Fdtxjr

Multi tool use

The IT department bottlenecks progress, how should I handle this?

How much theory knowledge is actually used while playing?

How many arrows is an archer expected to fire by the end of the Tyranny of Dragons pair of adventures?

Is this part of the description of the Archfey warlock's Misty Escape feature redundant?

Multiplicative persistence

Does an advisor owe his/her student anything? Will an advisor keep a PhD student only out of pity?

Biological Blimps: Propulsion

What does Apple's new App Store requirement mean

The Digit Triangles

Are Captain Marvel's powers affected by Thanos breaking the Tesseract and claiming the stone?

A variation to the phrase "hanging over my shoulders"

Pre-mixing cryogenic fuels and using only one fuel tank

"It doesn't matter" or "it won't matter"?

Can you use Vicious Mockery to win an argument or gain favours?

Which Article Helped Get Rid of Technobabble in RPGs?

Why is the "ls" command showing permissions of files in a FAT32 partition?

Can I say "fingers" when referring to toes?

Strong empirical falsification of quantum mechanics based on vacuum energy density?

Why do ¬, ∀ and ∃ have the same precedence?

How does electrical safety system work on ISS?

Does "he squandered his car on drink" sound natural?

Do we have to expect a queue for the shuttle from Watford Junction to Harry Potter Studio?

How to convince somebody that he is fit for something else, but not this job?

Stack Interview Code methods made from class Node and Smart Pointers

Left representation is isomorphic to group

Left regular action isomorphic to the right regular actionSurjectivity of a surjective restricted ring homomorphismGroup isomorphic to (Z,+)Group of conjugation mappings isomorphic to original groupInverse map of a group isomorphism is a group homomorphismProve that if G has a faithful complex irreducible representationUnderstanding Semidirect Products in Group-Theory through exercises.infinite cyclic group is isomorphic to the group of integers under addition.Is $mathbbZ[G^n]$ isomorphic to $bigoplus_nmathbbZ[G]$?Show that the quotient group $G/N$ contains a subgroup isomorphic to $H$.

1

$begingroup$

Let $G$ be a group and $G_L$ be its left representation, that is $G_L = g_L ; g_L(x)=gx$.

Show that $G$ is isomorphic to $G_L$.

Solution To show that $G$ is isomorphic to $G_L$ we need an isomorphism.

Let $$F: G rightarrow G_L$$ be a map defined as $$g mapsto g_L$$
Claim: F is an isomorphism.

Homomorphism: $$F(g+h) = (g+h)_L (x) = (g+h)x = gx+hx = F(g) + F(h)$$

1. Injective:
   
   Assume $F(g)= F(h) implies g_L(x) =h_L(x) forall x in G implies gx=hx implies g=h $




2. Now I am stuck, can someone please help how to prove surjectivity?

group-theory group-isomorphism

share|cite|improve this question

edited Mar 14 at 20:34

MarianD

1,4741616

asked Mar 14 at 17:41

MariyaKavMariyaKav

31510

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
  $endgroup$
  – Shaun
  Mar 14 at 19:10
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $G$ be a group and $G_L$ be its left representation, that is $G_L = g_L ; g_L(x)=gx$.

Show that $G$ is isomorphic to $G_L$.

Solution To show that $G$ is isomorphic to $G_L$ we need an isomorphism.

Let $$F: G rightarrow G_L$$ be a map defined as $$g mapsto g_L$$
Claim: F is an isomorphism.

Homomorphism: $$F(g+h) = (g+h)_L (x) = (g+h)x = gx+hx = F(g) + F(h)$$

1. Injective:
   
   Assume $F(g)= F(h) implies g_L(x) =h_L(x) forall x in G implies gx=hx implies g=h $




2. Now I am stuck, can someone please help how to prove surjectivity?

group-theory group-isomorphism

share|cite|improve this question

edited Mar 14 at 20:34

MarianD

1,4741616

asked Mar 14 at 17:41

MariyaKavMariyaKav

31510

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
  $endgroup$
  – Shaun
  Mar 14 at 19:10
  

  
  
  
  

add a comment | 

$begingroup$

Let $G$ be a group and $G_L$ be its left representation, that is $G_L = g_L ; g_L(x)=gx$.

Show that $G$ is isomorphic to $G_L$.

Solution To show that $G$ is isomorphic to $G_L$ we need an isomorphism.

Let $$F: G rightarrow G_L$$ be a map defined as $$g mapsto g_L$$
Claim: F is an isomorphism.

Homomorphism: $$F(g+h) = (g+h)_L (x) = (g+h)x = gx+hx = F(g) + F(h)$$

1. Injective:
   
   Assume $F(g)= F(h) implies g_L(x) =h_L(x) forall x in G implies gx=hx implies g=h $




2. Now I am stuck, can someone please help how to prove surjectivity?

group-theory group-isomorphism

share|cite|improve this question

edited Mar 14 at 20:34

MarianD

1,4741616

asked Mar 14 at 17:41

MariyaKavMariyaKav

31510

$endgroup$

share|cite|improve this question

edited Mar 14 at 20:34

MarianD

1,4741616

asked Mar 14 at 17:41

MariyaKavMariyaKav

31510

share|cite|improve this question

edited Mar 14 at 20:34

MarianD

1,4741616

asked Mar 14 at 17:41

MariyaKavMariyaKav

31510

edited Mar 14 at 20:34

MarianD

1,4741616

edited Mar 14 at 20:34

MarianD

1,4741616

asked Mar 14 at 17:41

MariyaKavMariyaKav

31510

asked Mar 14 at 17:41

MariyaKavMariyaKav

31510

1 Answer
1

active

oldest

votes

1

$begingroup$

Let $g_Lin G_L$. Then $g_L(x)=gx$, some $gin G$ and all $xin G$. But then $F(g)colorred(x)=g_L(x)$ for all $xin G$. Hence $g_L=F(g)$.

I think your confusion is that you forgot the $colorred(x)$.

share|cite|improve this answer

answered Mar 14 at 19:05

ShaunShaun

9,690113684

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Does that make sense to you, @MariyaKav?
  $endgroup$
  – Shaun
  Mar 14 at 19:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, thank you for your help!
  $endgroup$
  – MariyaKav
  Mar 14 at 19:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome, @MariyaKav :)
  $endgroup$
  – Shaun
  Mar 14 at 19:29
  
  

  
  
  
  

add a comment | 

Your Answer

StackExchange.ifUsing("editor", function ()
return StackExchange.using("mathjaxEditing", function ()
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix)
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
);
);
, "mathjax-editing");

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%2f3148296%2fleft-representation-is-isomorphic-to-group%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

1

$begingroup$

Let $g_Lin G_L$. Then $g_L(x)=gx$, some $gin G$ and all $xin G$. But then $F(g)colorred(x)=g_L(x)$ for all $xin G$. Hence $g_L=F(g)$.

I think your confusion is that you forgot the $colorred(x)$.

share|cite|improve this answer

answered Mar 14 at 19:05

ShaunShaun

9,690113684

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Does that make sense to you, @MariyaKav?
  $endgroup$
  – Shaun
  Mar 14 at 19:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, thank you for your help!
  $endgroup$
  – MariyaKav
  Mar 14 at 19:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome, @MariyaKav :)
  $endgroup$
  – Shaun
  Mar 14 at 19:29
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $g_Lin G_L$. Then $g_L(x)=gx$, some $gin G$ and all $xin G$. But then $F(g)colorred(x)=g_L(x)$ for all $xin G$. Hence $g_L=F(g)$.

I think your confusion is that you forgot the $colorred(x)$.

share|cite|improve this answer

answered Mar 14 at 19:05

ShaunShaun

9,690113684

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Does that make sense to you, @MariyaKav?
  $endgroup$
  – Shaun
  Mar 14 at 19:11
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, thank you for your help!
  $endgroup$
  – MariyaKav
  Mar 14 at 19:24
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  You're welcome, @MariyaKav :)
  $endgroup$
  – Shaun
  Mar 14 at 19:29
  
  

  
  
  
  

add a comment | 

$begingroup$

Let $g_Lin G_L$. Then $g_L(x)=gx$, some $gin G$ and all $xin G$. But then $F(g)colorred(x)=g_L(x)$ for all $xin G$. Hence $g_L=F(g)$.

I think your confusion is that you forgot the $colorred(x)$.

share|cite|improve this answer

answered Mar 14 at 19:05

ShaunShaun

9,690113684

$endgroup$

share|cite|improve this answer

answered Mar 14 at 19:05

ShaunShaun

9,690113684

answered Mar 14 at 19:05

ShaunShaun

9,690113684

answered Mar 14 at 19:05

ShaunShaun

9,690113684