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Infinite Abelian subgroup of infinite non Abelian group example

Conjugacy in Infinite GroupsUsing semiproduct to construct a non-abelian group.When can an infinite abelian group be embedded in the multiplicative group of a field?A group is generated by two elements of order $2$ is infinite and non-abelianProve that any subgroup of a free Abelian group of rank $r$ is free Abelian of rank at most $r$.Does there exist an infinite non-abelian group such that all of its proper subgroups become cyclic?Showing a group is Abelian using its presentation.Noncyclic (infinite) group with totally ordered subgroup latticeNon-split central extension of Z by a finite simple non-abelian groupA group with an infinite cyclic normal subgroup that has a finite cyclic quotient is abelian

7

$begingroup$

My thought is that we may take GL(2,F) as the group and this is obviously infinite and non abelian since matrix multiplication does not commute. Then I thought that if we make $langle grangle$, for some $g$ in $mathrmGL(2,F)$, which will be cyclic and hence Abelian, for instance:
$ g=
bigg[
beginmatrix
1&0\0&2
endmatrix
bigg]
$. Then $g^n$ will be in the form $ g^n=
bigg[
beginmatrix
1&0\0&2^n
endmatrix
bigg]
$. This is obviously infinite since $g^n=e Leftrightarrow n = 0$.
Would this example work? Much thanks in advance!

abstract-algebra group-theory

share|cite|improve this question

edited Apr 4 at 12:36

user1729

17.7k64294

asked Apr 4 at 12:18

JustWanderingJustWandering

692

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  If $A$ is an infinite abelian group and $H$ is a finite, non-abelian group then $Atimes H$ works. [Also, you could take $F=mathbbZ$ in your example to get something easy to work with, but which isn't a field :-) ]
  $endgroup$
  – user1729
  Apr 4 at 12:32
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A physical example: If you rotate a 3D object around the z-axis, those rotations are abelian. However, if you rotate it about both the z-axis and the x-axis, that's non-abelian.
  $endgroup$
  – Mateen Ulhaq
  2 days ago
  

  
  
  
  

add a comment | 

7

$begingroup$

My thought is that we may take GL(2,F) as the group and this is obviously infinite and non abelian since matrix multiplication does not commute. Then I thought that if we make $langle grangle$, for some $g$ in $mathrmGL(2,F)$, which will be cyclic and hence Abelian, for instance:
$ g=
bigg[
beginmatrix
1&0\0&2
endmatrix
bigg]
$. Then $g^n$ will be in the form $ g^n=
bigg[
beginmatrix
1&0\0&2^n
endmatrix
bigg]
$. This is obviously infinite since $g^n=e Leftrightarrow n = 0$.
Would this example work? Much thanks in advance!

abstract-algebra group-theory

share|cite|improve this question

edited Apr 4 at 12:36

user1729

17.7k64294

asked Apr 4 at 12:18

JustWanderingJustWandering

692

$endgroup$

• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  $begingroup$
  If $A$ is an infinite abelian group and $H$ is a finite, non-abelian group then $Atimes H$ works. [Also, you could take $F=mathbbZ$ in your example to get something easy to work with, but which isn't a field :-) ]
  $endgroup$
  – user1729
  Apr 4 at 12:32
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A physical example: If you rotate a 3D object around the z-axis, those rotations are abelian. However, if you rotate it about both the z-axis and the x-axis, that's non-abelian.
  $endgroup$
  – Mateen Ulhaq
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

My thought is that we may take GL(2,F) as the group and this is obviously infinite and non abelian since matrix multiplication does not commute. Then I thought that if we make $langle grangle$, for some $g$ in $mathrmGL(2,F)$, which will be cyclic and hence Abelian, for instance:
$ g=
bigg[
beginmatrix
1&0\0&2
endmatrix
bigg]
$. Then $g^n$ will be in the form $ g^n=
bigg[
beginmatrix
1&0\0&2^n
endmatrix
bigg]
$. This is obviously infinite since $g^n=e Leftrightarrow n = 0$.
Would this example work? Much thanks in advance!

abstract-algebra group-theory

share|cite|improve this question

edited Apr 4 at 12:36

user1729

17.7k64294

asked Apr 4 at 12:18

JustWanderingJustWandering

692

$endgroup$

share|cite|improve this question

edited Apr 4 at 12:36

user1729

17.7k64294

asked Apr 4 at 12:18

JustWanderingJustWandering

692

share|cite|improve this question

edited Apr 4 at 12:36

user1729

17.7k64294

asked Apr 4 at 12:18

JustWanderingJustWandering

692

edited Apr 4 at 12:36

user1729

17.7k64294

edited Apr 4 at 12:36

user1729

17.7k64294

asked Apr 4 at 12:18

JustWanderingJustWandering

692

asked Apr 4 at 12:18

JustWanderingJustWandering

692

5 Answers
5

active

oldest

votes

9

$begingroup$

The simplest example is $G=mathbb Z times S_3$ and $H=mathbb Z$.

share|cite|improve this answer

answered Apr 4 at 12:31

lhflhf

167k11172404

$endgroup$

• 
  
  
  

  
  11
  

  
  
  
  
  
  

  
  $begingroup$
  It's funny because the groups aren't simple.
  $endgroup$
  – Servaes
  Apr 4 at 12:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's math humor! :-)
  $endgroup$
  – alexis
  2 days ago
  

  
  
  
  

add a comment | 

8

$begingroup$

This example works indeed, if $F$ is infinite and $2^nneq1$ in $F$ for all non-zero $ninBbbZ$. This is satisfied for obvious candidates for $F$ such as $BbbR$, $BbbC$ and $BbbQ$, but fails for other candidates such as the finite fields $BbbF_q$, but also infinite fields of positive characteristic such as $BbbF_p(T)$.

Assuming $F$ is a field, the condition that $2^nneq1$ for all non-zero $ninBbbZ$ is equivalent to $operatornamecharF=0$, from which it follows that $F$ is infinite. So your example works if and only if $operatornamecharF=0$.

share|cite|improve this answer

answered Apr 4 at 12:22

ServaesServaes

30k342101

$endgroup$

add a comment | 

5

$begingroup$

Assuming that $Bbb F$ has characteristic $0,$ that definitely works. Nicely done!

It also allows you to prove an inclusion $Bbb Zhookrightarrow GL(2,Bbb F).$

share|cite|improve this answer

edited Apr 4 at 12:23

answered Apr 4 at 12:22

Cameron BuieCameron Buie

86.6k773161

$endgroup$

add a comment | 

1

$begingroup$

A simple example: let $G = S(mathbb Z)$, the group of all permutations of the integers. Let $A$ be the subgroup generated by the transpositions $(n,n+1) $. Since the generating transpositions are all pairwise disjoint, they trivially commute with each other.

share|cite|improve this answer

answered Apr 4 at 15:51

John ColemanJohn Coleman

4,00311224

$endgroup$

add a comment | 

-1

$begingroup$

Yes it works if you take $F$ to be an infinite field for example.

Although, as was pointed out by others even in this case you'll have to make some assumptions on $F$ to get your particular example working.

I guess it'd be more natural to consider the subset of all diagonal submatrices. It certainly is a subgroup as $mathrmdiag(x,y)^-1 = mathrmdiag(x^-1, y^-1)$.

This subgroup is isomorphic to $F^times oplus F^times$, which is abelian and infinite if $F$ is.

share|cite|improve this answer

edited Apr 4 at 12:27

answered Apr 4 at 12:21

lushlush

757116

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It does not work for infinite fields such as $BbbF_p(T)$.
  $endgroup$
  – Servaes
  Apr 4 at 12:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ah obviously yes.
  $endgroup$
  – lush
  Apr 4 at 12:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Changed it @Servaes, I had forgotten that he asked for a particular example to work.
  $endgroup$
  – lush
  Apr 4 at 12:34
  

  
  
  
  

add a comment | 

Your Answer

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9

$begingroup$

The simplest example is $G=mathbb Z times S_3$ and $H=mathbb Z$.

share|cite|improve this answer

answered Apr 4 at 12:31

lhflhf

167k11172404

$endgroup$

• 
  
  
  

  
  11
  

  
  
  
  
  
  

  
  $begingroup$
  It's funny because the groups aren't simple.
  $endgroup$
  – Servaes
  Apr 4 at 12:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's math humor! :-)
  $endgroup$
  – alexis
  2 days ago
  

  
  
  
  

add a comment | 

9

$begingroup$

The simplest example is $G=mathbb Z times S_3$ and $H=mathbb Z$.

share|cite|improve this answer

answered Apr 4 at 12:31

lhflhf

167k11172404

$endgroup$

• 
  
  
  

  
  11
  

  
  
  
  
  
  

  
  $begingroup$
  It's funny because the groups aren't simple.
  $endgroup$
  – Servaes
  Apr 4 at 12:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  That's math humor! :-)
  $endgroup$
  – alexis
  2 days ago
  

  
  
  
  

add a comment | 

$begingroup$

The simplest example is $G=mathbb Z times S_3$ and $H=mathbb Z$.

share|cite|improve this answer

answered Apr 4 at 12:31

lhflhf

167k11172404

$endgroup$

share|cite|improve this answer

answered Apr 4 at 12:31

lhflhf

167k11172404

answered Apr 4 at 12:31

lhflhf

167k11172404

answered Apr 4 at 12:31

lhflhf

167k11172404

8

$begingroup$

This example works indeed, if $F$ is infinite and $2^nneq1$ in $F$ for all non-zero $ninBbbZ$. This is satisfied for obvious candidates for $F$ such as $BbbR$, $BbbC$ and $BbbQ$, but fails for other candidates such as the finite fields $BbbF_q$, but also infinite fields of positive characteristic such as $BbbF_p(T)$.

Assuming $F$ is a field, the condition that $2^nneq1$ for all non-zero $ninBbbZ$ is equivalent to $operatornamecharF=0$, from which it follows that $F$ is infinite. So your example works if and only if $operatornamecharF=0$.

share|cite|improve this answer

answered Apr 4 at 12:22

ServaesServaes

30k342101

$endgroup$

add a comment | 

8

$begingroup$

This example works indeed, if $F$ is infinite and $2^nneq1$ in $F$ for all non-zero $ninBbbZ$. This is satisfied for obvious candidates for $F$ such as $BbbR$, $BbbC$ and $BbbQ$, but fails for other candidates such as the finite fields $BbbF_q$, but also infinite fields of positive characteristic such as $BbbF_p(T)$.

Assuming $F$ is a field, the condition that $2^nneq1$ for all non-zero $ninBbbZ$ is equivalent to $operatornamecharF=0$, from which it follows that $F$ is infinite. So your example works if and only if $operatornamecharF=0$.

share|cite|improve this answer

answered Apr 4 at 12:22

ServaesServaes

30k342101

$endgroup$

add a comment | 

$begingroup$

This example works indeed, if $F$ is infinite and $2^nneq1$ in $F$ for all non-zero $ninBbbZ$. This is satisfied for obvious candidates for $F$ such as $BbbR$, $BbbC$ and $BbbQ$, but fails for other candidates such as the finite fields $BbbF_q$, but also infinite fields of positive characteristic such as $BbbF_p(T)$.

Assuming $F$ is a field, the condition that $2^nneq1$ for all non-zero $ninBbbZ$ is equivalent to $operatornamecharF=0$, from which it follows that $F$ is infinite. So your example works if and only if $operatornamecharF=0$.

share|cite|improve this answer

answered Apr 4 at 12:22

ServaesServaes

30k342101

$endgroup$

share|cite|improve this answer

answered Apr 4 at 12:22

ServaesServaes

30k342101

answered Apr 4 at 12:22

ServaesServaes

30k342101

answered Apr 4 at 12:22

ServaesServaes

30k342101

5

$begingroup$

Assuming that $Bbb F$ has characteristic $0,$ that definitely works. Nicely done!

It also allows you to prove an inclusion $Bbb Zhookrightarrow GL(2,Bbb F).$

share|cite|improve this answer

edited Apr 4 at 12:23

answered Apr 4 at 12:22

Cameron BuieCameron Buie

86.6k773161

$endgroup$

add a comment | 

5

$begingroup$

Assuming that $Bbb F$ has characteristic $0,$ that definitely works. Nicely done!

It also allows you to prove an inclusion $Bbb Zhookrightarrow GL(2,Bbb F).$

share|cite|improve this answer

edited Apr 4 at 12:23

answered Apr 4 at 12:22

Cameron BuieCameron Buie

86.6k773161

$endgroup$

add a comment | 

$begingroup$

Assuming that $Bbb F$ has characteristic $0,$ that definitely works. Nicely done!

It also allows you to prove an inclusion $Bbb Zhookrightarrow GL(2,Bbb F).$

share|cite|improve this answer

edited Apr 4 at 12:23

answered Apr 4 at 12:22

Cameron BuieCameron Buie

86.6k773161

$endgroup$

share|cite|improve this answer

edited Apr 4 at 12:23

answered Apr 4 at 12:22

Cameron BuieCameron Buie

86.6k773161

answered Apr 4 at 12:22

Cameron BuieCameron Buie

86.6k773161

answered Apr 4 at 12:22

Cameron BuieCameron Buie

86.6k773161

1

$begingroup$

A simple example: let $G = S(mathbb Z)$, the group of all permutations of the integers. Let $A$ be the subgroup generated by the transpositions $(n,n+1) $. Since the generating transpositions are all pairwise disjoint, they trivially commute with each other.

share|cite|improve this answer

answered Apr 4 at 15:51

John ColemanJohn Coleman

4,00311224

$endgroup$

add a comment | 

1

$begingroup$

A simple example: let $G = S(mathbb Z)$, the group of all permutations of the integers. Let $A$ be the subgroup generated by the transpositions $(n,n+1) $. Since the generating transpositions are all pairwise disjoint, they trivially commute with each other.

share|cite|improve this answer

answered Apr 4 at 15:51

John ColemanJohn Coleman

4,00311224

$endgroup$

add a comment | 

$begingroup$

A simple example: let $G = S(mathbb Z)$, the group of all permutations of the integers. Let $A$ be the subgroup generated by the transpositions $(n,n+1) $. Since the generating transpositions are all pairwise disjoint, they trivially commute with each other.

share|cite|improve this answer

answered Apr 4 at 15:51

John ColemanJohn Coleman

4,00311224

$endgroup$

share|cite|improve this answer

answered Apr 4 at 15:51

John ColemanJohn Coleman

4,00311224

answered Apr 4 at 15:51

John ColemanJohn Coleman

4,00311224

answered Apr 4 at 15:51

John ColemanJohn Coleman

4,00311224

-1

$begingroup$

Yes it works if you take $F$ to be an infinite field for example.

Although, as was pointed out by others even in this case you'll have to make some assumptions on $F$ to get your particular example working.

I guess it'd be more natural to consider the subset of all diagonal submatrices. It certainly is a subgroup as $mathrmdiag(x,y)^-1 = mathrmdiag(x^-1, y^-1)$.

This subgroup is isomorphic to $F^times oplus F^times$, which is abelian and infinite if $F$ is.

share|cite|improve this answer

edited Apr 4 at 12:27

answered Apr 4 at 12:21

lushlush

757116

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It does not work for infinite fields such as $BbbF_p(T)$.
  $endgroup$
  – Servaes
  Apr 4 at 12:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ah obviously yes.
  $endgroup$
  – lush
  Apr 4 at 12:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Changed it @Servaes, I had forgotten that he asked for a particular example to work.
  $endgroup$
  – lush
  Apr 4 at 12:34
  

  
  
  
  

add a comment | 

-1

$begingroup$

Yes it works if you take $F$ to be an infinite field for example.

Although, as was pointed out by others even in this case you'll have to make some assumptions on $F$ to get your particular example working.

I guess it'd be more natural to consider the subset of all diagonal submatrices. It certainly is a subgroup as $mathrmdiag(x,y)^-1 = mathrmdiag(x^-1, y^-1)$.

This subgroup is isomorphic to $F^times oplus F^times$, which is abelian and infinite if $F$ is.

share|cite|improve this answer

edited Apr 4 at 12:27

answered Apr 4 at 12:21

lushlush

757116

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  It does not work for infinite fields such as $BbbF_p(T)$.
  $endgroup$
  – Servaes
  Apr 4 at 12:22
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  ah obviously yes.
  $endgroup$
  – lush
  Apr 4 at 12:23
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Changed it @Servaes, I had forgotten that he asked for a particular example to work.
  $endgroup$
  – lush
  Apr 4 at 12:34
  

  
  
  
  

add a comment | 

$begingroup$

Yes it works if you take $F$ to be an infinite field for example.

Although, as was pointed out by others even in this case you'll have to make some assumptions on $F$ to get your particular example working.

I guess it'd be more natural to consider the subset of all diagonal submatrices. It certainly is a subgroup as $mathrmdiag(x,y)^-1 = mathrmdiag(x^-1, y^-1)$.

This subgroup is isomorphic to $F^times oplus F^times$, which is abelian and infinite if $F$ is.

share|cite|improve this answer

edited Apr 4 at 12:27

answered Apr 4 at 12:21

lushlush

757116

$endgroup$

share|cite|improve this answer

edited Apr 4 at 12:27

answered Apr 4 at 12:21

lushlush

757116

answered Apr 4 at 12:21

lushlush

757116

answered Apr 4 at 12:21

lushlush

757116