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How do I show that $mathbbQ$ and $mathbbQtimesmathbbQ$ are not group isomorphic? [duplicate]

Are the groups $mathbbC$ and $mathbbR$ isomorphic?To which group is the $mathbb Z_20^*$ isomorphic?Determine whether $mathbbZtimes mathbbZ$ and $mathbbZtimes mathbbZtimes mathbbZ$ are isomorphic groups or not.Prove that $mathbbQ^times$ not isomorphic to $mathbbZ^n$Show that $C_3 times C_3$ is not isomorphic to $C_9$How to prove that $(mathbb Q,+)$ is not isomorphic to $(mathbb Q^*,times)$?Quotient group of invertible matrices isomorphic to $mathbbRtimes$Is $mathbbC^times/mathbbR^times$ isomorphic to $mathbbC^times/pm 1$Heisenberg Group over $mathbbF_3$ is not isomorphic to $C_3 times C_3 times C_3$Showing that $2$ of the following groups are not isomorphicShow that $mathbbZ/n^2mathbbZ$ is not isomorphic to other group

2

1

$begingroup$

This question already has an answer here:

• 
  Are the groups $mathbbC$ and $mathbbR$ isomorphic?
  
  4 answers
  
  

It's easy to show that $mathbbQ$ and $mathbbQtimesmathbbQ$ are not ring isomorphic as the first one has no zero divisors where as the second one has zero divisors. But I can't find any solution in case of group isomorphism.

group-theory

share|cite|improve this question

edited Mar 17 at 16:05

Fabio Lucchini

9,51111426

asked Mar 17 at 13:08

A. B.A. B.

262

$endgroup$

marked as duplicate by GEdgar, Derek Holt, Cesareo, Eevee Trainer, Alex Provost Mar 19 at 4:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What are the operations?
  $endgroup$
  – Aniruddha Deshmukh
  Mar 17 at 13:10
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Are the groups $mathbbC$ and $mathbbR$ isomorphic? The additional question there about $mathbb Q$ and $mathbb Q[i]$ answers this question.
  $endgroup$
  – GEdgar
  Mar 17 at 13:16
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  More generally there's the notion of $mathbfQ$-rank of an abelian group $G$, which is obviously an isomorphism invariant. Namely, this is the maximal $k$ such that $G$ has a subgroup isomorphic to $mathbfZ^k$ ($infty$ if there's no max). Exercise: what's the $mathbfQ$-rank of $mathbfQ^n$?
  $endgroup$
  – YCor
  Mar 17 at 14:37
  
  

  
  
  
  

add a comment | 

2

1

$begingroup$

This question already has an answer here:

• 
  Are the groups $mathbbC$ and $mathbbR$ isomorphic?
  
  4 answers
  
  

It's easy to show that $mathbbQ$ and $mathbbQtimesmathbbQ$ are not ring isomorphic as the first one has no zero divisors where as the second one has zero divisors. But I can't find any solution in case of group isomorphism.

group-theory

share|cite|improve this question

edited Mar 17 at 16:05

Fabio Lucchini

9,51111426

asked Mar 17 at 13:08

A. B.A. B.

262

$endgroup$

marked as duplicate by GEdgar, Derek Holt, Cesareo, Eevee Trainer, Alex Provost Mar 19 at 4:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What are the operations?
  $endgroup$
  – Aniruddha Deshmukh
  Mar 17 at 13:10
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Possible duplicate of Are the groups $mathbbC$ and $mathbbR$ isomorphic? The additional question there about $mathbb Q$ and $mathbb Q[i]$ answers this question.
  $endgroup$
  – GEdgar
  Mar 17 at 13:16
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  More generally there's the notion of $mathbfQ$-rank of an abelian group $G$, which is obviously an isomorphism invariant. Namely, this is the maximal $k$ such that $G$ has a subgroup isomorphic to $mathbfZ^k$ ($infty$ if there's no max). Exercise: what's the $mathbfQ$-rank of $mathbfQ^n$?
  $endgroup$
  – YCor
  Mar 17 at 14:37
  
  

  
  
  
  

add a comment | 

$begingroup$

This question already has an answer here:

• 
  Are the groups $mathbbC$ and $mathbbR$ isomorphic?
  
  4 answers
  
  

It's easy to show that $mathbbQ$ and $mathbbQtimesmathbbQ$ are not ring isomorphic as the first one has no zero divisors where as the second one has zero divisors. But I can't find any solution in case of group isomorphism.

group-theory

share|cite|improve this question

edited Mar 17 at 16:05

Fabio Lucchini

9,51111426

asked Mar 17 at 13:08

A. B.A. B.

262

$endgroup$

share|cite|improve this question

edited Mar 17 at 16:05

Fabio Lucchini

9,51111426

asked Mar 17 at 13:08

A. B.A. B.

262

share|cite|improve this question

edited Mar 17 at 16:05

Fabio Lucchini

9,51111426

asked Mar 17 at 13:08

A. B.A. B.

262

edited Mar 17 at 16:05

Fabio Lucchini

9,51111426

edited Mar 17 at 16:05

Fabio Lucchini

9,51111426

asked Mar 17 at 13:08

A. B.A. B.

262

asked Mar 17 at 13:08

A. B.A. B.

262

2 Answers
2

active

oldest

votes

12

$begingroup$

In $mathbbQ$, as a group under addition, any two elements are commensurable; that is, for any $p,q$, there is some $r$ such that both $p$ and $q$ are multiples of $r$. Consequently, every finitely generated subgroup is cyclic.

The same is not true of $mathbbQtimesmathbbQ$; $(1,0)$ and $(0,1)$ are incommensurable.

share|cite|improve this answer

answered Mar 17 at 13:14

jmerryjmerry

16.4k11633

$endgroup$

add a comment | 

0

$begingroup$

Since $Bbb ZhookrightarrowBbb Q $ is a ring epimorphism, the scalar restriction functor $operatornameMod_Bbb QtooperatornameMod_Bbb Z $ is fully faithful, hence every isomorphism of abelian groups $Bbb QtoBbb Q^2$ lifts to a $Bbb Q $-vector space isomorphism.
Since they have different dimension as vector space, they cannot be isomorphic as abelian groups.

share|cite|improve this answer

answered Mar 17 at 16:01

Fabio LucchiniFabio Lucchini

9,51111426

$endgroup$

add a comment | 

12

$begingroup$

In $mathbbQ$, as a group under addition, any two elements are commensurable; that is, for any $p,q$, there is some $r$ such that both $p$ and $q$ are multiples of $r$. Consequently, every finitely generated subgroup is cyclic.

The same is not true of $mathbbQtimesmathbbQ$; $(1,0)$ and $(0,1)$ are incommensurable.

share|cite|improve this answer

answered Mar 17 at 13:14

jmerryjmerry

16.4k11633

$endgroup$

add a comment | 

12

$begingroup$

In $mathbbQ$, as a group under addition, any two elements are commensurable; that is, for any $p,q$, there is some $r$ such that both $p$ and $q$ are multiples of $r$. Consequently, every finitely generated subgroup is cyclic.

The same is not true of $mathbbQtimesmathbbQ$; $(1,0)$ and $(0,1)$ are incommensurable.

share|cite|improve this answer

answered Mar 17 at 13:14

jmerryjmerry

16.4k11633

$endgroup$

add a comment | 

$begingroup$

In $mathbbQ$, as a group under addition, any two elements are commensurable; that is, for any $p,q$, there is some $r$ such that both $p$ and $q$ are multiples of $r$. Consequently, every finitely generated subgroup is cyclic.

The same is not true of $mathbbQtimesmathbbQ$; $(1,0)$ and $(0,1)$ are incommensurable.

share|cite|improve this answer

answered Mar 17 at 13:14

jmerryjmerry

16.4k11633

$endgroup$

share|cite|improve this answer

answered Mar 17 at 13:14

jmerryjmerry

16.4k11633

answered Mar 17 at 13:14

jmerryjmerry

16.4k11633

answered Mar 17 at 13:14

jmerryjmerry

16.4k11633

0

$begingroup$

Since $Bbb ZhookrightarrowBbb Q $ is a ring epimorphism, the scalar restriction functor $operatornameMod_Bbb QtooperatornameMod_Bbb Z $ is fully faithful, hence every isomorphism of abelian groups $Bbb QtoBbb Q^2$ lifts to a $Bbb Q $-vector space isomorphism.
Since they have different dimension as vector space, they cannot be isomorphic as abelian groups.

share|cite|improve this answer

answered Mar 17 at 16:01

Fabio LucchiniFabio Lucchini

9,51111426

$endgroup$

add a comment | 

0

$begingroup$

Since $Bbb ZhookrightarrowBbb Q $ is a ring epimorphism, the scalar restriction functor $operatornameMod_Bbb QtooperatornameMod_Bbb Z $ is fully faithful, hence every isomorphism of abelian groups $Bbb QtoBbb Q^2$ lifts to a $Bbb Q $-vector space isomorphism.
Since they have different dimension as vector space, they cannot be isomorphic as abelian groups.

share|cite|improve this answer

answered Mar 17 at 16:01

Fabio LucchiniFabio Lucchini

9,51111426

$endgroup$

add a comment | 

$begingroup$

Since $Bbb ZhookrightarrowBbb Q $ is a ring epimorphism, the scalar restriction functor $operatornameMod_Bbb QtooperatornameMod_Bbb Z $ is fully faithful, hence every isomorphism of abelian groups $Bbb QtoBbb Q^2$ lifts to a $Bbb Q $-vector space isomorphism.
Since they have different dimension as vector space, they cannot be isomorphic as abelian groups.

share|cite|improve this answer

answered Mar 17 at 16:01

Fabio LucchiniFabio Lucchini

9,51111426

$endgroup$

share|cite|improve this answer

answered Mar 17 at 16:01

Fabio LucchiniFabio Lucchini

9,51111426

answered Mar 17 at 16:01

Fabio LucchiniFabio Lucchini

9,51111426

answered Mar 17 at 16:01

Fabio LucchiniFabio Lucchini

9,51111426