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.

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?

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.