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$SL_n(mathbb R)$ is a subgroup of kernel of $fin$ Hom$(GL_n(mathbb R), A)$, $A$ is an abelian group.
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Try to prove the following proposition.
$SL_n(mathbb R)$ is a subgroup of kernel of $fin$ 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 $ABA^-1B^-1$, then it's clear that $ABA^-1B^-1$ 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?
group-theory abelian-groups group-homomorphism
asked Jun 5 '17 at 23:06
Hamio Jiang
18310
add a comment |Â
up vote
4
down vote
favorite
Try to prove the following proposition.
$SL_n(mathbb R)$ is a subgroup of kernel of $fin$ 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 $ABA^-1B^-1$, then it's clear that $ABA^-1B^-1$ 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?
group-theory abelian-groups group-homomorphism
asked Jun 5 '17 at 23:06
Hamio Jiang
18310
2
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: $textSL_n(k)$ is its own commutator ($k$ a field), except if $n=2$ and the cardinality of $kle 3$. 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
1
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
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Try to prove the following proposition.
$SL_n(mathbb R)$ is a subgroup of kernel of $fin$ 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 $ABA^-1B^-1$, then it's clear that $ABA^-1B^-1$ 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?
group-theory abelian-groups group-homomorphism
asked Jun 5 '17 at 23:06
Hamio Jiang
18310
Try to prove the following proposition.
$SL_n(mathbb R)$ is a subgroup of kernel of $fin$ 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 $ABA^-1B^-1$, then it's clear that $ABA^-1B^-1$ 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?
group-theory abelian-groups group-homomorphism
asked Jun 5 '17 at 23:06
Hamio Jiang
18310
asked Jun 5 '17 at 23:06
Hamio Jiang
18310
asked Jun 5 '17 at 23:06
Hamio Jiang
18310
asked Jun 5 '17 at 23:06
Hamio Jiang
18310
18310
2
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: $textSL_n(k)$ is its own commutator ($k$ a field), except if $n=2$ and the cardinality of $kle 3$. 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
1
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
add a comment |Â
2
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: $textSL_n(k)$ is its own commutator ($k$ a field), except if $n=2$ and the cardinality of $kle 3$. 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
1
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
2
2
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: $textSL_n(k)$ is its own commutator ($k$ a field), except if $n=2$ and the cardinality of $kle 3$. I bet other standard algebra books (Lang?) must do this too.
– peter a g
Jun 6 '17 at 0:07
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: $textSL_n(k)$ is its own commutator ($k$ a field), except if $n=2$ and the cardinality of $kle 3$. 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
@peterag Thanks Peter. I will read the proof.
– Hamio Jiang
Jun 6 '17 at 1:21
1
1
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
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
@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
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
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: $[a,b]:=aba^-1b^-1$, called commutator. $k$ is a field.
Recall that $GL'_n(k):= a,bin GL_n(k)$. Then it is clear that any $Ain GL'_n(k), A=[a,b]$ for some $a,bin GL_n(k)$. Then $detA=det([a,b])=det(aba^-1b^-1)=1$. Hence, for any element in $GL'_n(k)$, it is in $SL_n(k)$$Rightarrow GL'_n(k)subseteq SL_n(k)$.
On the other hand, notice $SL_n(k)$ is generated by all elementary matrices, say $E_ij(alpha):= I+alpha e_ij$ where $alpha in k$ and $1le i neq j le n$. Then it suffices to show that any elementary matrix is in $GL'_n(k)$. Here are two cases:
Case 1: $nge 3$. Notice that $E_ij(alpha beta)=[E_ir(alpha),E_rj(beta)]$, since we could let $rneq ineq j$ in the case $nge 3$. Hence, for $nge 3$, $SL_n(k)subseteq GL'_n(k)$.
Case 2: $n=2$ and $|k| gt 3$. Then the equation $x(x^2-1)=0$ has at most three solutions in the field $k$, and since $|k|gt 3$, we can pick another non-zero element such that $x^2neq 1$, say $gamma$. Thus $gamma^2-1$ is invertible in $k$. Now given $alpha in k$, let $beta_1=alpha (gamma^2-1)^-1$ and $beta_2=alpha gamma (1-gamma^2)^-1$. Let
$$A=beginpmatrix
gamma & \
& gamma^-1\
endpmatrix$$
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 $nge 3$.
Hence, $SL_n(k)=GL'_n(k)$ unless $n=2$ and $|k|le 3$.
Here, $k=mathbb R$, $|mathbb R|=2^aleph_0gt 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,bin GL_n(mathbb R)$. Then $f([a,b])=f(aba^-1b^-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
answered Jun 6 '17 at 15:04
Hamio Jiang
18310
1
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
@peterag Thanks. You revealed the hidden truth I ignored. I just keep the notation from the link you gave. For general linear groups, this notation is fine, but others may not work.
– Hamio Jiang
Jun 7 '17 at 13:16
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
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^-1b^-1$, called commutator. $k$ is a field.
Recall that $GL'_n(k):= a,bin GL_n(k)$. Then it is clear that any $Ain GL'_n(k), A=[a,b]$ for some $a,bin GL_n(k)$. Then $detA=det([a,b])=det(aba^-1b^-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 $1le i neq j le n$. Then it suffices to show that any elementary matrix is in $GL'_n(k)$. Here are two cases:
Case 1: $nge 3$. Notice that $E_ij(alpha beta)=[E_ir(alpha),E_rj(beta)]$, since we could let $rneq ineq j$ in the case $nge 3$. Hence, for $nge 3$, $SL_n(k)subseteq GL'_n(k)$.
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 $x^2neq 1$, 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
$$A=beginpmatrix
gamma & \
& gamma^-1\
endpmatrix$$
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 $nge 3$.
Hence, $SL_n(k)=GL'_n(k)$ unless $n=2$ and $|k|le 3$.
Here, $k=mathbb R$, $|mathbb R|=2^aleph_0gt 3$. 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 $a,bin GL_n(mathbb R)$. Then $f([a,b])=f(aba^-1b^-1)=f(a)f(a^-1)f(b)f(b^-1)=1$. 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
answered Jun 6 '17 at 15:04
Hamio Jiang
18310
1
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
@peterag Thanks. You revealed the hidden truth I ignored. I just keep the notation from the link you gave. For general linear groups, this notation is fine, but others may not work.
– Hamio Jiang
Jun 7 '17 at 13:16
add a comment |Â
up vote
1
down vote
accepted
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: $[a,b]:=aba^-1b^-1$, called commutator. $k$ is a field.
Recall that $GL'_n(k):= a,bin GL_n(k)$. Then it is clear that any $Ain GL'_n(k), A=[a,b]$ for some $a,bin GL_n(k)$. Then $detA=det([a,b])=det(aba^-1b^-1)=1$. Hence, for any element in $GL'_n(k)$, it is in $SL_n(k)$$Rightarrow GL'_n(k)subseteq SL_n(k)$.
On the other hand, notice $SL_n(k)$ is generated by all elementary matrices, say $E_ij(alpha):= I+alpha e_ij$ where $alpha in k$ and $1le i neq j le n$. Then it suffices to show that any elementary matrix is in $GL'_n(k)$. Here are two cases:
Case 1: $nge 3$. Notice that $E_ij(alpha beta)=[E_ir(alpha),E_rj(beta)]$, since we could let $rneq ineq j$ in the case $nge 3$. Hence, for $nge 3$, $SL_n(k)subseteq GL'_n(k)$.
Case 2: $n=2$ and $|k| gt 3$. Then the equation $x(x^2-1)=0$ has at most three solutions in the field $k$, and since $|k|gt 3$, we can pick another non-zero element such that $x^2neq 1$, say $gamma$. Thus $gamma^2-1$ is invertible in $k$. Now given $alpha in k$, let $beta_1=alpha (gamma^2-1)^-1$ and $beta_2=alpha gamma (1-gamma^2)^-1$. Let
$$A=beginpmatrix
gamma & \
& gamma^-1\
endpmatrix$$
Then $E_12(alpha)=[A,E_12(beta_1)]$, and $E_21(alpha)=[A,E_21(beta_2)]$.
Then from case 1 and 2, we can conclude that $SL_n(k)subseteq GL'_n(k)$ if $n=2$ and $|k|gt 3$, or $nge 3$.
Hence, $SL_n(k)=GL'_n(k)$ unless $n=2$ and $|k|le 3$.
Here, $k=mathbb R$, $|mathbb R|=2^aleph_0gt 3$. Hence, $SL_n(mathbb R)=GL'_n(mathbb R)$. In other words, any element in $SL_n(mathbb R)$ can be expressed by $[a,b]$for some $a,bin GL_n(mathbb R)$. Then $f([a,b])=f(aba^-1b^-1)=f(a)f(a^-1)f(b)f(b^-1)=1$. Hence, for any element in $SL_n(mathbb R)$, it is also in the kernel of homomorphism from $GL_n(mathbb R)$ to any abelian group$Rightarrow$ $SL_n(mathbb R)$ is subgroup of $ker (f)$.
Q.E.D
answered Jun 6 '17 at 15:04
Hamio Jiang
18310
1
Just a little, pedantic remark, in reference to your paragraph starting with "Recall that ..." If $G$ is a group, and $G'$ its commutator subgroup, then $G'$ is not usually the set $C$ of commutators of $G$; rather, it is the subgroup generated by $C$: $G' = langle C rangle $.
– peter a g
Jun 7 '17 at 12:42
@peterag Thanks. You revealed the hidden truth I ignored. I just keep the notation from the link you gave. For general linear groups, this notation is fine, but others may not work.
– Hamio Jiang
Jun 7 '17 at 13:16
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Commutator Subgp. of GL(n,k)
The main idea of the proof in the above link is to show the commutator subgroup of $GL_n(k)$ is $SL_n(k)$ unless $n=2$ and $|k|le 3$, we need to show the commutator subgroup of $GL_n(k)$, denoted by $GL'_n(k)$, satisfies $GL'_n(k)subseteq SL_n(k)$ and $SL_n(k)subseteq GL'_n(k)$, then $GL'_n(k)=SL_n(k)$.
Notation: $[a,b]:=aba^-1b^-1$, called commutator. $k$ is a field.
Recall that $GL'_n(k):= a,bin GL_n(k)$. Then it is clear that any $Ain GL'_n(k), A=[a,b]$ for some $a,bin GL_n(k)$. Then $detA=det([a,b])=det(aba^-1b^-1)=1$. Hence, for any element in $GL'_n(k)$, it is in $SL_n(k)$$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 $1le i neq j le n$. Then it suffices to show that any elementary matrix is in $GL'_n(k)$. Here are two cases:
Case 1: $nge 3$. Notice that $E_ij(alpha beta)=[E_ir(alpha),E_rj(beta)]$, since we could let $rneq ineq j$ in the case $nge 3$. Hence, for $nge 3$, $SL_n(k)subseteq GL'_n(k)$.
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 $x^2neq 1$, 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
$$A=beginpmatrix
gamma & \
& gamma^-1\
endpmatrix$$
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 $nge 3$.
Hence, $SL_n(k)=GL'_n(k)$ unless $n=2$ and $|k|le 3$.
Here, $k=mathbb R$, $|mathbb R|=2^aleph_0gt 3$. 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 $a,bin GL_n(mathbb R)$. Then $f([a,b])=f(aba^-1b^-1)=f(a)f(a^-1)f(b)f(b^-1)=1$. 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
answered Jun 6 '17 at 15:04
Hamio Jiang
18310
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: $[a,b]:=aba^-1b^-1$, called commutator. $k$ is a field.
Recall that $GL'_n(k):= a,bin GL_n(k)$. Then it is clear that any $Ain GL'_n(k), A=[a,b]$ for some $a,bin GL_n(k)$. Then $detA=det([a,b])=det(aba^-1b^-1)=1$. Hence, for any element in $GL'_n(k)$, it is in $SL_n(k)$$Rightarrow GL'_n(k)subseteq SL_n(k)$.
On the other hand, notice $SL_n(k)$ is generated by all elementary matrices, say $E_ij(alpha):= I+alpha e_ij$ where $alpha in k$ and $1le i neq j le n$. Then it suffices to show that any elementary matrix is in $GL'_n(k)$. Here are two cases:
Case 1: $nge 3$. Notice that $E_ij(alpha beta)=[E_ir(alpha),E_rj(beta)]$, since we could let $rneq ineq j$ in the case $nge 3$. Hence, for $nge 3$, $SL_n(k)subseteq GL'_n(k)$.
Case 2: $n=2$ and $|k| gt 3$. Then the equation $x(x^2-1)=0$ has at most three solutions in the field $k$, and since $|k|gt 3$, we can pick another non-zero element such that $x^2neq 1$, say $gamma$. Thus $gamma^2-1$ is invertible in $k$. Now given $alpha in k$, let $beta_1=alpha (gamma^2-1)^-1$ and $beta_2=alpha gamma (1-gamma^2)^-1$. Let
$$A=beginpmatrix
gamma & \
& gamma^-1\
endpmatrix$$
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 $nge 3$.
Hence, $SL_n(k)=GL'_n(k)$ unless $n=2$ and $|k|le 3$.
Here, $k=mathbb R$, $|mathbb R|=2^aleph_0gt 3$. 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 $a,bin GL_n(mathbb R)$. Then $f([a,b])=f(aba^-1b^-1)=f(a)f(a^-1)f(b)f(b^-1)=1$. 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
answered Jun 6 '17 at 15:04
Hamio Jiang
18310
answered Jun 6 '17 at 15:04
Hamio Jiang
18310
answered Jun 6 '17 at 15:04
Hamio Jiang
18310
answered Jun 6 '17 at 15:04
Hamio Jiang
18310
18310
1
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
@peterag Thanks. You revealed the hidden truth I ignored. I just keep the notation from the link you gave. For general linear groups, this notation is fine, but others may not work.
– Hamio Jiang
Jun 7 '17 at 13:16
add a comment |Â
1
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
@peterag Thanks. You revealed the hidden truth I ignored. I just keep the notation from the link you gave. For general linear groups, this notation is fine, but others may not work.
– Hamio Jiang
Jun 7 '17 at 13:16
1
1
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
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
@peterag Thanks. You revealed the hidden truth I ignored. I just keep the notation from the link you gave. For general linear groups, this notation is fine, but others may not work.
– Hamio Jiang
Jun 7 '17 at 13:16
@peterag Thanks. You revealed the hidden truth I ignored. I just keep the notation from the link you gave. For general linear groups, this notation is fine, but others may not work.
– Hamio Jiang
Jun 7 '17 at 13:16
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2
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: $textSL_n(k)$ is its own commutator ($k$ a field), except if $n=2$ and the cardinality of $kle 3$. 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
1
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