$\DeclareMathOperator\GL{GL}\DeclareMathOperator\SL{SL}$What is the abelianization of $\GL_n(\mathbb{Z})$? I know the abelianzation of $\GL_n(\mathbb{F})$ where $\mathbb{F}$ is a field and the abelianization of $ \SL_n(\mathbb{Z})$. However I can't find the abelianization of $\GL_n(\mathbb{Z})$ and I suspect this is something that should be wellknown.
1 Answer
For $n=1$ and $n\ge 3$ $\mathrm{SL}_n(\mathbf{Z})$ is perfect and hence the abelianization has order 2, given by the $\pm 1$ determinant.
For $n=2$, $I_2$ is a commutator so this is the same as the abelianization of $\mathrm{PGL}_2(\mathbf{Z})$, which is an amalgam $(C_2\ltimes C_3)\ast (C_2\times C_2)$ (with the order 6 dihedral group on the left) and its abelianization is the Klein group $C_2\times C_2$.
For $n=2$ one can explicitly define the abelianization map onto $\{\pm 1\}\times\mathbf{Z}/2\mathbf{Z}$ as $A\mapsto (\det(A),\psi(A))$, where $\psi\left(\begin{pmatrix}a & b \\ c & d\end{pmatrix}\right)=bc\pmod 2$. The latter just reflects the abelianization of $\mathrm{GL}_2(\mathbf{Z}/2\mathbf{Z})$, which is dihedral of order 6. By the way this is also the abelianization of the quotient $\mathrm{GL}_2(\mathbf{Z}/4\mathbf{Z})$ of $\mathrm{GL}_2(\mathbf{Z})$.

5$\begingroup$ Another way to describe $\psi$ is that $\text{GL}_2(\mathbb{Z})$ acts on the three element set $\mathbb{P}^1(\mathbb{F}_2)$, so we get a map $\text{GL}_2(\mathbb{Z}) \to S_3$, and $\psi$ is the composite of this with the sign map $S_3 \to \mathbb{Z}/2 \mathbb{Z}$. Only works for $p=2$; for every other prime, $g \in \text{GL}_2(\mathbb{F}_p)$ acts on $\mathbb{P}^1(\mathbb{F}_p)$ by a permutation with sign $\left( \frac{\det(g)}{p} \right)$ (this is the Legendre symbol), so it is redundant with $\det$. $\endgroup$ Commented May 24 at 10:51