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Subgroups of finitely generated groups are not necessarily finitely generated
Clash Royale CLAN TAG#URR8PPP
up vote
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I was wondering this today, and my algebra professor didn't know the answer.
Are subgroups of finitely generated groups finitely generated?
I suppose it is necessarily true for finitely generated abelian groups, but is it true in general?
And if not, is there a simple example of a finitely generated group with a non-finitely generated subgroup?
NOTE: This question has been merged with another question, asked by an undergraduate. For an example not involving free groups, please see Andreas Caranti's answer, which was the accepted answer on the merged question.
abstract-algebra group-theory
edited Feb 19 '13 at 18:19
Tara B
4,8101436
asked Oct 26 '10 at 10:13
crasic
2,01322028
add a comment |Â
up vote
27
down vote
favorite
I was wondering this today, and my algebra professor didn't know the answer.
Are subgroups of finitely generated groups finitely generated?
I suppose it is necessarily true for finitely generated abelian groups, but is it true in general?
And if not, is there a simple example of a finitely generated group with a non-finitely generated subgroup?
NOTE: This question has been merged with another question, asked by an undergraduate. For an example not involving free groups, please see Andreas Caranti's answer, which was the accepted answer on the merged question.
abstract-algebra group-theory
edited Feb 19 '13 at 18:19
Tara B
4,8101436
asked Oct 26 '10 at 10:13
crasic
2,01322028
2
Like you noted, this is indeed true for finitely generated abelian groups. Put another way, this is the same as the statement that every finitely generated Z-module is Noetherian, which holds since Z is Noetherian.
– Jason Juett
Jul 9 '16 at 11:33
Since there are Noetherian groups, you can guess the answer is no.
– TheoYou
Jul 31 '16 at 15:54
add a comment |Â
up vote
27
down vote
favorite
up vote
27
down vote
favorite
I was wondering this today, and my algebra professor didn't know the answer.
Are subgroups of finitely generated groups finitely generated?
I suppose it is necessarily true for finitely generated abelian groups, but is it true in general?
And if not, is there a simple example of a finitely generated group with a non-finitely generated subgroup?
NOTE: This question has been merged with another question, asked by an undergraduate. For an example not involving free groups, please see Andreas Caranti's answer, which was the accepted answer on the merged question.
abstract-algebra group-theory
edited Feb 19 '13 at 18:19
Tara B
4,8101436
asked Oct 26 '10 at 10:13
crasic
2,01322028
I was wondering this today, and my algebra professor didn't know the answer.
Are subgroups of finitely generated groups finitely generated?
I suppose it is necessarily true for finitely generated abelian groups, but is it true in general?
And if not, is there a simple example of a finitely generated group with a non-finitely generated subgroup?
NOTE: This question has been merged with another question, asked by an undergraduate. For an example not involving free groups, please see Andreas Caranti's answer, which was the accepted answer on the merged question.
abstract-algebra group-theory
edited Feb 19 '13 at 18:19
Tara B
4,8101436
asked Oct 26 '10 at 10:13
crasic
2,01322028
edited Feb 19 '13 at 18:19
Tara B
4,8101436
edited Feb 19 '13 at 18:19
Tara B
4,8101436
edited Feb 19 '13 at 18:19
Tara B
4,8101436
4,8101436
asked Oct 26 '10 at 10:13
crasic
2,01322028
asked Oct 26 '10 at 10:13
crasic
2,01322028
asked Oct 26 '10 at 10:13
crasic
2,01322028
2,01322028
2
Like you noted, this is indeed true for finitely generated abelian groups. Put another way, this is the same as the statement that every finitely generated Z-module is Noetherian, which holds since Z is Noetherian.
– Jason Juett
Jul 9 '16 at 11:33
Since there are Noetherian groups, you can guess the answer is no.
– TheoYou
Jul 31 '16 at 15:54
add a comment |Â
2
Like you noted, this is indeed true for finitely generated abelian groups. Put another way, this is the same as the statement that every finitely generated Z-module is Noetherian, which holds since Z is Noetherian.
– Jason Juett
Jul 9 '16 at 11:33
Since there are Noetherian groups, you can guess the answer is no.
– TheoYou
Jul 31 '16 at 15:54
2
2
Like you noted, this is indeed true for finitely generated abelian groups. Put another way, this is the same as the statement that every finitely generated Z-module is Noetherian, which holds since Z is Noetherian.
– Jason Juett
Jul 9 '16 at 11:33
Like you noted, this is indeed true for finitely generated abelian groups. Put another way, this is the same as the statement that every finitely generated Z-module is Noetherian, which holds since Z is Noetherian.
– Jason Juett
Jul 9 '16 at 11:33
Since there are Noetherian groups, you can guess the answer is no.
– TheoYou
Jul 31 '16 at 15:54
Since there are Noetherian groups, you can guess the answer is no.
– TheoYou
Jul 31 '16 at 15:54
add a comment |Â
5 Answers
5
active
oldest
votes
up vote
14
down vote
accepted
It is well-known that the free group $F_2$ on two generators
has as a subgroup a group isomorphic to a free group on a countably
infinite set of generators. See Qiaochu's example.
However a finite index subgroup of a finitely generated
group is finitely generated.
answered Oct 26 '10 at 10:20
Robin Chapman
19.1k24470
add a comment |Â
up vote
16
down vote
No. The example given on Wikipedia is that the free group $F_2$ contains a subgroup generated by $y^n x y^-n, n ge 1$, which is free on countably many generators.
answered Oct 26 '10 at 10:18
Qiaochu Yuan
269k32563899
But wouldn't that subgroup still be generated by $x,y$?
– Guacho Perez
Jun 29 '17 at 15:56
2
@Guacho: no. $y$ is not an element of the subgroup.
– Qiaochu Yuan
Jun 29 '17 at 23:32
add a comment |Â
up vote
15
down vote
A theorem of Higman, Neumann, and Neumann says that every countable group (no matter what horrible properties it might have) can be embedded as a subgroup of a group generated by $2$ elements. Thus subgroups of finitely generated groups can be pretty much anything.
add a comment |Â
up vote
9
down vote
Perhaps an elementary example can be provided by the wreath product of two copies of (the additive group of) $mathbfZ$.
Take copies $G_i$ of $mathbfZ$, for $i in mathbfZ$, and let
$$
B = coprod_i in mathbfZ G_i
$$
be the direct sum (coproduct).
Now let another copy $H = langle h rangle$ of $mathbfZ$ act on $B$ by
$$
G_i^h = G_i+1.
$$
More precisely, conjugation by $h$ takes a generator $g_i$ in the copy $G_i$ of $mathbfZ$ to a generator $g_i+1$ of the $(i+1)$-th copy.
Then the semidirect product $G = B rtimes H$ is generated by $g_0$ and $h$, but its subgroup $B$ requires an infinite number of generators.
It is easy to see what is going on. $B$ requires an infinite number of generators $g_i$. Now $h$ takes one of these generators by conjugation to all others.
answered Feb 19 '13 at 17:14
Andreas Caranti
55k34090
1
OK, turns out that this one works! I wouldn't have thought someone would know wreath products before free groups, but I guess it's understandable they might.
– Tara B
Feb 19 '13 at 17:15
1
@TaraB, thanks! I was exposed to this very construction (in the elementary terms I have tried to use above, and without any mention of the term wreath product) before I learned about free groups. The key concept is that of a semidirect product.
– Andreas Caranti
Feb 19 '13 at 17:21
Yes. I don't think I even encountered semidirect products until after free groups, myself, though. In fact I know I didn't, because I definitely knew about presentations by then.
– Tara B
Feb 19 '13 at 17:23
Note that this answer and these comments were made in a different context (this question was merged with another question asked by an undergraduate who hadn't encountered free groups).
– Tara B
Feb 19 '13 at 18:14
add a comment |Â
up vote
8
down vote
One of the easiest (counter)example is in Hungerford's Algebra.
Let $G$ be the multiplicative group generated by the real matrices
$$a = left(beginarrayl l
1 & 1\
0 & 1
endarrayright),
b = left(beginarrayl l
2 & 0\
0 & 1
endarrayright)
$$
Let $H$ be the subgroup of $G$ consisting of matrices that have $1$s on the main diagonal. Then $H$ is not finitely generated.
answered May 1 '14 at 19:18
mez
4,99032768
"H have 1s" means what?why H is not finitely generated.
– David Chan
Jul 8 '14 at 13:02
@DavidChan It means the diagonal is (1,1)
– mez
Jul 8 '14 at 18:34
1
isn't H just generated by a?
– Auburn
Jul 12 '16 at 19:20
add a comment |Â
5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
14
down vote
accepted
It is well-known that the free group $F_2$ on two generators
has as a subgroup a group isomorphic to a free group on a countably
infinite set of generators. See Qiaochu's example.
However a finite index subgroup of a finitely generated
group is finitely generated.
answered Oct 26 '10 at 10:20
Robin Chapman
19.1k24470
add a comment |Â
up vote
14
down vote
accepted
It is well-known that the free group $F_2$ on two generators
has as a subgroup a group isomorphic to a free group on a countably
infinite set of generators. See Qiaochu's example.
However a finite index subgroup of a finitely generated
group is finitely generated.
answered Oct 26 '10 at 10:20
Robin Chapman
19.1k24470
add a comment |Â
up vote
14
down vote
accepted
up vote
14
down vote
accepted
It is well-known that the free group $F_2$ on two generators
has as a subgroup a group isomorphic to a free group on a countably
infinite set of generators. See Qiaochu's example.
However a finite index subgroup of a finitely generated
group is finitely generated.
answered Oct 26 '10 at 10:20
Robin Chapman
19.1k24470
It is well-known that the free group $F_2$ on two generators
has as a subgroup a group isomorphic to a free group on a countably
infinite set of generators. See Qiaochu's example.
However a finite index subgroup of a finitely generated
group is finitely generated.
answered Oct 26 '10 at 10:20
Robin Chapman
19.1k24470
edited Oct 26 '10 at 10:27
answered Oct 26 '10 at 10:20
Robin Chapman
19.1k24470
answered Oct 26 '10 at 10:20
Robin Chapman
19.1k24470
answered Oct 26 '10 at 10:20
Robin Chapman
19.1k24470
19.1k24470
add a comment |Â
add a comment |Â
up vote
16
down vote
No. The example given on Wikipedia is that the free group $F_2$ contains a subgroup generated by $y^n x y^-n, n ge 1$, which is free on countably many generators.
answered Oct 26 '10 at 10:18
Qiaochu Yuan
269k32563899
But wouldn't that subgroup still be generated by $x,y$?
– Guacho Perez
Jun 29 '17 at 15:56
2
@Guacho: no. $y$ is not an element of the subgroup.
– Qiaochu Yuan
Jun 29 '17 at 23:32
add a comment |Â
up vote
16
down vote
No. The example given on Wikipedia is that the free group $F_2$ contains a subgroup generated by $y^n x y^-n, n ge 1$, which is free on countably many generators.
answered Oct 26 '10 at 10:18
Qiaochu Yuan
269k32563899
But wouldn't that subgroup still be generated by $x,y$?
– Guacho Perez
Jun 29 '17 at 15:56
2
@Guacho: no. $y$ is not an element of the subgroup.
– Qiaochu Yuan
Jun 29 '17 at 23:32
add a comment |Â
up vote
16
down vote
up vote
16
down vote
No. The example given on Wikipedia is that the free group $F_2$ contains a subgroup generated by $y^n x y^-n, n ge 1$, which is free on countably many generators.
answered Oct 26 '10 at 10:18
Qiaochu Yuan
269k32563899
No. The example given on Wikipedia is that the free group $F_2$ contains a subgroup generated by $y^n x y^-n, n ge 1$, which is free on countably many generators.
answered Oct 26 '10 at 10:18
Qiaochu Yuan
269k32563899
answered Oct 26 '10 at 10:18
Qiaochu Yuan
269k32563899
answered Oct 26 '10 at 10:18
Qiaochu Yuan
269k32563899
answered Oct 26 '10 at 10:18
Qiaochu Yuan
269k32563899
269k32563899
But wouldn't that subgroup still be generated by $x,y$?
– Guacho Perez
Jun 29 '17 at 15:56
2
@Guacho: no. $y$ is not an element of the subgroup.
– Qiaochu Yuan
Jun 29 '17 at 23:32
add a comment |Â
But wouldn't that subgroup still be generated by $x,y$?
– Guacho Perez
Jun 29 '17 at 15:56
2
@Guacho: no. $y$ is not an element of the subgroup.
– Qiaochu Yuan
Jun 29 '17 at 23:32
But wouldn't that subgroup still be generated by $x,y$?
– Guacho Perez
Jun 29 '17 at 15:56
But wouldn't that subgroup still be generated by $x,y$?
– Guacho Perez
Jun 29 '17 at 15:56
2
2
@Guacho: no. $y$ is not an element of the subgroup.
– Qiaochu Yuan
Jun 29 '17 at 23:32
@Guacho: no. $y$ is not an element of the subgroup.
– Qiaochu Yuan
Jun 29 '17 at 23:32
add a comment |Â
up vote
15
down vote
A theorem of Higman, Neumann, and Neumann says that every countable group (no matter what horrible properties it might have) can be embedded as a subgroup of a group generated by $2$ elements. Thus subgroups of finitely generated groups can be pretty much anything.
add a comment |Â
up vote
15
down vote
A theorem of Higman, Neumann, and Neumann says that every countable group (no matter what horrible properties it might have) can be embedded as a subgroup of a group generated by $2$ elements. Thus subgroups of finitely generated groups can be pretty much anything.
add a comment |Â
up vote
15
down vote
up vote
15
down vote
A theorem of Higman, Neumann, and Neumann says that every countable group (no matter what horrible properties it might have) can be embedded as a subgroup of a group generated by $2$ elements. Thus subgroups of finitely generated groups can be pretty much anything.
A theorem of Higman, Neumann, and Neumann says that every countable group (no matter what horrible properties it might have) can be embedded as a subgroup of a group generated by $2$ elements. Thus subgroups of finitely generated groups can be pretty much anything.
answered Feb 15 '11 at 7:12
user5067
add a comment |Â
add a comment |Â
up vote
9
down vote
Perhaps an elementary example can be provided by the wreath product of two copies of (the additive group of) $mathbfZ$.
Take copies $G_i$ of $mathbfZ$, for $i in mathbfZ$, and let
$$
B = coprod_i in mathbfZ G_i
$$
be the direct sum (coproduct).
Now let another copy $H = langle h rangle$ of $mathbfZ$ act on $B$ by
$$
G_i^h = G_i+1.
$$
More precisely, conjugation by $h$ takes a generator $g_i$ in the copy $G_i$ of $mathbfZ$ to a generator $g_i+1$ of the $(i+1)$-th copy.
Then the semidirect product $G = B rtimes H$ is generated by $g_0$ and $h$, but its subgroup $B$ requires an infinite number of generators.
It is easy to see what is going on. $B$ requires an infinite number of generators $g_i$. Now $h$ takes one of these generators by conjugation to all others.
answered Feb 19 '13 at 17:14
Andreas Caranti
55k34090
1
OK, turns out that this one works! I wouldn't have thought someone would know wreath products before free groups, but I guess it's understandable they might.
– Tara B
Feb 19 '13 at 17:15
1
@TaraB, thanks! I was exposed to this very construction (in the elementary terms I have tried to use above, and without any mention of the term wreath product) before I learned about free groups. The key concept is that of a semidirect product.
– Andreas Caranti
Feb 19 '13 at 17:21
Yes. I don't think I even encountered semidirect products until after free groups, myself, though. In fact I know I didn't, because I definitely knew about presentations by then.
– Tara B
Feb 19 '13 at 17:23
Note that this answer and these comments were made in a different context (this question was merged with another question asked by an undergraduate who hadn't encountered free groups).
– Tara B
Feb 19 '13 at 18:14
add a comment |Â
up vote
9
down vote
Perhaps an elementary example can be provided by the wreath product of two copies of (the additive group of) $mathbfZ$.
Take copies $G_i$ of $mathbfZ$, for $i in mathbfZ$, and let
$$
B = coprod_i in mathbfZ G_i
$$
be the direct sum (coproduct).
Now let another copy $H = langle h rangle$ of $mathbfZ$ act on $B$ by
$$
G_i^h = G_i+1.
$$
More precisely, conjugation by $h$ takes a generator $g_i$ in the copy $G_i$ of $mathbfZ$ to a generator $g_i+1$ of the $(i+1)$-th copy.
Then the semidirect product $G = B rtimes H$ is generated by $g_0$ and $h$, but its subgroup $B$ requires an infinite number of generators.
It is easy to see what is going on. $B$ requires an infinite number of generators $g_i$. Now $h$ takes one of these generators by conjugation to all others.
answered Feb 19 '13 at 17:14
Andreas Caranti
55k34090
1
OK, turns out that this one works! I wouldn't have thought someone would know wreath products before free groups, but I guess it's understandable they might.
– Tara B
Feb 19 '13 at 17:15
1
@TaraB, thanks! I was exposed to this very construction (in the elementary terms I have tried to use above, and without any mention of the term wreath product) before I learned about free groups. The key concept is that of a semidirect product.
– Andreas Caranti
Feb 19 '13 at 17:21
Yes. I don't think I even encountered semidirect products until after free groups, myself, though. In fact I know I didn't, because I definitely knew about presentations by then.
– Tara B
Feb 19 '13 at 17:23
Note that this answer and these comments were made in a different context (this question was merged with another question asked by an undergraduate who hadn't encountered free groups).
– Tara B
Feb 19 '13 at 18:14
add a comment |Â
up vote
9
down vote
up vote
9
down vote
Perhaps an elementary example can be provided by the wreath product of two copies of (the additive group of) $mathbfZ$.
Take copies $G_i$ of $mathbfZ$, for $i in mathbfZ$, and let
$$
B = coprod_i in mathbfZ G_i
$$
be the direct sum (coproduct).
Now let another copy $H = langle h rangle$ of $mathbfZ$ act on $B$ by
$$
G_i^h = G_i+1.
$$
More precisely, conjugation by $h$ takes a generator $g_i$ in the copy $G_i$ of $mathbfZ$ to a generator $g_i+1$ of the $(i+1)$-th copy.
Then the semidirect product $G = B rtimes H$ is generated by $g_0$ and $h$, but its subgroup $B$ requires an infinite number of generators.
It is easy to see what is going on. $B$ requires an infinite number of generators $g_i$. Now $h$ takes one of these generators by conjugation to all others.
answered Feb 19 '13 at 17:14
Andreas Caranti
55k34090
Perhaps an elementary example can be provided by the wreath product of two copies of (the additive group of) $mathbfZ$.
Take copies $G_i$ of $mathbfZ$, for $i in mathbfZ$, and let
$$
B = coprod_i in mathbfZ G_i
$$
be the direct sum (coproduct).
Now let another copy $H = langle h rangle$ of $mathbfZ$ act on $B$ by
$$
G_i^h = G_i+1.
$$
More precisely, conjugation by $h$ takes a generator $g_i$ in the copy $G_i$ of $mathbfZ$ to a generator $g_i+1$ of the $(i+1)$-th copy.
Then the semidirect product $G = B rtimes H$ is generated by $g_0$ and $h$, but its subgroup $B$ requires an infinite number of generators.
It is easy to see what is going on. $B$ requires an infinite number of generators $g_i$. Now $h$ takes one of these generators by conjugation to all others.
answered Feb 19 '13 at 17:14
Andreas Caranti
55k34090
answered Feb 19 '13 at 17:14
Andreas Caranti
55k34090
answered Feb 19 '13 at 17:14
Andreas Caranti
55k34090
answered Feb 19 '13 at 17:14
Andreas Caranti
55k34090
55k34090
1
OK, turns out that this one works! I wouldn't have thought someone would know wreath products before free groups, but I guess it's understandable they might.
– Tara B
Feb 19 '13 at 17:15
1
@TaraB, thanks! I was exposed to this very construction (in the elementary terms I have tried to use above, and without any mention of the term wreath product) before I learned about free groups. The key concept is that of a semidirect product.
– Andreas Caranti
Feb 19 '13 at 17:21
Yes. I don't think I even encountered semidirect products until after free groups, myself, though. In fact I know I didn't, because I definitely knew about presentations by then.
– Tara B
Feb 19 '13 at 17:23
Note that this answer and these comments were made in a different context (this question was merged with another question asked by an undergraduate who hadn't encountered free groups).
– Tara B
Feb 19 '13 at 18:14
add a comment |Â
1
OK, turns out that this one works! I wouldn't have thought someone would know wreath products before free groups, but I guess it's understandable they might.
– Tara B
Feb 19 '13 at 17:15
1
@TaraB, thanks! I was exposed to this very construction (in the elementary terms I have tried to use above, and without any mention of the term wreath product) before I learned about free groups. The key concept is that of a semidirect product.
– Andreas Caranti
Feb 19 '13 at 17:21
Yes. I don't think I even encountered semidirect products until after free groups, myself, though. In fact I know I didn't, because I definitely knew about presentations by then.
– Tara B
Feb 19 '13 at 17:23
Note that this answer and these comments were made in a different context (this question was merged with another question asked by an undergraduate who hadn't encountered free groups).
– Tara B
Feb 19 '13 at 18:14
1
1
OK, turns out that this one works! I wouldn't have thought someone would know wreath products before free groups, but I guess it's understandable they might.
– Tara B
Feb 19 '13 at 17:15
OK, turns out that this one works! I wouldn't have thought someone would know wreath products before free groups, but I guess it's understandable they might.
– Tara B
Feb 19 '13 at 17:15
1
1
@TaraB, thanks! I was exposed to this very construction (in the elementary terms I have tried to use above, and without any mention of the term wreath product) before I learned about free groups. The key concept is that of a semidirect product.
– Andreas Caranti
Feb 19 '13 at 17:21
@TaraB, thanks! I was exposed to this very construction (in the elementary terms I have tried to use above, and without any mention of the term wreath product) before I learned about free groups. The key concept is that of a semidirect product.
– Andreas Caranti
Feb 19 '13 at 17:21
Yes. I don't think I even encountered semidirect products until after free groups, myself, though. In fact I know I didn't, because I definitely knew about presentations by then.
– Tara B
Feb 19 '13 at 17:23
Yes. I don't think I even encountered semidirect products until after free groups, myself, though. In fact I know I didn't, because I definitely knew about presentations by then.
– Tara B
Feb 19 '13 at 17:23
Note that this answer and these comments were made in a different context (this question was merged with another question asked by an undergraduate who hadn't encountered free groups).
– Tara B
Feb 19 '13 at 18:14
Note that this answer and these comments were made in a different context (this question was merged with another question asked by an undergraduate who hadn't encountered free groups).
– Tara B
Feb 19 '13 at 18:14
add a comment |Â
up vote
8
down vote
One of the easiest (counter)example is in Hungerford's Algebra.
Let $G$ be the multiplicative group generated by the real matrices
$$a = left(beginarrayl l
1 & 1\
0 & 1
endarrayright),
b = left(beginarrayl l
2 & 0\
0 & 1
endarrayright)
$$
Let $H$ be the subgroup of $G$ consisting of matrices that have $1$s on the main diagonal. Then $H$ is not finitely generated.
answered May 1 '14 at 19:18
mez
4,99032768
"H have 1s" means what?why H is not finitely generated.
– David Chan
Jul 8 '14 at 13:02
@DavidChan It means the diagonal is (1,1)
– mez
Jul 8 '14 at 18:34
1
isn't H just generated by a?
– Auburn
Jul 12 '16 at 19:20
add a comment |Â
up vote
8
down vote
One of the easiest (counter)example is in Hungerford's Algebra.
Let $G$ be the multiplicative group generated by the real matrices
$$a = left(beginarrayl l
1 & 1\
0 & 1
endarrayright),
b = left(beginarrayl l
2 & 0\
0 & 1
endarrayright)
$$
Let $H$ be the subgroup of $G$ consisting of matrices that have $1$s on the main diagonal. Then $H$ is not finitely generated.
answered May 1 '14 at 19:18
mez
4,99032768
"H have 1s" means what?why H is not finitely generated.
– David Chan
Jul 8 '14 at 13:02
@DavidChan It means the diagonal is (1,1)
– mez
Jul 8 '14 at 18:34
1
isn't H just generated by a?
– Auburn
Jul 12 '16 at 19:20
add a comment |Â
up vote
8
down vote
up vote
8
down vote
One of the easiest (counter)example is in Hungerford's Algebra.
Let $G$ be the multiplicative group generated by the real matrices
$$a = left(beginarrayl l
1 & 1\
0 & 1
endarrayright),
b = left(beginarrayl l
2 & 0\
0 & 1
endarrayright)
$$
Let $H$ be the subgroup of $G$ consisting of matrices that have $1$s on the main diagonal. Then $H$ is not finitely generated.
answered May 1 '14 at 19:18
mez
4,99032768
One of the easiest (counter)example is in Hungerford's Algebra.
Let $G$ be the multiplicative group generated by the real matrices
$$a = left(beginarrayl l
1 & 1\
0 & 1
endarrayright),
b = left(beginarrayl l
2 & 0\
0 & 1
endarrayright)
$$
Let $H$ be the subgroup of $G$ consisting of matrices that have $1$s on the main diagonal. Then $H$ is not finitely generated.
answered May 1 '14 at 19:18
mez
4,99032768
edited May 1 '14 at 19:31
answered May 1 '14 at 19:18
mez
4,99032768
answered May 1 '14 at 19:18
mez
4,99032768
answered May 1 '14 at 19:18
mez
4,99032768
4,99032768
"H have 1s" means what?why H is not finitely generated.
– David Chan
Jul 8 '14 at 13:02
@DavidChan It means the diagonal is (1,1)
– mez
Jul 8 '14 at 18:34
1
isn't H just generated by a?
– Auburn
Jul 12 '16 at 19:20
add a comment |Â
"H have 1s" means what?why H is not finitely generated.
– David Chan
Jul 8 '14 at 13:02
@DavidChan It means the diagonal is (1,1)
– mez
Jul 8 '14 at 18:34
1
isn't H just generated by a?
– Auburn
Jul 12 '16 at 19:20
"H have 1s" means what?why H is not finitely generated.
– David Chan
Jul 8 '14 at 13:02
"H have 1s" means what?why H is not finitely generated.
– David Chan
Jul 8 '14 at 13:02
@DavidChan It means the diagonal is (1,1)
– mez
Jul 8 '14 at 18:34
@DavidChan It means the diagonal is (1,1)
– mez
Jul 8 '14 at 18:34
1
1
isn't H just generated by a?
– Auburn
Jul 12 '16 at 19:20
isn't H just generated by a?
– Auburn
Jul 12 '16 at 19:20
add a comment |Â
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2
Like you noted, this is indeed true for finitely generated abelian groups. Put another way, this is the same as the statement that every finitely generated Z-module is Noetherian, which holds since Z is Noetherian.
– Jason Juett
Jul 9 '16 at 11:33
Since there are Noetherian groups, you can guess the answer is no.
– TheoYou
Jul 31 '16 at 15:54