Proving the given $mathbb R^3/H$ $cong$ $mathbb R^2$ where $H$ = y in mathbb R$ Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Group of units of End($mathbbZ/_2 oplus mathbbZ/_2) cong S_3$?Show $SO_2(mathbbR) congA in GL_2(mathbbR) : A^tA = I, det A = 1$.prove $mathbbZ(Gtimes C_2)cong (mathbbZG)C_2$Show that $GL_n(mathbbR)/GL_n^+(mathbbR) cong mathbbZ_2$Show that $mathbbT^1 cong mathbbC^*/mathbbR_>0$Show $mathbb C^*congmathbb R^*timesmathbb Tcongmathbb C/mathbb Z$.Is $mathbbZ[x] / (x - a) cong mathbbZ$ where $a in mathbbZ$?Prove $G/H cong (mathbbR-0)times(mathbbR-0)$Is $frac mathbbC[x,y] langle x^m+ y^n -1rangle cong frac mathbbC[y]langle y^n-1rangle $?Question about the proof of Second Isomorphism Theorem
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Proving the given $mathbb R^3/H$ $cong$ $mathbb R^2$ where $H$ = $(y,0,0)
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Group of units of End($mathbbZ/_2 oplus mathbbZ/_2) cong S_3$?Show $SO_2(mathbbR) congA in GL_2(mathbbR) : A^tA = I, det A = 1$.prove $mathbbZ(Gtimes C_2)cong (mathbbZG)C_2$Show that $GL_n(mathbbR)/GL_n^+(mathbbR) cong mathbbZ_2$Show that $mathbbT^1 cong mathbbC^*/mathbbR_>0$Show $mathbb C^*congmathbb R^*timesmathbb Tcongmathbb C/mathbb Z$.Is $mathbbZ[x] / (x - a) cong mathbbZ$ where $a in mathbbZ$?Prove $G/H cong (mathbbR-0)times(mathbbR-0)$Is $frac mathbbC[x,y] langle x^m+ y^n -1rangle cong frac mathbbC[y]langle y^n-1rangle $?Question about the proof of Second Isomorphism Theorem
$begingroup$
So I am given a group $mathbb R^3$ and a group $H$ = y in mathbb R$. I have to prove that that $mathbb R^3/H$ $cong$ $mathbb R^2$. I am not sure how to even begin. My difficulty is coming up with a map between the two sets. I have already verified that $H unlhd mathbb R^3$. So all I know is $mathbb R^3/H$ is a group. Also, from first isomorphism theorem, I know that the group is isomorphic to the image of the map $f: mathbb R^3 to A$, and I do not know what that $A$ is supposed to be. Today is the first day I learned about isomorphism, and I am very confused about what is going on. Can anyone provide some help on this?
abstract-algebra group-isomorphism
edited Apr 13 at 5:33
YuiTo Cheng
2,54841037
asked Apr 13 at 4:09
UfomammutUfomammut
411414
$endgroup$
add a comment |
$begingroup$
So I am given a group $mathbb R^3$ and a group $H$ = y in mathbb R$. I have to prove that that $mathbb R^3/H$ $cong$ $mathbb R^2$. I am not sure how to even begin. My difficulty is coming up with a map between the two sets. I have already verified that $H unlhd mathbb R^3$. So all I know is $mathbb R^3/H$ is a group. Also, from first isomorphism theorem, I know that the group is isomorphic to the image of the map $f: mathbb R^3 to A$, and I do not know what that $A$ is supposed to be. Today is the first day I learned about isomorphism, and I am very confused about what is going on. Can anyone provide some help on this?
abstract-algebra group-isomorphism
edited Apr 13 at 5:33
YuiTo Cheng
2,54841037
asked Apr 13 at 4:09
UfomammutUfomammut
411414
$endgroup$
add a comment |
$begingroup$
So I am given a group $mathbb R^3$ and a group $H$ = y in mathbb R$. I have to prove that that $mathbb R^3/H$ $cong$ $mathbb R^2$. I am not sure how to even begin. My difficulty is coming up with a map between the two sets. I have already verified that $H unlhd mathbb R^3$. So all I know is $mathbb R^3/H$ is a group. Also, from first isomorphism theorem, I know that the group is isomorphic to the image of the map $f: mathbb R^3 to A$, and I do not know what that $A$ is supposed to be. Today is the first day I learned about isomorphism, and I am very confused about what is going on. Can anyone provide some help on this?
abstract-algebra group-isomorphism
edited Apr 13 at 5:33
YuiTo Cheng
2,54841037
asked Apr 13 at 4:09
UfomammutUfomammut
411414
$endgroup$
So I am given a group $mathbb R^3$ and a group $H$ = y in mathbb R$. I have to prove that that $mathbb R^3/H$ $cong$ $mathbb R^2$. I am not sure how to even begin. My difficulty is coming up with a map between the two sets. I have already verified that $H unlhd mathbb R^3$. So all I know is $mathbb R^3/H$ is a group. Also, from first isomorphism theorem, I know that the group is isomorphic to the image of the map $f: mathbb R^3 to A$, and I do not know what that $A$ is supposed to be. Today is the first day I learned about isomorphism, and I am very confused about what is going on. Can anyone provide some help on this?
abstract-algebra group-isomorphism
abstract-algebra group-isomorphism
edited Apr 13 at 5:33
YuiTo Cheng
2,54841037
asked Apr 13 at 4:09
UfomammutUfomammut
411414
edited Apr 13 at 5:33
YuiTo Cheng
2,54841037
asked Apr 13 at 4:09
UfomammutUfomammut
411414
edited Apr 13 at 5:33
YuiTo Cheng
2,54841037
edited Apr 13 at 5:33
YuiTo Cheng
2,54841037
edited Apr 13 at 5:33
YuiTo Cheng
2,54841037
2,54841037
asked Apr 13 at 4:09
UfomammutUfomammut
411414
asked Apr 13 at 4:09
UfomammutUfomammut
411414
asked Apr 13 at 4:09
UfomammutUfomammut
411414
411414
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
We can use the first isomorphism theorem for groups here as you indicated. Consider the map $f : mathbbR^3 longrightarrow mathbbR^2$ as follows, $f(x,y,z) = (y,y+z)$. First, we show that this map is a well-defined group homomorphism and next show that $H = (y,0,0) $ is its kernel.
If $(x_1,y_1,z_1) = (x_2,y_2,z_2)$ then $f(x_1,y_1,z_1) = f(x_2,y_2,z_2)$ hence map is well defined.
Next we show this map is homomorphism. $f((x_1,y_1,z_1) + (x_2,y_2,z_2)) = f(x_1+x_2,y_1+y_2,z_1+z_2) = (y_1+y_2, (y_1+y_2)+(z_1+z_2)) = (y_1+y_2, (y_1+z_1)+(y_2+z_2) = (y_1,y_1+z_1) + (y_2,y_2+z_2) = f(x_1,y_1,z_1) +f(x_2,y_2,z_2). textMoreover, f(0,0,0) = (0,0)$
The kernel of this map is seen to be all $(x,y,z) in mathbbR$ such that $y,z$ are $0$ , i.e., $H$.
Hence first isomorphism theorem applies and $ mathbbR^3/H equiv mathbbR^2.$
answered Apr 13 at 4:46
Mayank MishraMayank Mishra
1489
$endgroup$
1
$begingroup$
I came up with something similar, but what about the map $f((x,y,z)) = (y,z)$? I think this one should be fine too, right?
$endgroup$
– Ufomammut
Apr 13 at 4:48
$begingroup$
Yes, that will also work.
$endgroup$
– Mayank Mishra
Apr 13 at 4:51
add a comment |
$begingroup$
The first isomorphism theorem asserts that, if $varphi: Ato B$ is a surjective homomorphism, then $Bcong A/kervarphi$. In your problem, you wish to show $mathbb R^2congmathbb R^3/H $, so a natural guess would be to take $A=mathbb R^3$ and $B=mathbb R^2$. Now it remains to construct the homomorphism so that $kervarphi=H$. I will leave the rest to you.
answered Apr 13 at 4:40
lEmlEm
3,4971921
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
We can use the first isomorphism theorem for groups here as you indicated. Consider the map $f : mathbbR^3 longrightarrow mathbbR^2$ as follows, $f(x,y,z) = (y,y+z)$. First, we show that this map is a well-defined group homomorphism and next show that $H = (y,0,0) $ is its kernel.
If $(x_1,y_1,z_1) = (x_2,y_2,z_2)$ then $f(x_1,y_1,z_1) = f(x_2,y_2,z_2)$ hence map is well defined.
Next we show this map is homomorphism. $f((x_1,y_1,z_1) + (x_2,y_2,z_2)) = f(x_1+x_2,y_1+y_2,z_1+z_2) = (y_1+y_2, (y_1+y_2)+(z_1+z_2)) = (y_1+y_2, (y_1+z_1)+(y_2+z_2) = (y_1,y_1+z_1) + (y_2,y_2+z_2) = f(x_1,y_1,z_1) +f(x_2,y_2,z_2). textMoreover, f(0,0,0) = (0,0)$
The kernel of this map is seen to be all $(x,y,z) in mathbbR$ such that $y,z$ are $0$ , i.e., $H$.
Hence first isomorphism theorem applies and $ mathbbR^3/H equiv mathbbR^2.$
answered Apr 13 at 4:46
Mayank MishraMayank Mishra
1489
$endgroup$
1
$begingroup$
I came up with something similar, but what about the map $f((x,y,z)) = (y,z)$? I think this one should be fine too, right?
$endgroup$
– Ufomammut
Apr 13 at 4:48
$begingroup$
Yes, that will also work.
$endgroup$
– Mayank Mishra
Apr 13 at 4:51
add a comment |
$begingroup$
We can use the first isomorphism theorem for groups here as you indicated. Consider the map $f : mathbbR^3 longrightarrow mathbbR^2$ as follows, $f(x,y,z) = (y,y+z)$. First, we show that this map is a well-defined group homomorphism and next show that $H = (y,0,0) $ is its kernel.
If $(x_1,y_1,z_1) = (x_2,y_2,z_2)$ then $f(x_1,y_1,z_1) = f(x_2,y_2,z_2)$ hence map is well defined.
Next we show this map is homomorphism. $f((x_1,y_1,z_1) + (x_2,y_2,z_2)) = f(x_1+x_2,y_1+y_2,z_1+z_2) = (y_1+y_2, (y_1+y_2)+(z_1+z_2)) = (y_1+y_2, (y_1+z_1)+(y_2+z_2) = (y_1,y_1+z_1) + (y_2,y_2+z_2) = f(x_1,y_1,z_1) +f(x_2,y_2,z_2). textMoreover, f(0,0,0) = (0,0)$
The kernel of this map is seen to be all $(x,y,z) in mathbbR$ such that $y,z$ are $0$ , i.e., $H$.
Hence first isomorphism theorem applies and $ mathbbR^3/H equiv mathbbR^2.$
answered Apr 13 at 4:46
Mayank MishraMayank Mishra
1489
$endgroup$
1
$begingroup$
I came up with something similar, but what about the map $f((x,y,z)) = (y,z)$? I think this one should be fine too, right?
$endgroup$
– Ufomammut
Apr 13 at 4:48
$begingroup$
Yes, that will also work.
$endgroup$
– Mayank Mishra
Apr 13 at 4:51
add a comment |
$begingroup$
We can use the first isomorphism theorem for groups here as you indicated. Consider the map $f : mathbbR^3 longrightarrow mathbbR^2$ as follows, $f(x,y,z) = (y,y+z)$. First, we show that this map is a well-defined group homomorphism and next show that $H = (y,0,0) $ is its kernel.
If $(x_1,y_1,z_1) = (x_2,y_2,z_2)$ then $f(x_1,y_1,z_1) = f(x_2,y_2,z_2)$ hence map is well defined.
Next we show this map is homomorphism. $f((x_1,y_1,z_1) + (x_2,y_2,z_2)) = f(x_1+x_2,y_1+y_2,z_1+z_2) = (y_1+y_2, (y_1+y_2)+(z_1+z_2)) = (y_1+y_2, (y_1+z_1)+(y_2+z_2) = (y_1,y_1+z_1) + (y_2,y_2+z_2) = f(x_1,y_1,z_1) +f(x_2,y_2,z_2). textMoreover, f(0,0,0) = (0,0)$
The kernel of this map is seen to be all $(x,y,z) in mathbbR$ such that $y,z$ are $0$ , i.e., $H$.
Hence first isomorphism theorem applies and $ mathbbR^3/H equiv mathbbR^2.$
answered Apr 13 at 4:46
Mayank MishraMayank Mishra
1489
$endgroup$
We can use the first isomorphism theorem for groups here as you indicated. Consider the map $f : mathbbR^3 longrightarrow mathbbR^2$ as follows, $f(x,y,z) = (y,y+z)$. First, we show that this map is a well-defined group homomorphism and next show that $H = (y,0,0) $ is its kernel.
If $(x_1,y_1,z_1) = (x_2,y_2,z_2)$ then $f(x_1,y_1,z_1) = f(x_2,y_2,z_2)$ hence map is well defined.
Next we show this map is homomorphism. $f((x_1,y_1,z_1) + (x_2,y_2,z_2)) = f(x_1+x_2,y_1+y_2,z_1+z_2) = (y_1+y_2, (y_1+y_2)+(z_1+z_2)) = (y_1+y_2, (y_1+z_1)+(y_2+z_2) = (y_1,y_1+z_1) + (y_2,y_2+z_2) = f(x_1,y_1,z_1) +f(x_2,y_2,z_2). textMoreover, f(0,0,0) = (0,0)$
The kernel of this map is seen to be all $(x,y,z) in mathbbR$ such that $y,z$ are $0$ , i.e., $H$.
Hence first isomorphism theorem applies and $ mathbbR^3/H equiv mathbbR^2.$
answered Apr 13 at 4:46
Mayank MishraMayank Mishra
1489
edited Apr 13 at 5:05
answered Apr 13 at 4:46
Mayank MishraMayank Mishra
1489
answered Apr 13 at 4:46
Mayank MishraMayank Mishra
1489
answered Apr 13 at 4:46
Mayank MishraMayank Mishra
1489
1489
1
$begingroup$
I came up with something similar, but what about the map $f((x,y,z)) = (y,z)$? I think this one should be fine too, right?
$endgroup$
– Ufomammut
Apr 13 at 4:48
$begingroup$
Yes, that will also work.
$endgroup$
– Mayank Mishra
Apr 13 at 4:51
add a comment |
1
$begingroup$
I came up with something similar, but what about the map $f((x,y,z)) = (y,z)$? I think this one should be fine too, right?
$endgroup$
– Ufomammut
Apr 13 at 4:48
$begingroup$
Yes, that will also work.
$endgroup$
– Mayank Mishra
Apr 13 at 4:51
1
1
$begingroup$
I came up with something similar, but what about the map $f((x,y,z)) = (y,z)$? I think this one should be fine too, right?
$endgroup$
– Ufomammut
Apr 13 at 4:48
$begingroup$
I came up with something similar, but what about the map $f((x,y,z)) = (y,z)$? I think this one should be fine too, right?
$endgroup$
– Ufomammut
Apr 13 at 4:48
$begingroup$
Yes, that will also work.
$endgroup$
– Mayank Mishra
Apr 13 at 4:51
$begingroup$
Yes, that will also work.
$endgroup$
– Mayank Mishra
Apr 13 at 4:51
add a comment |
$begingroup$
The first isomorphism theorem asserts that, if $varphi: Ato B$ is a surjective homomorphism, then $Bcong A/kervarphi$. In your problem, you wish to show $mathbb R^2congmathbb R^3/H $, so a natural guess would be to take $A=mathbb R^3$ and $B=mathbb R^2$. Now it remains to construct the homomorphism so that $kervarphi=H$. I will leave the rest to you.
answered Apr 13 at 4:40
lEmlEm
3,4971921
$endgroup$
add a comment |
$begingroup$
The first isomorphism theorem asserts that, if $varphi: Ato B$ is a surjective homomorphism, then $Bcong A/kervarphi$. In your problem, you wish to show $mathbb R^2congmathbb R^3/H $, so a natural guess would be to take $A=mathbb R^3$ and $B=mathbb R^2$. Now it remains to construct the homomorphism so that $kervarphi=H$. I will leave the rest to you.
answered Apr 13 at 4:40
lEmlEm
3,4971921
$endgroup$
add a comment |
$begingroup$
The first isomorphism theorem asserts that, if $varphi: Ato B$ is a surjective homomorphism, then $Bcong A/kervarphi$. In your problem, you wish to show $mathbb R^2congmathbb R^3/H $, so a natural guess would be to take $A=mathbb R^3$ and $B=mathbb R^2$. Now it remains to construct the homomorphism so that $kervarphi=H$. I will leave the rest to you.
answered Apr 13 at 4:40
lEmlEm
3,4971921
$endgroup$
The first isomorphism theorem asserts that, if $varphi: Ato B$ is a surjective homomorphism, then $Bcong A/kervarphi$. In your problem, you wish to show $mathbb R^2congmathbb R^3/H $, so a natural guess would be to take $A=mathbb R^3$ and $B=mathbb R^2$. Now it remains to construct the homomorphism so that $kervarphi=H$. I will leave the rest to you.
answered Apr 13 at 4:40
lEmlEm
3,4971921
answered Apr 13 at 4:40
lEmlEm
3,4971921
answered Apr 13 at 4:40
lEmlEm
3,4971921
answered Apr 13 at 4:40
lEmlEm
3,4971921
3,4971921
add a comment |
add a comment |
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