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Let $G$ be a group of $2 \times 2$ matrices and $N$ be the subgroup of $G$ of matrices with determinant 1. Prove $N=G'$ where $G'$ is commutator subgroup of $G$.

I was able to show $G'$ is subset of $N$ but having difficulty proving converse.

My first approach was to break any $n\in N$ into $g_1g_2g_1^{-1}g_2^{-1}$. Then I tried to use normal subgroup property. For $n\in N$, $n=gn_1g^{-1}$ for some $g\in G$ and $n_1\in N$ as $N$ is normal. Now $nn_1^{-1}=gn_1g^{-1}n_1^{-1}$, this shows $nn_1^-1 \in G'$.

What should be the next step from here?

Aaron Hendrickson
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Arun
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    The commutator subgroup is not the set of commutators, it is the subgroup generates by that set. That is, you may not be able to write every element of $G'$ as a commutator, only as a product of commutators. – paul blart math cop Jun 17 '21 at 19:48
  • Can you please give more hint on how to proceed? – Arun Jun 17 '21 at 20:02
  • You could just say $n=ene^{-1}$. Doesn't really give you anything. – David P Jun 17 '21 at 20:02
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    You won't be able to prove this in complete generality because it fails when the field has 2 elements. That aside, I think the general approach is to use row reduction to decompose a matrix into a product of elementary matrices. – paul blart math cop Jun 17 '21 at 20:06
  • You will have to express elements of $N$ in terms of commutators: Try to find commutators that give you matrices such as [[1,1],[0,1]] – ahulpke Jun 17 '21 at 20:08

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