Try to prove the following proposition.
$SL_n(\mathbb R)$ is a subgroup of kernel of $f\in$ Hom($GL_n(\mathbb R), A)$, where $A$ is an abelian group.
My first attempt is to show that $SL_n(\mathbb R)$ is in form of $ABA^{-1}B^{-1}$, then it's clear that $ABA^{-1}B^{-1}$ is in $ker(f)$. However, there exists an elementary row reduction matrix, i.e. $E_{ij}(m)$ adding m times $j$-th row to $i$-th row. Then $E_{ij}(m)\in SL_n(\mathbb R)$. Now, it is possible that $E_{ij}(m)$ is not in $ker(f)$.
Then I try to find any relation between $\mathbb R^{\times}$ and $A$, but, except knowing they are both abelian groups, I cannot figure out what else connects those two group.
Could you give me some hints?
Commutator Subgp. of GL(n,k)
The main idea of the proof in the above link is to show the commutator subgroup of $GL_n(k)$ is $SL_n(k)$ unless $n=2$ and $|k|\le 3$, we need to show the commutator subgroup of $GL_n(k)$, denoted by $GL'_n(k)$, satisfies $GL'_n(k)\subseteq SL_n(k)$ and $SL_n(k)\subseteq GL'_n(k)$, then $GL'_n(k)=SL_n(k)$.
Notation: $[a,b]:=aba^{-1}b^{-1}$, called commutator. $k$ is a field.
Recall that $GL'_n(k):=\{[a,b]|\ a,b\in GL_n(k)\}$. Then it is clear that any $A\in GL'_n(k),\ A=[a,b]$ for some $a,b\in GL_n(k)$. Then $detA=det([a,b])=det(aba^{-1}b^{-1})=1$. Hence, for any element in $GL'_n(k)$, it is in $SL_n(k)$$\Rightarrow\ GL'_n(k)\subseteq SL_n(k)$.
On the other hand, notice $SL_n(k)$ is generated by all elementary matrices, say $E_{ij}(\alpha):= I+\alpha e_{ij}$ where $\alpha \in k$ and $1\le i \neq j \le n$. Then it suffices to show that any elementary matrix is in $GL'_n(k)$. Here are two cases:
Case 1: $n\ge 3$. Notice that $E_{ij}(\alpha \beta)=[E_{ir}(\alpha),E_{rj}(\beta)]$, since we could let $r\neq i\neq j$ in the case $n\ge 3$. Hence, for $n\ge 3$, $SL_n(k)\subseteq GL'_n(k)$.
Case 2: $n=2$ and $|k| \gt 3$. Then the equation $x(x^2-1)=0$ has at most three solutions in the field $k$, and since $|k|\gt 3$, we can pick another non-zero element such that $x^2\neq 1$, say $\gamma$. Thus $\gamma^2-1$ is invertible in $k$. Now given $\alpha \in k$, let $\beta_1=\alpha (\gamma^2-1)^{-1}$ and $\beta_2=\alpha \gamma (1-\gamma^2)^{-1}$. Let $$A=\begin{pmatrix} \gamma & \\ & \gamma^{-1}\\ \end{pmatrix}$$
Then $E_{12}(\alpha)=[A,E_{12}(\beta_1)]$, and $E_{21}(\alpha)=[A,E_{21}(\beta_2)]$.
Then from case 1 and 2, we can conclude that $SL_n(k)\subseteq GL'_n(k)$ if $n=2$ and $|k|\gt 3$, or $n\ge 3$.
Hence, $SL_n(k)=GL'_n(k)$ unless $n=2$ and $|k|\le 3$.
Here, $k=\mathbb R$, $|\mathbb R|=2^{\aleph_0}\gt 3$. Hence, $SL_n(\mathbb R)=GL'_n(\mathbb R)$. In other words, any element in $SL_n(\mathbb R)$ can be expressed by $[a,b]$for some $a,b\in GL_n(\mathbb R)$. Then $f([a,b])=f(aba^{-1}b^{-1})=f(a)f(a^{-1})f(b)f(b^{-1})=1$. Hence, for any element in $SL_n(\mathbb R)$, it is also in the kernel of homomorphism from $GL_n(\mathbb R)$ to any abelian group$\Rightarrow$ $SL_n(\mathbb R)$ is subgroup of $ker (f)$.
Q.E.D