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An application of the first (group) isomorphism theorem

2

1

$begingroup$

Suppose that $G$ is a group and $N$ is a normal subgroup of $G$. How do we find out what $G/N$ is isomorphic to? For example, $C_m$ is a normal subgroup of $D_2n$ generated by a rotation of angle $2pi/m$. We can partiotion the group $D_2n$ using the normal subgrouop $C_m$. We want to find a surjective homomorphism $phi: Gto G'$ such that $ker(phi) = C_m$? What surjective homomorphism $phi: Gto G'$ should we have so that $D_2n/C_m$ is isomorphic to $G'$?

abstract-algebra group-theory normal-subgroups group-homomorphism

share|cite|improve this question

edited Apr 2 at 16:26

user546106

asked Apr 2 at 16:05

user546106user546106

1748

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Its difficult to work out what your question is. So far as I can tell, any answer is just going to re-state the first isomorphism theorem (and the current two answers do this).
  $endgroup$
  – user1729
  Apr 2 at 16:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (On the other hand, the answer to the abstract question "how do we find out what $G/N$ is isomorphic to?" does have a concrete answer. The answer is: "you don't; in fact you can't even determine if $G/N$ is the trivial group".)
  $endgroup$
  – user1729
  Apr 2 at 16:41
  

  
  
  
  

add a comment | 

2

1

$begingroup$

Suppose that $G$ is a group and $N$ is a normal subgroup of $G$. How do we find out what $G/N$ is isomorphic to? For example, $C_m$ is a normal subgroup of $D_2n$ generated by a rotation of angle $2pi/m$. We can partiotion the group $D_2n$ using the normal subgrouop $C_m$. We want to find a surjective homomorphism $phi: Gto G'$ such that $ker(phi) = C_m$? What surjective homomorphism $phi: Gto G'$ should we have so that $D_2n/C_m$ is isomorphic to $G'$?

abstract-algebra group-theory normal-subgroups group-homomorphism

share|cite|improve this question

edited Apr 2 at 16:26

user546106

asked Apr 2 at 16:05

user546106user546106

1748

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Its difficult to work out what your question is. So far as I can tell, any answer is just going to re-state the first isomorphism theorem (and the current two answers do this).
  $endgroup$
  – user1729
  Apr 2 at 16:37
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  (On the other hand, the answer to the abstract question "how do we find out what $G/N$ is isomorphic to?" does have a concrete answer. The answer is: "you don't; in fact you can't even determine if $G/N$ is the trivial group".)
  $endgroup$
  – user1729
  Apr 2 at 16:41
  

  
  
  
  

add a comment | 

$begingroup$

Suppose that $G$ is a group and $N$ is a normal subgroup of $G$. How do we find out what $G/N$ is isomorphic to? For example, $C_m$ is a normal subgroup of $D_2n$ generated by a rotation of angle $2pi/m$. We can partiotion the group $D_2n$ using the normal subgrouop $C_m$. We want to find a surjective homomorphism $phi: Gto G'$ such that $ker(phi) = C_m$? What surjective homomorphism $phi: Gto G'$ should we have so that $D_2n/C_m$ is isomorphic to $G'$?

abstract-algebra group-theory normal-subgroups group-homomorphism

share|cite|improve this question

edited Apr 2 at 16:26

user546106

asked Apr 2 at 16:05

user546106user546106

1748

$endgroup$

share|cite|improve this question

edited Apr 2 at 16:26

user546106

asked Apr 2 at 16:05

user546106user546106

1748

share|cite|improve this question

edited Apr 2 at 16:26

user546106

asked Apr 2 at 16:05

user546106user546106

1748

asked Apr 2 at 16:05

user546106user546106

1748

asked Apr 2 at 16:05

user546106user546106

1748

3 Answers
3

active

oldest

votes

2

$begingroup$

I assume $m|n$, so $n=md$ and $C_mconglangle r^drangle$. In this case, $|D_2n/C_m|=2d$. Working with cosets you can check that the quotient is generated by $[r]=rC_m$ and $[j]=jC_m$ satisfying the relations $[r]^d=[1]=C_m$, $[j]^2=[1]=C_m$ and $[j][r]=[r^-1][j]$. The correct guess is that $$D_2n/C_mcong D_2d=langle R,Jmid R^d=J^2=1,JR=R^-1Jrangle.$$

Now, define a homomorphism $phi:D_2nto D_2d$ by $phi(r)=R$ and $phi(j)=J$ (and extend multiplicatively). Check that this map is well defined (this involves the fact that $d|n$) and surjective. By the first isomorphism theorem $$D_2n/kerphicong D_2d.$$ Finally, check that $kerphi=langle r^drangle$.

share|cite|improve this answer

edited Apr 2 at 16:43

answered Apr 2 at 16:34

David HillDavid Hill

9,6561619

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is it possible that $D_2n/C_m cong$ some cyclic group?
  $endgroup$
  – user546106
  Apr 2 at 16:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  When $m=n$, it is isomorphic to $D_2=C_2$. Otherwise, the quotient is non-abelian.
  $endgroup$
  – David Hill
  Apr 2 at 16:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why is it a natural choice to pick $[r]=rC_m$ as a generator? What about $r^2C_m$?
  $endgroup$
  – user546106
  Apr 2 at 17:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Because $r$ is a generator of $D_2n$ while $r^2$ is not, in general.
  $endgroup$
  – David Hill
  Apr 2 at 17:08
  

  
  
  
  

add a comment | 

1

$begingroup$

Define the natural homomorphism $gamma: G to G/N$ by $g mapsto gN.$ Then $N$ is the kernel of $gamma$ and we have $$G/Ncong operatornameIm(gamma).$$

See Theorem 10.4 of Gallian's "Contemporary Abstract Algebra."

share|cite|improve this answer

edited Apr 2 at 16:28

answered Apr 2 at 16:19

ShaunShaun

11.1k113688

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you not saying something like $G/Ncong G/N$? I think $operatornameIm(gamma) = G/N$
  $endgroup$
  – user546106
  Apr 2 at 16:28
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No, since I'm applying the first isomorphism theorem, @user546106.
  $endgroup$
  – Shaun
  Apr 2 at 16:29
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Your $gamma$ is the same as the $pi$ in my picture. How do you define $varphi$ and $G'$?
  $endgroup$
  – user546106
  Apr 2 at 16:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I've taken the liberty of having $G'=operatornameIm(gamma)$, @user546106. It's probably best if you work on what $phi$ is yourself.
  $endgroup$
  – Shaun
  Apr 2 at 16:34
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Usually, the first isomorphism theorem is easily satisfied by the projection map $pi:Gto G/N$ defined by $$ pi (g) = gN $$ for example see this question. Then in your specific dihedral group example, you can explicitly calculate $pi(G)$ and get that $$G/N = G/ker(pi)congpi(G) $$

share|cite|improve this answer

edited Apr 2 at 16:30

Shaun

11.1k113688

answered Apr 2 at 16:22

NazimJNazimJ

890110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think your answer and Shaun's just restate the first isomorphism theorem.
  $endgroup$
  – user546106
  Apr 2 at 16:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Well to answer your question "How to find out what $G/N$ is isomorphic to?", it is equivalent to answer "how do you compute the right hand side of the first isomorphism theorem?". And we both suggest that you can compute the image of $G$ under the projection map to answer your question. Is this clearer?
  $endgroup$
  – NazimJ
  Apr 2 at 16:30
  
  

  
  
  
  

add a comment | 

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2

$begingroup$

I assume $m|n$, so $n=md$ and $C_mconglangle r^drangle$. In this case, $|D_2n/C_m|=2d$. Working with cosets you can check that the quotient is generated by $[r]=rC_m$ and $[j]=jC_m$ satisfying the relations $[r]^d=[1]=C_m$, $[j]^2=[1]=C_m$ and $[j][r]=[r^-1][j]$. The correct guess is that $$D_2n/C_mcong D_2d=langle R,Jmid R^d=J^2=1,JR=R^-1Jrangle.$$

Now, define a homomorphism $phi:D_2nto D_2d$ by $phi(r)=R$ and $phi(j)=J$ (and extend multiplicatively). Check that this map is well defined (this involves the fact that $d|n$) and surjective. By the first isomorphism theorem $$D_2n/kerphicong D_2d.$$ Finally, check that $kerphi=langle r^drangle$.

share|cite|improve this answer

edited Apr 2 at 16:43

answered Apr 2 at 16:34

David HillDavid Hill

9,6561619

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is it possible that $D_2n/C_m cong$ some cyclic group?
  $endgroup$
  – user546106
  Apr 2 at 16:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  When $m=n$, it is isomorphic to $D_2=C_2$. Otherwise, the quotient is non-abelian.
  $endgroup$
  – David Hill
  Apr 2 at 16:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why is it a natural choice to pick $[r]=rC_m$ as a generator? What about $r^2C_m$?
  $endgroup$
  – user546106
  Apr 2 at 17:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Because $r$ is a generator of $D_2n$ while $r^2$ is not, in general.
  $endgroup$
  – David Hill
  Apr 2 at 17:08
  

  
  
  
  

add a comment | 

2

$begingroup$

I assume $m|n$, so $n=md$ and $C_mconglangle r^drangle$. In this case, $|D_2n/C_m|=2d$. Working with cosets you can check that the quotient is generated by $[r]=rC_m$ and $[j]=jC_m$ satisfying the relations $[r]^d=[1]=C_m$, $[j]^2=[1]=C_m$ and $[j][r]=[r^-1][j]$. The correct guess is that $$D_2n/C_mcong D_2d=langle R,Jmid R^d=J^2=1,JR=R^-1Jrangle.$$

Now, define a homomorphism $phi:D_2nto D_2d$ by $phi(r)=R$ and $phi(j)=J$ (and extend multiplicatively). Check that this map is well defined (this involves the fact that $d|n$) and surjective. By the first isomorphism theorem $$D_2n/kerphicong D_2d.$$ Finally, check that $kerphi=langle r^drangle$.

share|cite|improve this answer

edited Apr 2 at 16:43

answered Apr 2 at 16:34

David HillDavid Hill

9,6561619

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is it possible that $D_2n/C_m cong$ some cyclic group?
  $endgroup$
  – user546106
  Apr 2 at 16:49
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  When $m=n$, it is isomorphic to $D_2=C_2$. Otherwise, the quotient is non-abelian.
  $endgroup$
  – David Hill
  Apr 2 at 16:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Why is it a natural choice to pick $[r]=rC_m$ as a generator? What about $r^2C_m$?
  $endgroup$
  – user546106
  Apr 2 at 17:04
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Because $r$ is a generator of $D_2n$ while $r^2$ is not, in general.
  $endgroup$
  – David Hill
  Apr 2 at 17:08
  

  
  
  
  

add a comment | 

$begingroup$

I assume $m|n$, so $n=md$ and $C_mconglangle r^drangle$. In this case, $|D_2n/C_m|=2d$. Working with cosets you can check that the quotient is generated by $[r]=rC_m$ and $[j]=jC_m$ satisfying the relations $[r]^d=[1]=C_m$, $[j]^2=[1]=C_m$ and $[j][r]=[r^-1][j]$. The correct guess is that $$D_2n/C_mcong D_2d=langle R,Jmid R^d=J^2=1,JR=R^-1Jrangle.$$

Now, define a homomorphism $phi:D_2nto D_2d$ by $phi(r)=R$ and $phi(j)=J$ (and extend multiplicatively). Check that this map is well defined (this involves the fact that $d|n$) and surjective. By the first isomorphism theorem $$D_2n/kerphicong D_2d.$$ Finally, check that $kerphi=langle r^drangle$.

share|cite|improve this answer

edited Apr 2 at 16:43

answered Apr 2 at 16:34

David HillDavid Hill

9,6561619

$endgroup$

share|cite|improve this answer

edited Apr 2 at 16:43

answered Apr 2 at 16:34

David HillDavid Hill

9,6561619

answered Apr 2 at 16:34

David HillDavid Hill

9,6561619

answered Apr 2 at 16:34

David HillDavid Hill

9,6561619

1

$begingroup$

Define the natural homomorphism $gamma: G to G/N$ by $g mapsto gN.$ Then $N$ is the kernel of $gamma$ and we have $$G/Ncong operatornameIm(gamma).$$

See Theorem 10.4 of Gallian's "Contemporary Abstract Algebra."

share|cite|improve this answer

edited Apr 2 at 16:28

answered Apr 2 at 16:19

ShaunShaun

11.1k113688

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you not saying something like $G/Ncong G/N$? I think $operatornameIm(gamma) = G/N$
  $endgroup$
  – user546106
  Apr 2 at 16:28
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No, since I'm applying the first isomorphism theorem, @user546106.
  $endgroup$
  – Shaun
  Apr 2 at 16:29
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Your $gamma$ is the same as the $pi$ in my picture. How do you define $varphi$ and $G'$?
  $endgroup$
  – user546106
  Apr 2 at 16:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I've taken the liberty of having $G'=operatornameIm(gamma)$, @user546106. It's probably best if you work on what $phi$ is yourself.
  $endgroup$
  – Shaun
  Apr 2 at 16:34
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Define the natural homomorphism $gamma: G to G/N$ by $g mapsto gN.$ Then $N$ is the kernel of $gamma$ and we have $$G/Ncong operatornameIm(gamma).$$

See Theorem 10.4 of Gallian's "Contemporary Abstract Algebra."

share|cite|improve this answer

edited Apr 2 at 16:28

answered Apr 2 at 16:19

ShaunShaun

11.1k113688

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Are you not saying something like $G/Ncong G/N$? I think $operatornameIm(gamma) = G/N$
  $endgroup$
  – user546106
  Apr 2 at 16:28
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  No, since I'm applying the first isomorphism theorem, @user546106.
  $endgroup$
  – Shaun
  Apr 2 at 16:29
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Your $gamma$ is the same as the $pi$ in my picture. How do you define $varphi$ and $G'$?
  $endgroup$
  – user546106
  Apr 2 at 16:30
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I've taken the liberty of having $G'=operatornameIm(gamma)$, @user546106. It's probably best if you work on what $phi$ is yourself.
  $endgroup$
  – Shaun
  Apr 2 at 16:34
  
  

  
  
  
  

add a comment | 

$begingroup$

Define the natural homomorphism $gamma: G to G/N$ by $g mapsto gN.$ Then $N$ is the kernel of $gamma$ and we have $$G/Ncong operatornameIm(gamma).$$

See Theorem 10.4 of Gallian's "Contemporary Abstract Algebra."

share|cite|improve this answer

edited Apr 2 at 16:28

answered Apr 2 at 16:19

ShaunShaun

11.1k113688

$endgroup$

share|cite|improve this answer

edited Apr 2 at 16:28

answered Apr 2 at 16:19

ShaunShaun

11.1k113688

answered Apr 2 at 16:19

ShaunShaun

11.1k113688

answered Apr 2 at 16:19

ShaunShaun

11.1k113688

1

$begingroup$

Usually, the first isomorphism theorem is easily satisfied by the projection map $pi:Gto G/N$ defined by $$ pi (g) = gN $$ for example see this question. Then in your specific dihedral group example, you can explicitly calculate $pi(G)$ and get that $$G/N = G/ker(pi)congpi(G) $$

share|cite|improve this answer

edited Apr 2 at 16:30

Shaun

11.1k113688

answered Apr 2 at 16:22

NazimJNazimJ

890110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think your answer and Shaun's just restate the first isomorphism theorem.
  $endgroup$
  – user546106
  Apr 2 at 16:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Well to answer your question "How to find out what $G/N$ is isomorphic to?", it is equivalent to answer "how do you compute the right hand side of the first isomorphism theorem?". And we both suggest that you can compute the image of $G$ under the projection map to answer your question. Is this clearer?
  $endgroup$
  – NazimJ
  Apr 2 at 16:30
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Usually, the first isomorphism theorem is easily satisfied by the projection map $pi:Gto G/N$ defined by $$ pi (g) = gN $$ for example see this question. Then in your specific dihedral group example, you can explicitly calculate $pi(G)$ and get that $$G/N = G/ker(pi)congpi(G) $$

share|cite|improve this answer

edited Apr 2 at 16:30

Shaun

11.1k113688

answered Apr 2 at 16:22

NazimJNazimJ

890110

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think your answer and Shaun's just restate the first isomorphism theorem.
  $endgroup$
  – user546106
  Apr 2 at 16:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Well to answer your question "How to find out what $G/N$ is isomorphic to?", it is equivalent to answer "how do you compute the right hand side of the first isomorphism theorem?". And we both suggest that you can compute the image of $G$ under the projection map to answer your question. Is this clearer?
  $endgroup$
  – NazimJ
  Apr 2 at 16:30
  
  

  
  
  
  

add a comment | 

$begingroup$

Usually, the first isomorphism theorem is easily satisfied by the projection map $pi:Gto G/N$ defined by $$ pi (g) = gN $$ for example see this question. Then in your specific dihedral group example, you can explicitly calculate $pi(G)$ and get that $$G/N = G/ker(pi)congpi(G) $$

share|cite|improve this answer

edited Apr 2 at 16:30

Shaun

11.1k113688

answered Apr 2 at 16:22

NazimJNazimJ

890110

$endgroup$

share|cite|improve this answer

edited Apr 2 at 16:30

Shaun

11.1k113688

answered Apr 2 at 16:22

NazimJNazimJ

890110

edited Apr 2 at 16:30

Shaun

11.1k113688

edited Apr 2 at 16:30

Shaun

11.1k113688

answered Apr 2 at 16:22

NazimJNazimJ

890110

answered Apr 2 at 16:22

NazimJNazimJ

890110