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 11 hours ago
user1729
17.6k64294
asked 12 hours ago
JustWanderingJustWandering
562
$endgroup$
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 11 hours ago
user1729
17.6k64294
asked 12 hours ago
JustWanderingJustWandering
562
$endgroup$
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
12 hours 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 11 hours ago
user1729
17.6k64294
asked 12 hours ago
JustWanderingJustWandering
562
$endgroup$
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 11 hours ago
user1729
17.6k64294
asked 12 hours ago
JustWanderingJustWandering
562
edited 11 hours ago
user1729
17.6k64294
asked 12 hours ago
JustWanderingJustWandering
562
edited 11 hours ago
user1729
17.6k64294
edited 11 hours ago
user1729
17.6k64294
edited 11 hours ago
user1729
17.6k64294
17.6k64294
asked 12 hours ago
JustWanderingJustWandering
562
asked 12 hours ago
JustWanderingJustWandering
562
asked 12 hours ago
JustWanderingJustWandering
562
562
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
12 hours 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
12 hours 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
12 hours ago
$begingroup$
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
12 hours 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 12 hours ago
lhflhf
167k11172404
$endgroup$
7
$begingroup$
It's funny because the groups aren't simple.
$endgroup$
– Servaes
11 hours 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 12 hours ago
ServaesServaes
29.8k342101
$endgroup$
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 12 hours ago
Cameron BuieCameron Buie
86.4k773161
$endgroup$
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 in mathbb Z, n , rm even $. Since the generating transpositions are all pairwise disjoint, they trivially commute with each other.
answered 8 hours ago
John ColemanJohn Coleman
3,99311224
$endgroup$
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 12 hours ago
lushlush
732115
$endgroup$
$begingroup$
It does not work for infinite fields such as $BbbF_p(T)$.
$endgroup$
– Servaes
12 hours ago
$begingroup$
ah obviously yes.
$endgroup$
– lush
12 hours ago
$begingroup$
Changed it @Servaes, I had forgotten that he asked for a particular example to work.
$endgroup$
– lush
12 hours ago
add a comment |
Your Answer
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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 12 hours ago
lhflhf
167k11172404
$endgroup$
7
$begingroup$
It's funny because the groups aren't simple.
$endgroup$
– Servaes
11 hours ago
add a comment |
$begingroup$
The simplest example is $G=mathbb Z times S_3$ and $H=mathbb Z$.
answered 12 hours ago
lhflhf
167k11172404
$endgroup$
7
$begingroup$
It's funny because the groups aren't simple.
$endgroup$
– Servaes
11 hours ago
add a comment |
$begingroup$
The simplest example is $G=mathbb Z times S_3$ and $H=mathbb Z$.
answered 12 hours ago
lhflhf
167k11172404
$endgroup$
The simplest example is $G=mathbb Z times S_3$ and $H=mathbb Z$.
answered 12 hours ago
lhflhf
167k11172404
answered 12 hours ago
lhflhf
167k11172404
answered 12 hours ago
lhflhf
167k11172404
answered 12 hours ago
lhflhf
167k11172404
167k11172404
7
$begingroup$
It's funny because the groups aren't simple.
$endgroup$
– Servaes
11 hours ago
add a comment |
7
$begingroup$
It's funny because the groups aren't simple.
$endgroup$
– Servaes
11 hours ago
7
7
$begingroup$
It's funny because the groups aren't simple.
$endgroup$
– Servaes
11 hours ago
$begingroup$
It's funny because the groups aren't simple.
$endgroup$
– Servaes
11 hours 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 12 hours ago
ServaesServaes
29.8k342101
$endgroup$
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 12 hours ago
ServaesServaes
29.8k342101
$endgroup$
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 12 hours ago
ServaesServaes
29.8k342101
$endgroup$
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 12 hours ago
ServaesServaes
29.8k342101
answered 12 hours ago
ServaesServaes
29.8k342101
answered 12 hours ago
ServaesServaes
29.8k342101
answered 12 hours ago
ServaesServaes
29.8k342101
29.8k342101
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 12 hours ago
Cameron BuieCameron Buie
86.4k773161
$endgroup$
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 12 hours ago
Cameron BuieCameron Buie
86.4k773161
$endgroup$
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 12 hours ago
Cameron BuieCameron Buie
86.4k773161
$endgroup$
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 12 hours ago
Cameron BuieCameron Buie
86.4k773161
edited 12 hours ago
answered 12 hours ago
Cameron BuieCameron Buie
86.4k773161
answered 12 hours ago
Cameron BuieCameron Buie
86.4k773161
answered 12 hours ago
Cameron BuieCameron Buie
86.4k773161
86.4k773161
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 in mathbb Z, n , rm even $. Since the generating transpositions are all pairwise disjoint, they trivially commute with each other.
answered 8 hours ago
John ColemanJohn Coleman
3,99311224
$endgroup$
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 in mathbb Z, n , rm even $. Since the generating transpositions are all pairwise disjoint, they trivially commute with each other.
answered 8 hours ago
John ColemanJohn Coleman
3,99311224
$endgroup$
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 in mathbb Z, n , rm even $. Since the generating transpositions are all pairwise disjoint, they trivially commute with each other.
answered 8 hours ago
John ColemanJohn Coleman
3,99311224
$endgroup$
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 in mathbb Z, n , rm even $. Since the generating transpositions are all pairwise disjoint, they trivially commute with each other.
answered 8 hours ago
John ColemanJohn Coleman
3,99311224
answered 8 hours ago
John ColemanJohn Coleman
3,99311224
answered 8 hours ago
John ColemanJohn Coleman
3,99311224
answered 8 hours ago
John ColemanJohn Coleman
3,99311224
3,99311224
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 12 hours ago
lushlush
732115
$endgroup$
$begingroup$
It does not work for infinite fields such as $BbbF_p(T)$.
$endgroup$
– Servaes
12 hours ago
$begingroup$
ah obviously yes.
$endgroup$
– lush
12 hours ago
$begingroup$
Changed it @Servaes, I had forgotten that he asked for a particular example to work.
$endgroup$
– lush
12 hours ago
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 12 hours ago
lushlush
732115
$endgroup$
$begingroup$
It does not work for infinite fields such as $BbbF_p(T)$.
$endgroup$
– Servaes
12 hours ago
$begingroup$
ah obviously yes.
$endgroup$
– lush
12 hours ago
$begingroup$
Changed it @Servaes, I had forgotten that he asked for a particular example to work.
$endgroup$
– lush
12 hours ago
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 12 hours ago
lushlush
732115
$endgroup$
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 12 hours ago
lushlush
732115
edited 12 hours ago
answered 12 hours ago
lushlush
732115
answered 12 hours ago
lushlush
732115
answered 12 hours ago
lushlush
732115
732115
$begingroup$
It does not work for infinite fields such as $BbbF_p(T)$.
$endgroup$
– Servaes
12 hours ago
$begingroup$
ah obviously yes.
$endgroup$
– lush
12 hours ago
$begingroup$
Changed it @Servaes, I had forgotten that he asked for a particular example to work.
$endgroup$
– lush
12 hours ago
add a comment |
$begingroup$
It does not work for infinite fields such as $BbbF_p(T)$.
$endgroup$
– Servaes
12 hours ago
$begingroup$
ah obviously yes.
$endgroup$
– lush
12 hours ago
$begingroup$
Changed it @Servaes, I had forgotten that he asked for a particular example to work.
$endgroup$
– lush
12 hours ago
$begingroup$
It does not work for infinite fields such as $BbbF_p(T)$.
$endgroup$
– Servaes
12 hours ago
$begingroup$
It does not work for infinite fields such as $BbbF_p(T)$.
$endgroup$
– Servaes
12 hours ago
$begingroup$
ah obviously yes.
$endgroup$
– lush
12 hours ago
$begingroup$
ah obviously yes.
$endgroup$
– lush
12 hours ago
$begingroup$
Changed it @Servaes, I had forgotten that he asked for a particular example to work.
$endgroup$
– lush
12 hours ago
$begingroup$
Changed it @Servaes, I had forgotten that he asked for a particular example to work.
$endgroup$
– lush
12 hours ago
add a comment |
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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
12 hours ago