Ttyjukl

should truth entail possible truth

60's-70's movie: home appliances revolting against the owners

Accepted by European university, rejected by all American ones I applied to? Possible reasons?

What happens to a Warlock's expended Spell Slots when they gain a Level?

Proving the given two groups are isomorphic

Why are PDP-7-style microprogrammed instructions out of vogue?

What do I do when my TA workload is more than expected?

Are there continuous functions who are the same in an interval but differ in at least one other point?

Circular reasoning in L'Hopital's rule

Single author papers against my advisor's will?

What is the role of 'For' here?

Why doesn't a hydraulic lever violate conservation of energy?

"is" operation returns false even though two objects have same id

Can we generate random numbers using irrational numbers like π and e?

different output for groups and groups USERNAME after adding a username to a group

Drawing arrows from one table cell reference to another

Sort list of array linked objects by keys and values

Drawing vertical/oblique lines in Metrical tree (tikz-qtree, tipa)

Deal with toxic manager when you can't quit

Identify 80s or 90s comics with ripped creatures (not dwarves)

What information about me do stores get via my credit card?

Homework question about an engine pulling a train

Is every episode of "Where are my Pants?" identical?

How do I design a circuit to convert a 100 mV and 50 Hz sine wave to a square wave?

Proving the given two groups are isomorphic

1

$begingroup$

So I am given a group $mathbb R^3$ and a group $H$ = {$(y,0,0)|y in mathbb R$}. 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 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 what is going on. Can anyone provide some help on this?

abstract-algebra group-isomorphism

share|cite|improve this question

asked 1 hour ago

UfomammutUfomammut

391314

$endgroup$

add a comment | 

1

$begingroup$

So I am given a group $mathbb R^3$ and a group $H$ = {$(y,0,0)|y in mathbb R$}. 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 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 what is going on. Can anyone provide some help on this?

abstract-algebra group-isomorphism

share|cite|improve this question

asked 1 hour ago

UfomammutUfomammut

391314

$endgroup$

add a comment | 

$begingroup$

So I am given a group $mathbb R^3$ and a group $H$ = {$(y,0,0)|y in mathbb R$}. 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 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 what is going on. Can anyone provide some help on this?

abstract-algebra group-isomorphism

share|cite|improve this question

asked 1 hour ago

UfomammutUfomammut

391314

$endgroup$

share|cite|improve this question

asked 1 hour ago

UfomammutUfomammut

391314

share|cite|improve this question

asked 1 hour ago

UfomammutUfomammut

391314

asked 1 hour ago

UfomammutUfomammut

391314

asked 1 hour ago

UfomammutUfomammut

391314

2 Answers
2

active

oldest

votes

2

$begingroup$

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.

share|cite|improve this answer

answered 29 mins ago

lEmlEm

3,4521921

$endgroup$

add a comment | 

2

$begingroup$

We can use the first isomorphism theorem for groups here as you indicated. Consider the map $f : mathbb{R}^3 longrightarrow mathbb{R}^2$ 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 $H = {(y,0,0) | y in mathbb{R} }$ 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. $f((x_1,y_1,z_1) + (x_2,y_2,z_2)) = f(x_1+x_2,y_1+y_2,z_1+z_2) = (y_1+y_2, (y_1+y_2)+(z_1+z_2)) = (y_1+y_2, (y_1+z_1)+(y_2+z_2) = (y_1,y_1+z_1) + (y_2,y_2+z_2) = f(x_1,y_1,z_1) +f(x_2,y_2,z_2). text{Moreover,} f(0,0,0) = (0,0)$
The kernel of this map is seen to be all $(x,y,z) in mathbb{R}$ such that $y,z$ are $0$ , i.e., $H$.
Hence first isomorphism theorem applies and $ mathbb{R}^3/H equiv mathbb{R}^2.$

share|cite|improve this answer

edited 3 mins ago

answered 23 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
  21 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, that will also work.
  $endgroup$
  – Mayank Mishra
  18 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
});


}
});

draft saved

draft discarded

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

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3185816%2fproving-the-given-two-groups-are-isomorphic%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

2

$begingroup$

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.

share|cite|improve this answer

answered 29 mins ago

lEmlEm

3,4521921

$endgroup$

add a comment | 

2

$begingroup$

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.

share|cite|improve this answer

answered 29 mins ago

lEmlEm

3,4521921

$endgroup$

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.

share|cite|improve this answer

answered 29 mins ago

lEmlEm

3,4521921

$endgroup$

share|cite|improve this answer

answered 29 mins ago

lEmlEm

3,4521921

answered 29 mins ago

lEmlEm

3,4521921

answered 29 mins ago

lEmlEm

3,4521921

2

$begingroup$

We can use the first isomorphism theorem for groups here as you indicated. Consider the map $f : mathbb{R}^3 longrightarrow mathbb{R}^2$ 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 $H = {(y,0,0) | y in mathbb{R} }$ 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. $f((x_1,y_1,z_1) + (x_2,y_2,z_2)) = f(x_1+x_2,y_1+y_2,z_1+z_2) = (y_1+y_2, (y_1+y_2)+(z_1+z_2)) = (y_1+y_2, (y_1+z_1)+(y_2+z_2) = (y_1,y_1+z_1) + (y_2,y_2+z_2) = f(x_1,y_1,z_1) +f(x_2,y_2,z_2). text{Moreover,} f(0,0,0) = (0,0)$
The kernel of this map is seen to be all $(x,y,z) in mathbb{R}$ such that $y,z$ are $0$ , i.e., $H$.
Hence first isomorphism theorem applies and $ mathbb{R}^3/H equiv mathbb{R}^2.$

share|cite|improve this answer

edited 3 mins ago

answered 23 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
  21 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, that will also work.
  $endgroup$
  – Mayank Mishra
  18 mins ago
  

  
  
  
  

add a comment | 

2

$begingroup$

We can use the first isomorphism theorem for groups here as you indicated. Consider the map $f : mathbb{R}^3 longrightarrow mathbb{R}^2$ 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 $H = {(y,0,0) | y in mathbb{R} }$ 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. $f((x_1,y_1,z_1) + (x_2,y_2,z_2)) = f(x_1+x_2,y_1+y_2,z_1+z_2) = (y_1+y_2, (y_1+y_2)+(z_1+z_2)) = (y_1+y_2, (y_1+z_1)+(y_2+z_2) = (y_1,y_1+z_1) + (y_2,y_2+z_2) = f(x_1,y_1,z_1) +f(x_2,y_2,z_2). text{Moreover,} f(0,0,0) = (0,0)$
The kernel of this map is seen to be all $(x,y,z) in mathbb{R}$ such that $y,z$ are $0$ , i.e., $H$.
Hence first isomorphism theorem applies and $ mathbb{R}^3/H equiv mathbb{R}^2.$

share|cite|improve this answer

edited 3 mins ago

answered 23 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
  21 mins ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, that will also work.
  $endgroup$
  – Mayank Mishra
  18 mins ago
  

  
  
  
  

add a comment | 

$begingroup$

We can use the first isomorphism theorem for groups here as you indicated. Consider the map $f : mathbb{R}^3 longrightarrow mathbb{R}^2$ 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 $H = {(y,0,0) | y in mathbb{R} }$ 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. $f((x_1,y_1,z_1) + (x_2,y_2,z_2)) = f(x_1+x_2,y_1+y_2,z_1+z_2) = (y_1+y_2, (y_1+y_2)+(z_1+z_2)) = (y_1+y_2, (y_1+z_1)+(y_2+z_2) = (y_1,y_1+z_1) + (y_2,y_2+z_2) = f(x_1,y_1,z_1) +f(x_2,y_2,z_2). text{Moreover,} f(0,0,0) = (0,0)$
The kernel of this map is seen to be all $(x,y,z) in mathbb{R}$ such that $y,z$ are $0$ , i.e., $H$.
Hence first isomorphism theorem applies and $ mathbb{R}^3/H equiv mathbb{R}^2.$

share|cite|improve this answer

edited 3 mins ago

answered 23 mins ago

Mayank MishraMayank Mishra

1068

$endgroup$

share|cite|improve this answer

edited 3 mins ago

answered 23 mins ago

Mayank MishraMayank Mishra

1068

answered 23 mins ago

Mayank MishraMayank Mishra

1068

answered 23 mins ago

Mayank MishraMayank Mishra

1068