# Abelianization of $\mathrm{GL}_n(\mathbb{Z})$

$$\DeclareMathOperator\GL{GL}\DeclareMathOperator\SL{SL}$$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 well-known.

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 $$\mathrm{GL}_2(\mathbf{Z})$$.
• 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 $\det$. Commented May 24 at 10:51