a proof of the above fact.

Let $N$ be the commutator subgroup of the general linear group $GL(2,\mathbb R)$; i.e.,

$N=\langle ABA^{-1}B^{-1}:A,B\in GL(2,\mathbb R)\rangle$.

First, it is clear that $N$ is contained in the special linear group $SL(2,\mathbb R)$, since $\det(ABA^{-1}B^{-1})=1$ for any $A,B\in GL(2,\mathbb R)$. Next, we claim that $N$ contains all matrices

$\begin{pmatrix} 1 & b\\ 0 & 1\end{pmatrix}$.

This follows from noting that

$\begin{pmatrix} 1 & b\\ 0 & 1\end{pmatrix}=\begin{pmatrix} 1 & b\\ 0 & b\end{pmatrix}\begin{pmatrix} 1 & 1\\ 0 & 1\end{pmatrix}\begin{pmatrix} 1 & b\\ 0 & b\end{pmatrix}^{-1}\begin{pmatrix} 1 & 1\\ 0 & 1\end{pmatrix}^{-1}$.

By taking transposes, it also follows that $N$ contains all matrices

$\begin{pmatrix} 1 & 0\\ c & 1\end{pmatrix}$.

Further, $N$ contains all matrices

$\begin{pmatrix} a & 0\\ 0 & 1/a\end{pmatrix}$

since

$\begin{pmatrix} a & 0\\ 0 & 1/a\end{pmatrix}=\begin{pmatrix} a & 0\\ 0 & 1\end{pmatrix}\begin{pmatrix} 0 & 1\\ 1 & 0\end{pmatrix}\begin{pmatrix} a & 0\\ 0 & 1\end{pmatrix}^{-1}\begin{pmatrix} 0 & 1\\ 1 & 0\end{pmatrix}^{-1}$

for any $a\neq 0$.

Now let

$\begin{pmatrix} a & b\\ c & d\end{pmatrix}\in SL(2,\mathbb R)$.

Then $ad-bc=1$. Using the above results,

$\begin{pmatrix} a & b\\ c & d\end{pmatrix}=\begin{pmatrix} 1 & 0\\ c/a & 1\end{pmatrix}\begin{pmatrix} 1 & ab\\ 0 & 1\end{pmatrix}\begin{pmatrix} a & 0\\ 0 & 1/a\end{pmatrix}\in N$

if $a\neq 0$, and

$\begin{pmatrix} a & b\\ c & d\end{pmatrix}=\begin{pmatrix}0&1\\-1&0\end{pmatrix}\begin{pmatrix}1&-d/b\\ 0&1\end{pmatrix}\begin{pmatrix}1&0\\ ab&1\end{pmatrix}\begin{pmatrix}1/b&0\\ 0&b\end{pmatrix}\in N$

if $b\neq 0$, and the latter since

\begin{aligned}&\begin{pmatrix} 0 & -1\\ 1 & 0\end{pmatrix}\\=&\begin{pmatrix}x&y\\0&-x-y\end{pmatrix}\begin{pmatrix}-x-y&0\\ x&y\end{pmatrix}\begin{pmatrix}x&y\\0&-x-y\end{pmatrix}^{-1}\begin{pmatrix}-x-y&0\\ x&y\end{pmatrix}^{-1}\\ \in &\ N\end{aligned}

for any $x,y,x+y\neq 0$. Thus $SL(2,\mathbb R)\subseteq N$, i.e., $N=SL(2,\mathbb R)$.

2 Comments

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### 2 responses to “SL(2,IR) is the Commutator Subgroup of GL(2,IR)”

1. Your matrix product for $\begin{bmatrix} 1&b \\ 0&1 \end{bmatrix}$ has a typo in it (set $b=1$ to see this quickly). See https://math.stackexchange.com/a/2762248/448 for a variant that works.

• Thanks for pointing that out!