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Let $F$ a field. I want proves that $[GL_2(F),GL_2(F)]=SL_2(F)$. The inclusion $[GL_2(F),GL_2(F)]\subset SL_2(F)$ holds, by (Special linear group contains commutator subgroup of general linear group.)

$SL_2(F)\subset [GL_2(F),GL_2(F)]=SL_2(F)$? I need for any $A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}$ exists $B,C\in GL_2(F)$ such that $A=BCB^{-1}C^{-1}$ but I can't find $B,C$.

Actualization 1.

Let $0,1\in F$.

LEt $\begin{bmatrix} a & b\\ c & d \end{bmatrix}\in SL_2(F)$ then $ad-bc=1$.

for all $x\in F$, we have $$\begin{bmatrix} 1 & x\\ 0 & 1 \end{bmatrix}=\begin{bmatrix} 1+1 & 0\\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1 & x\\ 0 & 1 \end{bmatrix}\begin{bmatrix} 1+1 & 0\\ 0 & 1 \end{bmatrix}^{-1}\begin{bmatrix} 1 & x \\ 0 & 1 \end{bmatrix}^{-1}$$ where $$\begin{bmatrix} 1+1 & 0\\ 0 & 1 \end{bmatrix}^{-1}=\begin{bmatrix} (1+1)^{-1} & 0 \\ 0 & 1 \end{bmatrix}\text{ y }\begin{bmatrix} 1 & x\\ 0 & 1 \end{bmatrix}^{-1}=\begin{bmatrix} 1 & -x\\ 0 & 1 \end{bmatrix}$$ where $(1+1)^{-1}$ inverse multiplicative of $1+1$ y $-x$ aditive inverse of $x$.

therefore, $\begin{bmatrix} 1 & x\\ 0 & 1 \end{bmatrix}$ is a product of commutators. Analogously, $\begin{bmatrix} 1 & 0\\ x & 1 \end{bmatrix}$ is product of commutators.

Simillary $$\begin{bmatrix} x & 0\\ 0 & x^{-1} \end{bmatrix}=\begin{bmatrix} x & 0\\ 0 &1 \end{bmatrix}\begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}\begin{bmatrix} x & 0\\ 0 & 1 \end{bmatrix}^{-1}\begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix}^{-1}.$$

If $a\neq 0$, we have $$\begin{bmatrix} a & b\\ c & d \end{bmatrix}=\begin{bmatrix} 1 & 0\\ ca^{-1} & 1 \end{bmatrix}\begin{bmatrix} 1 & ab\\ 0 & 1 \end{bmatrix}\begin{bmatrix} a & 0\\ 0 & a^{-1} \end{bmatrix}$$ where each matrix is a product of commutators.

If $a=0$ then $b\neq 0$, so $$\begin{bmatrix} 0 & b\\ c & d \end{bmatrix}=\begin{bmatrix} 0 & 1\\ -1 & 0 \end{bmatrix}\begin{bmatrix} 1 & -db^{-1}\\ 0 & 1 \end{bmatrix}\begin{bmatrix} b^{-1} & 0\\ 0 & b \end{bmatrix}$$ where each matrix is a product of commutators.

And $\begin{bmatrix} 0 & 1\\ -1 & 0 \end{bmatrix}$ is a product of commutator, indeed

$$\begin{bmatrix} 0 & 1\\ -1 & 0 \end{bmatrix}=\begin{bmatrix} 1 & 1+1\\ 0 & 1 \end{bmatrix}\begin{bmatrix} -1 & 0\\ 1 & 1+1 \end{bmatrix}\begin{bmatrix} 1 & 1+1\\ 0 & 1 \end{bmatrix}^{-1}\begin{bmatrix} -1 & 0\\ 1 & 1+1 \end{bmatrix}^{-1}$$ therefore $SL_2(F)\subset [GL_2(F),GL_2(F)]$.

This is correct? This idea is by José Carlos Santos Commutator Group of $GL_2(R)$ is $SL_2(R)$

eraldcoil
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    usually commutator is written with a comma, not a colon – J. W. Tanner May 04 '20 at 02:39
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    If the result is true (maybe not over all fields though, see https://en.wikipedia.org/wiki/Special_linear_group), it is maybe easier to obtain existence of $B, C$ by a dimensional argument. – KeiOh May 04 '20 at 04:25
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    This is not true when $|F|=2$. It is true otherwise. – Derek Holt May 04 '20 at 07:30
  • with |F|=2 exists a counterexample? o.o – eraldcoil May 04 '20 at 07:53
  • When $|F|=2$, we have ${\rm GL}(2,F) = {\rm SL}(2,F) \cong S_3$, but $[{\rm GL}(2,F) ,{\rm GL}(2,F) ]$ is a proper subgroup of order $3$. – Derek Holt May 04 '20 at 10:37
  • I see. $GL_2(F)=SL_2(F)$ because $ad-bc\neq 0$ impllies $ad-bc=1$ (in $F$) Now, $SL_2(F)$ have 6 elements and $SL_2(F)$ is not abelian then $SL_2(F)\simeq S_3$ But, how prove that $[GL_2(F),GL_2(F)]$ has order 3. – eraldcoil May 08 '20 at 00:39
  • I see, $[S_3,S_3]=A_3$ order 3 then by isomorphism $[GL_2(F),GL_2(F)]$ order 3. – eraldcoil May 08 '20 at 01:12

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