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Suppose $k$ is a field.

  1. How to prove that the abelianization of $GL_2(k)$ is $GL_1(k)$ ?

  2. Ditto for $GL_n(k)$.

  3. Can we say the same thing, were we to replace $k$ with $\mathbb Z$ ?

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    $\det$ is a homomorphism from $GL_n(R)\rightarrow k^*=GL_1(R)$ for any commutative ring $R$. To see that it is really an isomorphism we have to show that every matrix with determinant $1$ can be written as a product of commutators. So we have to find a generating system for $SL_n(k)$ and show that each of them is a commutator. This should be possible for Euclidean rings. You might want to read the wikipedia article about algebraic K-theory.2011-09-30

2 Answers 2

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For any field $k$, the abelianization of $GL_n(k)$ is $k^{\ast}$ except for $n=2$ and $k=\mathbb{F}_2$ or $\mathbb{F}_3$. I think this is in Lang's Algebra, Chapter XIII sections 8 and 9. (I say I think because I am relying on google books here, and some key pages are missing.)

EDIT: I just looked back at this old answer, and the abelianization of $GL_2(\mathbb{F}_3)$ is $\mathbb{F}_3^{\times}$ as well, so $GL_2(\mathbb{F}_2)$ is the only counterexample. I think that what I was thinking when I wrote this is that the abelianization of $SL_2(\mathbb{F}_3)$ in nontrivial.

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    Sorry, I should have been clearer that this statement is for fields. As you say, if we knew that $SL_2(\mathbb{Z})$ was its own commutator, then it would follow that $SL_2(\mathbb{F}_p)$ was for every $p$; so this shows that $SL_2(\mathbb{Z})$ is NOT its own commutator.2011-10-01
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As HenrikRueping points out, the determinant map $\det:\textrm{GL}_n(k) \longrightarrow k^\ast$ factors through the abelianization $\textrm{GL}_n(k)_{ab}$ of $\textrm{GL}_n(k)$. The induced morphism $\textrm{GL}_n(k)_{ab} \longrightarrow k^\ast$ can be shown to be an isomorphism as follows. It's clearly surjective. Thus, we have to show injectivity. To do so, you have to show that each element of $\textrm{SL}_n(k)$ is a commutator.

For $n=2$, I thought one could generate $\textrm{SL}_2(k)$ with the two matrices

$\left( \begin{array}{cc} 1& 1 \\ 0 & 1 \end{array}\right)$

$\left( \begin{array}{cc} 0& 1 \\ -1 & 0 \end{array}\right)$

So if you just show that these are commutators you're done, right?

Jim points out that these matrices do not generate $\textrm{SL}_2(k)$ but $\textrm{SL}_2(\mathbf{Z})$.

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    These are generators for $SL_2(\mathbb Z)$, not $k$.2011-09-30