$SL_n(\mathbb R)$ is a subgroup of kernel of $f\in$ Hom$(GL_n(\mathbb R), A)$, $A$ is an abelian group.

Question

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?

Answer 1

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