Abelianization of GL_n Suppose $k$ is a field. 


How to prove that the abelianization of $GL_2(k)$ is $GL_1(k)$ ?
Ditto for $GL_n(k)$.
Can we say the same thing, were we to replace $k$ with $\mathbb Z$ ?
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.
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 

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

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

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 $\textrm{SL}_2(\mathbf{Z})$.