Show that $G/Hcongmathbb{R}^*$.
Multi tool use
$begingroup$
Let $G:=
bigg{left( begin{array}{ccc}
a & b \
0 & a \
end{array} right)Bigg| a,b in mathbb{R},ane 0bigg}$. Let $H:=
bigg{left( begin{array}{ccc}
1 & b \
0 & 1 \
end{array} right) Bigg| b in mathbb{R}bigg}$.
I know that $H$ is normal subgroup of $G$. I need to prove that $G/Hcongmathbb{R}^*$.
My attempt:
$$G/H={gHmid gin G}$$
$$
left( begin{array}{ccc}
a & b \
0 & a \
end{array} right)
$$
$$
left( begin{array}{ccc}
1 & b \
0 & 1
end{array} right)
=
left( begin{array}{ccc}
a & ab+b \
0 & a
end{array} right)
$$
And there are $infty$ solutions from the form :$
left( begin{array}{ccc}
a & ab \
0 & a
end{array} right)
$.
$$
begin{vmatrix}
left( begin{array}{ccc}
a & ab \
0 & a
end{array} right)end{vmatrix}=mathfrak{c}=|mathbb{R}^*|
$$
$$Rightarrow G/Ncong mathbb{R}^*$$
Is it correct? is there another way to solve this?
abstract-algebra matrices group-theory group-isomorphism infinite-groups
$endgroup$
add a comment |
$begingroup$
Let $G:=
bigg{left( begin{array}{ccc}
a & b \
0 & a \
end{array} right)Bigg| a,b in mathbb{R},ane 0bigg}$. Let $H:=
bigg{left( begin{array}{ccc}
1 & b \
0 & 1 \
end{array} right) Bigg| b in mathbb{R}bigg}$.
I know that $H$ is normal subgroup of $G$. I need to prove that $G/Hcongmathbb{R}^*$.
My attempt:
$$G/H={gHmid gin G}$$
$$
left( begin{array}{ccc}
a & b \
0 & a \
end{array} right)
$$
$$
left( begin{array}{ccc}
1 & b \
0 & 1
end{array} right)
=
left( begin{array}{ccc}
a & ab+b \
0 & a
end{array} right)
$$
And there are $infty$ solutions from the form :$
left( begin{array}{ccc}
a & ab \
0 & a
end{array} right)
$.
$$
begin{vmatrix}
left( begin{array}{ccc}
a & ab \
0 & a
end{array} right)end{vmatrix}=mathfrak{c}=|mathbb{R}^*|
$$
$$Rightarrow G/Ncong mathbb{R}^*$$
Is it correct? is there another way to solve this?
abstract-algebra matrices group-theory group-isomorphism infinite-groups
$endgroup$
1
$begingroup$
No it is not correct. Knowing that to groups have the same cardinality is not enough to prove that they are isomorphic (for example $(mathbb{R},+)$ and $(mathbb{R}^*,times)$ have the same cardinality, yet they are not isomorphic. By the way, there is a mistake in your matrix multiplication.
$endgroup$
– Roland
Jul 7 '15 at 13:40
$begingroup$
edited the matix multiplication
$endgroup$
– 3SAT
Jul 7 '15 at 13:44
$begingroup$
Is $R^*$ the group of real numbers under multiplication?
$endgroup$
– man_in_green_shirt
Jul 7 '15 at 13:47
1
$begingroup$
yes, multiplication without zero
$endgroup$
– 3SAT
Jul 7 '15 at 13:48
add a comment |
$begingroup$
Let $G:=
bigg{left( begin{array}{ccc}
a & b \
0 & a \
end{array} right)Bigg| a,b in mathbb{R},ane 0bigg}$. Let $H:=
bigg{left( begin{array}{ccc}
1 & b \
0 & 1 \
end{array} right) Bigg| b in mathbb{R}bigg}$.
I know that $H$ is normal subgroup of $G$. I need to prove that $G/Hcongmathbb{R}^*$.
My attempt:
$$G/H={gHmid gin G}$$
$$
left( begin{array}{ccc}
a & b \
0 & a \
end{array} right)
$$
$$
left( begin{array}{ccc}
1 & b \
0 & 1
end{array} right)
=
left( begin{array}{ccc}
a & ab+b \
0 & a
end{array} right)
$$
And there are $infty$ solutions from the form :$
left( begin{array}{ccc}
a & ab \
0 & a
end{array} right)
$.
$$
begin{vmatrix}
left( begin{array}{ccc}
a & ab \
0 & a
end{array} right)end{vmatrix}=mathfrak{c}=|mathbb{R}^*|
$$
$$Rightarrow G/Ncong mathbb{R}^*$$
Is it correct? is there another way to solve this?
abstract-algebra matrices group-theory group-isomorphism infinite-groups
$endgroup$
Let $G:=
bigg{left( begin{array}{ccc}
a & b \
0 & a \
end{array} right)Bigg| a,b in mathbb{R},ane 0bigg}$. Let $H:=
bigg{left( begin{array}{ccc}
1 & b \
0 & 1 \
end{array} right) Bigg| b in mathbb{R}bigg}$.
I know that $H$ is normal subgroup of $G$. I need to prove that $G/Hcongmathbb{R}^*$.
My attempt:
$$G/H={gHmid gin G}$$
$$
left( begin{array}{ccc}
a & b \
0 & a \
end{array} right)
$$
$$
left( begin{array}{ccc}
1 & b \
0 & 1
end{array} right)
=
left( begin{array}{ccc}
a & ab+b \
0 & a
end{array} right)
$$
And there are $infty$ solutions from the form :$
left( begin{array}{ccc}
a & ab \
0 & a
end{array} right)
$.
$$
begin{vmatrix}
left( begin{array}{ccc}
a & ab \
0 & a
end{array} right)end{vmatrix}=mathfrak{c}=|mathbb{R}^*|
$$
$$Rightarrow G/Ncong mathbb{R}^*$$
Is it correct? is there another way to solve this?
abstract-algebra matrices group-theory group-isomorphism infinite-groups
abstract-algebra matrices group-theory group-isomorphism infinite-groups
edited Dec 14 '18 at 20:47
Batominovski
33.1k33293
33.1k33293
asked Jul 7 '15 at 13:36
3SAT3SAT
6,35121334
6,35121334
1
$begingroup$
No it is not correct. Knowing that to groups have the same cardinality is not enough to prove that they are isomorphic (for example $(mathbb{R},+)$ and $(mathbb{R}^*,times)$ have the same cardinality, yet they are not isomorphic. By the way, there is a mistake in your matrix multiplication.
$endgroup$
– Roland
Jul 7 '15 at 13:40
$begingroup$
edited the matix multiplication
$endgroup$
– 3SAT
Jul 7 '15 at 13:44
$begingroup$
Is $R^*$ the group of real numbers under multiplication?
$endgroup$
– man_in_green_shirt
Jul 7 '15 at 13:47
1
$begingroup$
yes, multiplication without zero
$endgroup$
– 3SAT
Jul 7 '15 at 13:48
add a comment |
1
$begingroup$
No it is not correct. Knowing that to groups have the same cardinality is not enough to prove that they are isomorphic (for example $(mathbb{R},+)$ and $(mathbb{R}^*,times)$ have the same cardinality, yet they are not isomorphic. By the way, there is a mistake in your matrix multiplication.
$endgroup$
– Roland
Jul 7 '15 at 13:40
$begingroup$
edited the matix multiplication
$endgroup$
– 3SAT
Jul 7 '15 at 13:44
$begingroup$
Is $R^*$ the group of real numbers under multiplication?
$endgroup$
– man_in_green_shirt
Jul 7 '15 at 13:47
1
$begingroup$
yes, multiplication without zero
$endgroup$
– 3SAT
Jul 7 '15 at 13:48
1
1
$begingroup$
No it is not correct. Knowing that to groups have the same cardinality is not enough to prove that they are isomorphic (for example $(mathbb{R},+)$ and $(mathbb{R}^*,times)$ have the same cardinality, yet they are not isomorphic. By the way, there is a mistake in your matrix multiplication.
$endgroup$
– Roland
Jul 7 '15 at 13:40
$begingroup$
No it is not correct. Knowing that to groups have the same cardinality is not enough to prove that they are isomorphic (for example $(mathbb{R},+)$ and $(mathbb{R}^*,times)$ have the same cardinality, yet they are not isomorphic. By the way, there is a mistake in your matrix multiplication.
$endgroup$
– Roland
Jul 7 '15 at 13:40
$begingroup$
edited the matix multiplication
$endgroup$
– 3SAT
Jul 7 '15 at 13:44
$begingroup$
edited the matix multiplication
$endgroup$
– 3SAT
Jul 7 '15 at 13:44
$begingroup$
Is $R^*$ the group of real numbers under multiplication?
$endgroup$
– man_in_green_shirt
Jul 7 '15 at 13:47
$begingroup$
Is $R^*$ the group of real numbers under multiplication?
$endgroup$
– man_in_green_shirt
Jul 7 '15 at 13:47
1
1
$begingroup$
yes, multiplication without zero
$endgroup$
– 3SAT
Jul 7 '15 at 13:48
$begingroup$
yes, multiplication without zero
$endgroup$
– 3SAT
Jul 7 '15 at 13:48
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
I don't quite understand your solution, but here is another way to do it:
You can get the isomorphism from the First Isomorphism Theorem. For example, consider the map
$$
phi: G to mathbb{R}^times
$$
given by
$$
phipmatrix{a & b \ 0 & a} = a.
$$
You have to show that this map
- is a homomorphism,
- is surjective,
- has $H$ as kernel.
Then the First Isomorhpism Theorem will tell you that $G / H simeq mathbb{R}^times$.
If you don't know the First Isomorphism Theorem, you can define the map
$$
psi: G/H to mathbb{R}^times
$$
by
$$
psi pmatrix{a & b \ 0 & a}H = a
$$
and show that this map is
- well defined,
- a homomorphism,
3, surjective, - injective.
(Is you do this, then you pretty much prove the First Isomorphism Theorem in this special case.)
$endgroup$
add a comment |
$begingroup$
How about considering the map $varphi:Gtomathbb{R}^*$ sending $begin{bmatrix}a&b\0&aend{bmatrix}mapsto a$ for all $ainmathbb{R}^*$ and $binmathbb{R}$? Why is it a group homomorphism? What are the kernel and the image of $varphi$?
$endgroup$
add a comment |
$begingroup$
Here is the "backwards" proof:
Suppose $A = begin{bmatrix}a&b\0&aend{bmatrix},A' = begin{bmatrix}a'&b'\0&a'end{bmatrix}$.
Then $AH = A'H iff a = a'$, since $AA'^{-1} = begin{bmatrix}a/a'&(-ab')/a'^2+b/a'\0&a/a'end{bmatrix}$
(Note how we use the fact that $a' neq 0$).
Note as well that $AH = (aI)H$.
Thus $a mapsto (aI)H$ is the desired isomorphism.
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I don't quite understand your solution, but here is another way to do it:
You can get the isomorphism from the First Isomorphism Theorem. For example, consider the map
$$
phi: G to mathbb{R}^times
$$
given by
$$
phipmatrix{a & b \ 0 & a} = a.
$$
You have to show that this map
- is a homomorphism,
- is surjective,
- has $H$ as kernel.
Then the First Isomorhpism Theorem will tell you that $G / H simeq mathbb{R}^times$.
If you don't know the First Isomorphism Theorem, you can define the map
$$
psi: G/H to mathbb{R}^times
$$
by
$$
psi pmatrix{a & b \ 0 & a}H = a
$$
and show that this map is
- well defined,
- a homomorphism,
3, surjective, - injective.
(Is you do this, then you pretty much prove the First Isomorphism Theorem in this special case.)
$endgroup$
add a comment |
$begingroup$
I don't quite understand your solution, but here is another way to do it:
You can get the isomorphism from the First Isomorphism Theorem. For example, consider the map
$$
phi: G to mathbb{R}^times
$$
given by
$$
phipmatrix{a & b \ 0 & a} = a.
$$
You have to show that this map
- is a homomorphism,
- is surjective,
- has $H$ as kernel.
Then the First Isomorhpism Theorem will tell you that $G / H simeq mathbb{R}^times$.
If you don't know the First Isomorphism Theorem, you can define the map
$$
psi: G/H to mathbb{R}^times
$$
by
$$
psi pmatrix{a & b \ 0 & a}H = a
$$
and show that this map is
- well defined,
- a homomorphism,
3, surjective, - injective.
(Is you do this, then you pretty much prove the First Isomorphism Theorem in this special case.)
$endgroup$
add a comment |
$begingroup$
I don't quite understand your solution, but here is another way to do it:
You can get the isomorphism from the First Isomorphism Theorem. For example, consider the map
$$
phi: G to mathbb{R}^times
$$
given by
$$
phipmatrix{a & b \ 0 & a} = a.
$$
You have to show that this map
- is a homomorphism,
- is surjective,
- has $H$ as kernel.
Then the First Isomorhpism Theorem will tell you that $G / H simeq mathbb{R}^times$.
If you don't know the First Isomorphism Theorem, you can define the map
$$
psi: G/H to mathbb{R}^times
$$
by
$$
psi pmatrix{a & b \ 0 & a}H = a
$$
and show that this map is
- well defined,
- a homomorphism,
3, surjective, - injective.
(Is you do this, then you pretty much prove the First Isomorphism Theorem in this special case.)
$endgroup$
I don't quite understand your solution, but here is another way to do it:
You can get the isomorphism from the First Isomorphism Theorem. For example, consider the map
$$
phi: G to mathbb{R}^times
$$
given by
$$
phipmatrix{a & b \ 0 & a} = a.
$$
You have to show that this map
- is a homomorphism,
- is surjective,
- has $H$ as kernel.
Then the First Isomorhpism Theorem will tell you that $G / H simeq mathbb{R}^times$.
If you don't know the First Isomorphism Theorem, you can define the map
$$
psi: G/H to mathbb{R}^times
$$
by
$$
psi pmatrix{a & b \ 0 & a}H = a
$$
and show that this map is
- well defined,
- a homomorphism,
3, surjective, - injective.
(Is you do this, then you pretty much prove the First Isomorphism Theorem in this special case.)
answered Jul 7 '15 at 13:41
ThomasThomas
35.6k1058117
35.6k1058117
add a comment |
add a comment |
$begingroup$
How about considering the map $varphi:Gtomathbb{R}^*$ sending $begin{bmatrix}a&b\0&aend{bmatrix}mapsto a$ for all $ainmathbb{R}^*$ and $binmathbb{R}$? Why is it a group homomorphism? What are the kernel and the image of $varphi$?
$endgroup$
add a comment |
$begingroup$
How about considering the map $varphi:Gtomathbb{R}^*$ sending $begin{bmatrix}a&b\0&aend{bmatrix}mapsto a$ for all $ainmathbb{R}^*$ and $binmathbb{R}$? Why is it a group homomorphism? What are the kernel and the image of $varphi$?
$endgroup$
add a comment |
$begingroup$
How about considering the map $varphi:Gtomathbb{R}^*$ sending $begin{bmatrix}a&b\0&aend{bmatrix}mapsto a$ for all $ainmathbb{R}^*$ and $binmathbb{R}$? Why is it a group homomorphism? What are the kernel and the image of $varphi$?
$endgroup$
How about considering the map $varphi:Gtomathbb{R}^*$ sending $begin{bmatrix}a&b\0&aend{bmatrix}mapsto a$ for all $ainmathbb{R}^*$ and $binmathbb{R}$? Why is it a group homomorphism? What are the kernel and the image of $varphi$?
edited Dec 14 '18 at 20:51
answered Jul 7 '15 at 13:40
BatominovskiBatominovski
33.1k33293
33.1k33293
add a comment |
add a comment |
$begingroup$
Here is the "backwards" proof:
Suppose $A = begin{bmatrix}a&b\0&aend{bmatrix},A' = begin{bmatrix}a'&b'\0&a'end{bmatrix}$.
Then $AH = A'H iff a = a'$, since $AA'^{-1} = begin{bmatrix}a/a'&(-ab')/a'^2+b/a'\0&a/a'end{bmatrix}$
(Note how we use the fact that $a' neq 0$).
Note as well that $AH = (aI)H$.
Thus $a mapsto (aI)H$ is the desired isomorphism.
$endgroup$
add a comment |
$begingroup$
Here is the "backwards" proof:
Suppose $A = begin{bmatrix}a&b\0&aend{bmatrix},A' = begin{bmatrix}a'&b'\0&a'end{bmatrix}$.
Then $AH = A'H iff a = a'$, since $AA'^{-1} = begin{bmatrix}a/a'&(-ab')/a'^2+b/a'\0&a/a'end{bmatrix}$
(Note how we use the fact that $a' neq 0$).
Note as well that $AH = (aI)H$.
Thus $a mapsto (aI)H$ is the desired isomorphism.
$endgroup$
add a comment |
$begingroup$
Here is the "backwards" proof:
Suppose $A = begin{bmatrix}a&b\0&aend{bmatrix},A' = begin{bmatrix}a'&b'\0&a'end{bmatrix}$.
Then $AH = A'H iff a = a'$, since $AA'^{-1} = begin{bmatrix}a/a'&(-ab')/a'^2+b/a'\0&a/a'end{bmatrix}$
(Note how we use the fact that $a' neq 0$).
Note as well that $AH = (aI)H$.
Thus $a mapsto (aI)H$ is the desired isomorphism.
$endgroup$
Here is the "backwards" proof:
Suppose $A = begin{bmatrix}a&b\0&aend{bmatrix},A' = begin{bmatrix}a'&b'\0&a'end{bmatrix}$.
Then $AH = A'H iff a = a'$, since $AA'^{-1} = begin{bmatrix}a/a'&(-ab')/a'^2+b/a'\0&a/a'end{bmatrix}$
(Note how we use the fact that $a' neq 0$).
Note as well that $AH = (aI)H$.
Thus $a mapsto (aI)H$ is the desired isomorphism.
answered Jul 7 '15 at 15:34
David WheelerDavid Wheeler
12.7k11730
12.7k11730
add a comment |
add a comment |
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GuV9pl8uRtDXQ
1
$begingroup$
No it is not correct. Knowing that to groups have the same cardinality is not enough to prove that they are isomorphic (for example $(mathbb{R},+)$ and $(mathbb{R}^*,times)$ have the same cardinality, yet they are not isomorphic. By the way, there is a mistake in your matrix multiplication.
$endgroup$
– Roland
Jul 7 '15 at 13:40
$begingroup$
edited the matix multiplication
$endgroup$
– 3SAT
Jul 7 '15 at 13:44
$begingroup$
Is $R^*$ the group of real numbers under multiplication?
$endgroup$
– man_in_green_shirt
Jul 7 '15 at 13:47
1
$begingroup$
yes, multiplication without zero
$endgroup$
– 3SAT
Jul 7 '15 at 13:48