The abelianization of the general linear group $GL(n,\mathbb{R})$, defined by $$GL(n,\mathbb{R})^{ab} := GL(n,\mathbb{R})/[GL(n,\mathbb{R}), GL(n,\mathbb{R})],$$ is isomorphic to $\mathbb{R}^{\times}$. This follows from the fact that $[GL(n,\mathbb{R}),GL(n,\mathbb{R})] \cong SL(n,\mathbb{R})$, so that $GL(n,\mathbb{R})^{ab} \cong \mathbb{R}^{\times}$ by the first isomorphism theorem.


Since for any group $G$, the homology group $H\_1(G;\mathbb{Z}) \cong G^{ab}$, it turns out that $H\_1(GL(n,\mathbb{R});\mathbb{Z}) \cong \mathbb{R}^{\times}$.


I wonder if it possible to obtain this result by directly computing $H\_1(GL(n,\mathbb{R});\mathbb{Z})$ as the quotient $Z\_1(GL(n,\mathbb{R}))/B\_1(GL(n,\mathbb{R}))$ of 1-cycles by 1-boundaries.



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> What do $C\_1$, $Z\_1$ and $B\_1$ look like for $GL(n,\mathbb{R})$? Is a direct computation of $H\_1(GL(n,\mathbb{R}))$ possible?
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In case this question turns out to be too elementary for MO, I apologize in advance. In that case I will quickly move my post to MSE. 



First homology group of the general linear group
This more of a comment, but here it goes anyway.


Following Brown's book *[Cohomology of Groups](https://link.springer.com/book/10.1007/978-1-4684-9327-6)* on page 36, we see that with respect to the standard resolution and any group $G$: $$C\_2(G)\xrightarrow{\partial}C\_1(G)\xrightarrow{0}\mathbb{Z}\to 0,$$ where $\partial[g|h]=[h]-[gh]+[g]$.


Consequently, $Z\_1(G)=C\_1(G)=\{[g]\ |\ g\in G\} $ and $B\_1(G)=\{[h]-[gh]+[g]\ |\ h,g\in G\}$.


So $H\_1(G)\cong G/[G,G]$. I am not sure what more you could expect in any special case of this. 


In the case of a complex reductive (affine algebraic) group $G$, the quotient $G/[G,G]$ will always be isomorphic to an algebraic torus (isomorphic to a product of $\mathrm{GL}\_1$'s), but that has nothing to do with group cohomology (it's the central isogeny theorem; see Milne's book *[Algebraic Groups](https://www.cambridge.org/core/books/algebraic-groups/48D122BAEC63C2B70304E20F34482C8F)* for example). A special case of this general fact is $G=\mathrm{GL}\_n(\mathbb{C})$ where $G/[G,G]\cong \mathbb{C}^\*$. The fact that you get $\mathbb{R}^\*$ for $G=\mathrm{GL}\_n(\mathbb{R})$ is analogous to this case ([in fact it is true for any field](https://groupprops.subwiki.org/wiki/Derived_subgroup_of_general_linear_group_is_special_linear_group)).


