Proving the given $mathbb R^3/H$ $cong$ $mathbb R^2$ where $H$ = y in mathbb R$ The 2019 Stack Overflow Developer Survey Results Are In Unicorn Meta Zoo #1: Why another podcast? Announcing the arrival of Valued Associate #679: Cesar ManaraAre $(mathbbR,+)$ and $(mathbbC,+)$ isomorphic as additive groups?How do I show that these two presentations are isomorphic?Determine whether or not the two given groups are isomorphic.Surjective Homomorphisms of Isomorphic Abelian GroupsGroup isomorphism between two groups .How to use the first isomorphism theorem to show that two groups are isomorphic?Showing that these two groups are isomorphic?Showing that $2$ of the following groups are not isomorphicShow that the Two Given Groups are IsomorphicAre given groups isomorphic
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Proving the given $mathbb R^3/H$ $cong$ $mathbb R^2$ where $H$ = $(y,0,0)
The 2019 Stack Overflow Developer Survey Results Are In
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar ManaraAre $(mathbbR,+)$ and $(mathbbC,+)$ isomorphic as additive groups?How do I show that these two presentations are isomorphic?Determine whether or not the two given groups are isomorphic.Surjective Homomorphisms of Isomorphic Abelian GroupsGroup isomorphism between two groups .How to use the first isomorphism theorem to show that two groups are isomorphic?Showing that these two groups are isomorphic?Showing that $2$ of the following groups are not isomorphicShow that the Two Given Groups are IsomorphicAre given groups isomorphic
$begingroup$
So I am given a group $mathbb R^3$ and a group $H$ = $(y,0,0). 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 2 mins ago
YuiTo Cheng
2,4064937
asked 1 hour ago
UfomammutUfomammut
391314
$endgroup$
add a comment |
$begingroup$
So I am given a group $mathbb R^3$ and a group $H$ = $(y,0,0). 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 2 mins ago
YuiTo Cheng
2,4064937
asked 1 hour ago
UfomammutUfomammut
391314
$endgroup$
add a comment |
$begingroup$
So I am given a group $mathbb R^3$ and a group $H$ = $(y,0,0). 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 2 mins ago
YuiTo Cheng
2,4064937
asked 1 hour ago
UfomammutUfomammut
391314
$endgroup$
So I am given a group $mathbb R^3$ and a group $H$ = $(y,0,0). 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 2 mins ago
YuiTo Cheng
2,4064937
asked 1 hour ago
UfomammutUfomammut
391314
edited 2 mins ago
YuiTo Cheng
2,4064937
asked 1 hour ago
UfomammutUfomammut
391314
edited 2 mins ago
YuiTo Cheng
2,4064937
edited 2 mins ago
YuiTo Cheng
2,4064937
edited 2 mins ago
YuiTo Cheng
2,4064937
2,4064937
asked 1 hour ago
UfomammutUfomammut
391314
asked 1 hour ago
UfomammutUfomammut
391314
asked 1 hour ago
UfomammutUfomammut
391314
391314
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 55 mins ago
lEmlEm
3,4621921
$endgroup$
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 49 mins ago
Mayank MishraMayank Mishra
1068
$endgroup$
$begingroup$
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
47 mins ago
$begingroup$
Yes, that will also work.
$endgroup$
– Mayank Mishra
44 mins ago
add a comment |
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2 Answers
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2 Answers
2
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$begingroup$
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 55 mins ago
lEmlEm
3,4621921
$endgroup$
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 55 mins ago
lEmlEm
3,4621921
$endgroup$
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 55 mins ago
lEmlEm
3,4621921
$endgroup$
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 55 mins ago
lEmlEm
3,4621921
answered 55 mins ago
lEmlEm
3,4621921
answered 55 mins ago
lEmlEm
3,4621921
answered 55 mins ago
lEmlEm
3,4621921
3,4621921
add a comment |
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 49 mins ago
Mayank MishraMayank Mishra
1068
$endgroup$
$begingroup$
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
47 mins ago
$begingroup$
Yes, that will also work.
$endgroup$
– Mayank Mishra
44 mins ago
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 49 mins ago
Mayank MishraMayank Mishra
1068
$endgroup$
$begingroup$
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
47 mins ago
$begingroup$
Yes, that will also work.
$endgroup$
– Mayank Mishra
44 mins ago
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 49 mins ago
Mayank MishraMayank Mishra
1068
$endgroup$
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 49 mins ago
Mayank MishraMayank Mishra
1068
edited 30 mins ago
answered 49 mins ago
Mayank MishraMayank Mishra
1068
answered 49 mins ago
Mayank MishraMayank Mishra
1068
answered 49 mins ago
Mayank MishraMayank Mishra
1068
1068
$begingroup$
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
47 mins ago
$begingroup$
Yes, that will also work.
$endgroup$
– Mayank Mishra
44 mins ago
add a comment |
$begingroup$
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
47 mins ago
$begingroup$
Yes, that will also work.
$endgroup$
– Mayank Mishra
44 mins ago
$begingroup$
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
47 mins ago
$begingroup$
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
47 mins ago
$begingroup$
Yes, that will also work.
$endgroup$
– Mayank Mishra
44 mins ago
$begingroup$
Yes, that will also work.
$endgroup$
– Mayank Mishra
44 mins ago
add a comment |
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