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If $phi: Grightarrow G'$ be a group onto homomorphism then show that $|G'|$ divides $|G|$

If $G$ is a finitely generated abelian group, then $G cong mathbbZ^r/phi(mathbbZ^s)$Homomorphism theorem and group productLet $G=langle arangle$ and $G'=langle brangle$ of order 12 and 4. Prove that there is a onto homomorphism with $o(Ker~ phi)=3$.Prove that a homomorphism $phi$ must be trivial.Does there exist an onto group homomorphism from $mathbbRto mathbbZ$?No homomorphism from $Z_16oplus Z_2$ onto $Z_4oplus Z_4$.For ring homomorphism $phi:Rto S$, prove that $phi$ is an isomorphism iff $phi$ is onto and $ker(phi)=left 0_R right $.Prove that if H is a group of odd order, there exists no nontrivial homomorphism from D_2017 to H?Suppose there is homomorphism from finite group G onto $Z_10$ Then G has Normal Subgroup of index 5 and 2.If $phi:G to overline G$ is group homomorphism, then prove that $|phi(G)|$ divides $|G|$.

0

$begingroup$

If $phi: Grightarrow G'$ be a (finite) group onto homomorphism then show that $|G'|$ divides $|G|$

$phi: Grightarrow G'$ be a group onto homomorphism then by Isomorphism theorem,

$G/Ker~ phi simeq G'$ and then $|G/Ker~ phi|= |G'|$ i.e $$|Ker~ phi|=fracG $$

But $|G'|$ divides $|G|$ can be concluded only when $|Ker~ phi|$ exist finitely. What to do?

Is any other alternative method to solve?

abstract-algebra group-theory group-isomorphism group-homomorphism

share|cite|improve this question

edited Oct 10 '16 at 19:10

rama_ran

asked Oct 10 '16 at 18:58

rama_ranrama_ran

300314

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The statement only makes sense for finite groups $G$. Since $ker(phi)$ is a subgroup of $G$, any situations where the kernel is infinite are irrelevant.
  $endgroup$
  – Kaj Hansen
  Oct 10 '16 at 19:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $G$ is a finite group, then $ker phi subset G$ is also finite.
  $endgroup$
  – u1571372
  Oct 10 '16 at 19:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DietrichBurde Without using Isomorphism theorem directly.
  $endgroup$
  – rama_ran
  Oct 10 '16 at 19:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We could just show that the cardinalities $|G/ker(phi)|$ and $|G'|$ coincide. This is "less" than the "isomorphism theorem". But the proof would be suspiciously similar to the proof of the isomorphism theorem. So no real progress. Perhaps one would ask why you don't want the isomorphism theorem; it is so basic.
  $endgroup$
  – Dietrich Burde
  Oct 10 '16 at 19:25
  

  
  
  
  

add a comment | 

0

$begingroup$

If $phi: Grightarrow G'$ be a (finite) group onto homomorphism then show that $|G'|$ divides $|G|$

$phi: Grightarrow G'$ be a group onto homomorphism then by Isomorphism theorem,

$G/Ker~ phi simeq G'$ and then $|G/Ker~ phi|= |G'|$ i.e $$|Ker~ phi|=fracG $$

But $|G'|$ divides $|G|$ can be concluded only when $|Ker~ phi|$ exist finitely. What to do?

Is any other alternative method to solve?

abstract-algebra group-theory group-isomorphism group-homomorphism

share|cite|improve this question

edited Oct 10 '16 at 19:10

rama_ran

asked Oct 10 '16 at 18:58

rama_ranrama_ran

300314

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  The statement only makes sense for finite groups $G$. Since $ker(phi)$ is a subgroup of $G$, any situations where the kernel is infinite are irrelevant.
  $endgroup$
  – Kaj Hansen
  Oct 10 '16 at 19:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If $G$ is a finite group, then $ker phi subset G$ is also finite.
  $endgroup$
  – u1571372
  Oct 10 '16 at 19:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @DietrichBurde Without using Isomorphism theorem directly.
  $endgroup$
  – rama_ran
  Oct 10 '16 at 19:20
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  We could just show that the cardinalities $|G/ker(phi)|$ and $|G'|$ coincide. This is "less" than the "isomorphism theorem". But the proof would be suspiciously similar to the proof of the isomorphism theorem. So no real progress. Perhaps one would ask why you don't want the isomorphism theorem; it is so basic.
  $endgroup$
  – Dietrich Burde
  Oct 10 '16 at 19:25
  

  
  
  
  

add a comment | 

$begingroup$

If $phi: Grightarrow G'$ be a (finite) group onto homomorphism then show that $|G'|$ divides $|G|$

$phi: Grightarrow G'$ be a group onto homomorphism then by Isomorphism theorem,

$G/Ker~ phi simeq G'$ and then $|G/Ker~ phi|= |G'|$ i.e $$|Ker~ phi|=fracG $$

But $|G'|$ divides $|G|$ can be concluded only when $|Ker~ phi|$ exist finitely. What to do?

Is any other alternative method to solve?

abstract-algebra group-theory group-isomorphism group-homomorphism

share|cite|improve this question

edited Oct 10 '16 at 19:10

rama_ran

asked Oct 10 '16 at 18:58

rama_ranrama_ran

300314

$endgroup$

share|cite|improve this question

edited Oct 10 '16 at 19:10

rama_ran

asked Oct 10 '16 at 18:58

rama_ranrama_ran

300314

share|cite|improve this question

edited Oct 10 '16 at 19:10

rama_ran

asked Oct 10 '16 at 18:58

rama_ranrama_ran

300314

asked Oct 10 '16 at 18:58

rama_ranrama_ran

300314

asked Oct 10 '16 at 18:58

rama_ranrama_ran

300314

2 Answers
2

active

oldest

votes

1

$begingroup$

It is slightly more clear to stick to $|G'| |Ker phi|= |G|$, and not to divide by $|Ker phi|$.

This clearly shows the divisibility in case $|G|$ is finite. If $G$ is not finite, I'd say the question does not make sense. But if one were to make sense of it then it would be again the equality $|G'| |Ker phi|= |G|$ that is relevant.

share|cite|improve this answer

answered Oct 10 '16 at 19:05

quid♦quid

37.2k95193

$endgroup$

add a comment | 

0

$begingroup$

Since the problem asks us to prove $|G'|$ divides $|G|$ we assume that $G$ and $G'$ are already finite groups, which implies that $kerphi$ is also finite (as it is a subgroup of $G$)

share|cite|improve this answer

answered Oct 10 '16 at 19:02

Jorge Fernández HidalgoJorge Fernández Hidalgo

76.8k1394195

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok. Understood. Is any other alternative method to solve?
  $endgroup$
  – rama_ran
  Oct 10 '16 at 19:09
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is probably the simplest way to solve it.
  $endgroup$
  – Jorge Fernández Hidalgo
  Oct 10 '16 at 19:15
  

  
  
  
  

add a comment | 

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1

$begingroup$

It is slightly more clear to stick to $|G'| |Ker phi|= |G|$, and not to divide by $|Ker phi|$.

This clearly shows the divisibility in case $|G|$ is finite. If $G$ is not finite, I'd say the question does not make sense. But if one were to make sense of it then it would be again the equality $|G'| |Ker phi|= |G|$ that is relevant.

share|cite|improve this answer

answered Oct 10 '16 at 19:05

quid♦quid

37.2k95193

$endgroup$

add a comment | 

1

$begingroup$

It is slightly more clear to stick to $|G'| |Ker phi|= |G|$, and not to divide by $|Ker phi|$.

This clearly shows the divisibility in case $|G|$ is finite. If $G$ is not finite, I'd say the question does not make sense. But if one were to make sense of it then it would be again the equality $|G'| |Ker phi|= |G|$ that is relevant.

share|cite|improve this answer

answered Oct 10 '16 at 19:05

quid♦quid

37.2k95193

$endgroup$

add a comment | 

$begingroup$

It is slightly more clear to stick to $|G'| |Ker phi|= |G|$, and not to divide by $|Ker phi|$.

This clearly shows the divisibility in case $|G|$ is finite. If $G$ is not finite, I'd say the question does not make sense. But if one were to make sense of it then it would be again the equality $|G'| |Ker phi|= |G|$ that is relevant.

share|cite|improve this answer

answered Oct 10 '16 at 19:05

quid♦quid

37.2k95193

$endgroup$

share|cite|improve this answer

answered Oct 10 '16 at 19:05

quid♦quid

37.2k95193

answered Oct 10 '16 at 19:05

quid♦quid

37.2k95193

answered Oct 10 '16 at 19:05

quid♦quid

37.2k95193

0

$begingroup$

Since the problem asks us to prove $|G'|$ divides $|G|$ we assume that $G$ and $G'$ are already finite groups, which implies that $kerphi$ is also finite (as it is a subgroup of $G$)

share|cite|improve this answer

answered Oct 10 '16 at 19:02

Jorge Fernández HidalgoJorge Fernández Hidalgo

76.8k1394195

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok. Understood. Is any other alternative method to solve?
  $endgroup$
  – rama_ran
  Oct 10 '16 at 19:09
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is probably the simplest way to solve it.
  $endgroup$
  – Jorge Fernández Hidalgo
  Oct 10 '16 at 19:15
  

  
  
  
  

add a comment | 

0

$begingroup$

Since the problem asks us to prove $|G'|$ divides $|G|$ we assume that $G$ and $G'$ are already finite groups, which implies that $kerphi$ is also finite (as it is a subgroup of $G$)

share|cite|improve this answer

answered Oct 10 '16 at 19:02

Jorge Fernández HidalgoJorge Fernández Hidalgo

76.8k1394195

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok. Understood. Is any other alternative method to solve?
  $endgroup$
  – rama_ran
  Oct 10 '16 at 19:09
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  This is probably the simplest way to solve it.
  $endgroup$
  – Jorge Fernández Hidalgo
  Oct 10 '16 at 19:15
  

  
  
  
  

add a comment | 

$begingroup$

Since the problem asks us to prove $|G'|$ divides $|G|$ we assume that $G$ and $G'$ are already finite groups, which implies that $kerphi$ is also finite (as it is a subgroup of $G$)

share|cite|improve this answer

answered Oct 10 '16 at 19:02

Jorge Fernández HidalgoJorge Fernández Hidalgo

76.8k1394195

$endgroup$

share|cite|improve this answer

answered Oct 10 '16 at 19:02

Jorge Fernández HidalgoJorge Fernández Hidalgo

76.8k1394195

answered Oct 10 '16 at 19:02

Jorge Fernández HidalgoJorge Fernández Hidalgo

76.8k1394195

answered Oct 10 '16 at 19:02

Jorge Fernández HidalgoJorge Fernández Hidalgo

76.8k1394195