TITLE: Abelianization of GL_n
QUESTION [12 upvotes]: 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 $\mathbb Z$ ?

REPLY [5 votes]: 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.