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
Can the Right Ascension and Argument of Perigee of a spacecraft's orbit keep varying by themselves with time?
Can withdrawing asylum be illegal?
Visa regaring travelling European country
Student Loan from years ago pops up and is taking my salary
What is the padding with red substance inside of steak packaging?
What happens to a Warlock's expended Spell Slots when they gain a Level?
Single author papers against my advisor's will?
Why can't devices on different VLANs, but on the same subnet, communicate?
Are there continuous functions who are the same in an interval but differ in at least one other point?
Can the DM override racial traits?
What is the role of 'For' here?
Is it ethical to upload a automatically generated paper to a non peer-reviewed site as part of a larger research?
Example of compact Riemannian manifold with only one geodesic.
What do I do when my TA workload is more than expected?
Why don't hard Brexiteers insist on a hard border to prevent illegal immigration after Brexit?
Am I ethically obligated to go into work on an off day if the reason is sudden?
Didn't get enough time to take a Coding Test - what to do now?
Why doesn't a hydraulic lever violate conservation of energy?
US Healthcare consultation for visitors
Sub-subscripts in strings cause different spacings than subscripts
Mortgage adviser recommends a longer term than necessary combined with overpayments
First use of “packing” as in carrying a gun
How do spell lists change if the party levels up without taking a long rest?
Variable with quotation marks "$()"
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 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 5 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 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 5 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 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 5 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 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 5 mins ago
YuiTo Cheng
2,4064937
asked 1 hour ago
UfomammutUfomammut
391314
edited 5 mins ago
YuiTo Cheng
2,4064937
asked 1 hour ago
UfomammutUfomammut
391314
edited 5 mins ago
YuiTo Cheng
2,4064937
edited 5 mins ago
YuiTo Cheng
2,4064937
edited 5 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 59 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 52 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
51 mins ago
$begingroup$
Yes, that will also work.
$endgroup$
– Mayank Mishra
47 mins ago
add a comment |
Your Answer
StackExchange.ready(function()
var channelOptions =
tags: "".split(" "),
id: "69"
;
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function()
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled)
StackExchange.using("snippets", function()
createEditor();
);
else
createEditor();
);
function createEditor()
StackExchange.prepareEditor(
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader:
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
,
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
);
);
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3185816%2fproving-the-given-mathbb-r3-h-cong-mathbb-r2-where-h-y-0-0y%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
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 59 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 59 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 59 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 59 mins ago
lEmlEm
3,4621921
answered 59 mins ago
lEmlEm
3,4621921
answered 59 mins ago
lEmlEm
3,4621921
answered 59 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 52 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
51 mins ago
$begingroup$
Yes, that will also work.
$endgroup$
– Mayank Mishra
47 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 52 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
51 mins ago
$begingroup$
Yes, that will also work.
$endgroup$
– Mayank Mishra
47 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 52 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 52 mins ago
Mayank MishraMayank Mishra
1068
edited 33 mins ago
answered 52 mins ago
Mayank MishraMayank Mishra
1068
answered 52 mins ago
Mayank MishraMayank Mishra
1068
answered 52 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
51 mins ago
$begingroup$
Yes, that will also work.
$endgroup$
– Mayank Mishra
47 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
51 mins ago
$begingroup$
Yes, that will also work.
$endgroup$
– Mayank Mishra
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
51 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
51 mins ago
$begingroup$
Yes, that will also work.
$endgroup$
– Mayank Mishra
47 mins ago
$begingroup$
Yes, that will also work.
$endgroup$
– Mayank Mishra
47 mins ago
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3185816%2fproving-the-given-mathbb-r3-h-cong-mathbb-r2-where-h-y-0-0y%23new-answer', 'question_page');
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
