Infinite Abelian subgroup of infinite non Abelian group exampleConjugacy 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
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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
$begingroup$
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
edited Apr 4 at 12:36
user1729
17.7k64294
asked Apr 4 at 12:18
JustWanderingJustWandering
692
$endgroup$
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
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
edited Apr 4 at 12:36
user1729
17.7k64294
asked Apr 4 at 12:18
JustWanderingJustWandering
692
$endgroup$
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
abstract-algebra group-theory
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
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
edited Apr 4 at 12:36
user1729
17.7k64294
17.7k64294
asked Apr 4 at 12:18
JustWanderingJustWandering
692
asked Apr 4 at 12:18
JustWanderingJustWandering
692
asked Apr 4 at 12:18
JustWanderingJustWandering
692
692
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 |
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
3
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
$begingroup$
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
$begingroup$
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 |
5 Answers
5
active
oldest
votes
$begingroup$
The simplest example is $G=mathbb Z times S_3$ and $H=mathbb Z$.
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 |
$begingroup$
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 $operatornamecharF=0$.
answered Apr 4 at 12:22
ServaesServaes
30k342101
$endgroup$
add a comment |
$begingroup$
Assuming that $Bbb F$ has characteristic $0,$ that definitely works. Nicely done!
It also allows you to prove an inclusion $Bbb Zhookrightarrow GL(2,Bbb F).$
answered Apr 4 at 12:22
Cameron BuieCameron Buie
86.6k773161
$endgroup$
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.
answered Apr 4 at 15:51
John ColemanJohn Coleman
4,00311224
$endgroup$
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.
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 |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The simplest example is $G=mathbb Z times S_3$ and $H=mathbb Z$.
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 $H=mathbb Z$.
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 $H=mathbb Z$.
answered Apr 4 at 12:31
lhflhf
167k11172404
$endgroup$
The simplest example is $G=mathbb Z times S_3$ and $H=mathbb Z$.
answered Apr 4 at 12:31
lhflhf
167k11172404
answered Apr 4 at 12:31
lhflhf
167k11172404
answered Apr 4 at 12:31
lhflhf
167k11172404
answered Apr 4 at 12:31
lhflhf
167k11172404
167k11172404
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 |
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
11
11
$begingroup$
It's funny because the groups aren't simple.
$endgroup$
– Servaes
Apr 4 at 12:34
$begingroup$
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
$begingroup$
That's math humor! :-)
$endgroup$
– alexis
2 days ago
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 $operatornamecharF=0$.
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 $operatornamecharF=0$.
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 $operatornamecharF=0$.
answered Apr 4 at 12:22
ServaesServaes
30k342101
$endgroup$
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 $operatornamecharF=0$.
answered Apr 4 at 12:22
ServaesServaes
30k342101
answered Apr 4 at 12:22
ServaesServaes
30k342101
answered Apr 4 at 12:22
ServaesServaes
30k342101
answered Apr 4 at 12:22
ServaesServaes
30k342101
30k342101
add a comment |
add a comment |
$begingroup$
Assuming that $Bbb F$ has characteristic $0,$ that definitely works. Nicely done!
It also allows you to prove an inclusion $Bbb Zhookrightarrow GL(2,Bbb F).$
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 $Bbb Zhookrightarrow GL(2,Bbb F).$
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 $Bbb Zhookrightarrow GL(2,Bbb F).$
answered Apr 4 at 12:22
Cameron BuieCameron Buie
86.6k773161
$endgroup$
Assuming that $Bbb F$ has characteristic $0,$ that definitely works. Nicely done!
It also allows you to prove an inclusion $Bbb Zhookrightarrow GL(2,Bbb F).$
answered Apr 4 at 12:22
Cameron BuieCameron Buie
86.6k773161
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
86.6k773161
add a comment |
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.
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.
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.
answered Apr 4 at 15:51
John ColemanJohn Coleman
4,00311224
$endgroup$
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.
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
answered Apr 4 at 15:51
John ColemanJohn Coleman
4,00311224
4,00311224
add a comment |
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.
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.
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.
answered Apr 4 at 12:21
lushlush
757116
$endgroup$
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.
answered Apr 4 at 12:21
lushlush
757116
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
757116
$begingroup$
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 |
$begingroup$
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
$begingroup$
It does not work for infinite fields such as $BbbF_p(T)$.
$endgroup$
– Servaes
Apr 4 at 12:22
$begingroup$
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
$begingroup$
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
$begingroup$
Changed it @Servaes, I had forgotten that he asked for a particular example to work.
$endgroup$
– lush
Apr 4 at 12:34
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$begingroup$
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