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

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?

• ysharifi.wordpress.com/2011/01/29/… seems to be what you want? disclaimer - I haven't read the proof. See also Jacobson Basic Algebra I, 6.7, lemma 2: $\text{SL}_n(k)$ is its own commutator ($k$ a field), except if $n=2$ and the cardinality of $k\le 3$. I bet other standard algebra books (Lang?) must do this too. – peter a g Jun 6 '17 at 0:07
• @peterag Thanks Peter. I will read the proof. – Hamio Jiang Jun 6 '17 at 1:21
• It is important to understand that this question is equivalent to asking you to prove that ${\rm SL}_n({\mathbb R})$ is the commutators subgroup of ${\rm GL}_n({\mathbb R})$. And in fact this statement is true for all fields in place of ${\mathbb R}$, so there nothing specific about ${\mathbb R}$ involved. – Derek Holt Jun 6 '17 at 8:03
• @peterag Yes Peter, this is the proof I want. The proof is very neat. I think I will post a proof by referring to the material. – Hamio Jiang Jun 6 '17 at 13:47

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$\RightarrowSL_n(\mathbb R)$is subgroup of$ker (f)$. Q.E.D • Just a little, pedantic remark, in reference to your paragraph starting with "Recall that ..." If$G$is a group, and$G'$its commutator subgroup, then$G'$is not usually the set$C$of commutators of$G$; rather, it is the subgroup generated by$C$:$G' = \langle C \rangle \$. – peter a g Jun 7 '17 at 12:42