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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 | 

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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