Mixing
What is the difference between a free Abelian group (of say a finite basis) and a finitely generated Abelian...
$begingroup$
Is the difference that the generating set of the finitely generating Abelian group need not be linearly independent? I understand that both the basis and the finite generating set span the Abelian group. Thanks for clarifying.
abstract-algebra abelian-groups
edited Jan 30 at 9:24
YuiTo Cheng
2,1863937
asked Jan 30 at 5:51
manifoldedmanifolded
49519
$endgroup$
add a comment |
$begingroup$
Is the difference that the generating set of the finitely generating Abelian group need not be linearly independent? I understand that both the basis and the finite generating set span the Abelian group. Thanks for clarifying.
abstract-algebra abelian-groups
edited Jan 30 at 9:24
YuiTo Cheng
2,1863937
asked Jan 30 at 5:51
manifoldedmanifolded
49519
$endgroup$
3
$begingroup$
Every finite Abelian group is finitely generated, but unless it has order $1$, it is not free Abelian.
$endgroup$
– Lord Shark the Unknown
Jan 30 at 5:56
$begingroup$
I see, that is a good point.
$endgroup$
– manifolded
Jan 30 at 6:05
add a comment |
$begingroup$
Is the difference that the generating set of the finitely generating Abelian group need not be linearly independent? I understand that both the basis and the finite generating set span the Abelian group. Thanks for clarifying.
abstract-algebra abelian-groups
edited Jan 30 at 9:24
YuiTo Cheng
2,1863937
asked Jan 30 at 5:51
manifoldedmanifolded
49519
$endgroup$
Is the difference that the generating set of the finitely generating Abelian group need not be linearly independent? I understand that both the basis and the finite generating set span the Abelian group. Thanks for clarifying.
abstract-algebra abelian-groups
abstract-algebra abelian-groups
edited Jan 30 at 9:24
YuiTo Cheng
2,1863937
asked Jan 30 at 5:51
manifoldedmanifolded
49519
edited Jan 30 at 9:24
YuiTo Cheng
2,1863937
asked Jan 30 at 5:51
manifoldedmanifolded
49519
edited Jan 30 at 9:24
YuiTo Cheng
2,1863937
edited Jan 30 at 9:24
YuiTo Cheng
2,1863937
edited Jan 30 at 9:24
YuiTo Cheng
2,1863937
2,1863937
asked Jan 30 at 5:51
manifoldedmanifolded
49519
asked Jan 30 at 5:51
manifoldedmanifolded
49519
asked Jan 30 at 5:51
manifoldedmanifolded
49519
49519
3
$begingroup$
Every finite Abelian group is finitely generated, but unless it has order $1$, it is not free Abelian.
$endgroup$
– Lord Shark the Unknown
Jan 30 at 5:56
$begingroup$
I see, that is a good point.
$endgroup$
– manifolded
Jan 30 at 6:05
add a comment |
3
$begingroup$
Every finite Abelian group is finitely generated, but unless it has order $1$, it is not free Abelian.
$endgroup$
– Lord Shark the Unknown
Jan 30 at 5:56
$begingroup$
I see, that is a good point.
$endgroup$
– manifolded
Jan 30 at 6:05
3
3
$begingroup$
Every finite Abelian group is finitely generated, but unless it has order $1$, it is not free Abelian.
$endgroup$
– Lord Shark the Unknown
Jan 30 at 5:56
$begingroup$
Every finite Abelian group is finitely generated, but unless it has order $1$, it is not free Abelian.
$endgroup$
– Lord Shark the Unknown
Jan 30 at 5:56
$begingroup$
I see, that is a good point.
$endgroup$
– manifolded
Jan 30 at 6:05
$begingroup$
I see, that is a good point.
$endgroup$
– manifolded
Jan 30 at 6:05
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Every group $G$ is the quotient of a free group $F$. To say that $G$ is finitely generated means that $F$ has a finite set of generators. Saying basis and linearly independent is the right idea, but isn't quite proper, since we reserve those words for vector spaces. A finitely generated group has a finite generating set and possibly some relations among those generators.
To address your comment and expand a bit, every group $G$ has a presentation as a set of generators and relations. For example, this abelian group:
$$G = leftlangle a, b mid aba^{-1}b^{-1}, a^3, b^7 rightrangle$$
Now this group presentation is really shorthand for $G$ being the quotient of the free group $langle a,brangle$ by the normal subgroup generated by the elements $langle aba^{-1}b^{-1}, a^3, b^7 rangle$. If you want to get fancy and use more categorical language, that presentation of $G$ will be the cokernel of the map $langle x,y,z rangle to langle a, b rangle$, such that $x mapsto aba^{-1}b^{-1}$, $y mapsto a^3$, and $z mapsto b^7$.
answered Jan 30 at 5:57
Mike PierceMike Pierce
11.7k103585
$endgroup$
$begingroup$
May I please know what you mean by $G$ is the quotient of a group $F$? Quotient of which two groups is $G$? I know that any finitely generated abelian group can be written as the product of a torsion group and a free abelian group. Thanks
$endgroup$
– manifolded
Jan 30 at 6:02
add a comment |
$begingroup$
I personally find the following perspective useful. This also generalizes nicely to other situations.
An abelian group $G$ is free whenever there is an isomorphism
$$
alphacolonmathbf{Z}^I to G
$$
for some index set $I$.
The usual generators of $mathbf{Z}^I$ are then mapped to generators of $G$ by $alpha$.
An abelian group $H$ is finitely generated whenever there is a surjective homomorphism
$$
betacolonmathbf{Z}^n to H.
$$
for some natural number $ninmathbf{N}$.
Again, $beta$ maps generators to generators, but now they may be subject to relations. In other words, there is some subgroup $R$ of $mathbf{Z}^n$ such that $H$ is isomorphic to $mathbf{Z} ^n / R$.
In the situations above, if $I$ happens to be a finite set, or if $beta$ happens to be injective (or equivalently, $R=0$), then $G$ or $H$ respectively are both free and finitely generated.
answered Jan 30 at 7:00
ffffforallffffforall
36028
$endgroup$
add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
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$begingroup$
Every group $G$ is the quotient of a free group $F$. To say that $G$ is finitely generated means that $F$ has a finite set of generators. Saying basis and linearly independent is the right idea, but isn't quite proper, since we reserve those words for vector spaces. A finitely generated group has a finite generating set and possibly some relations among those generators.
To address your comment and expand a bit, every group $G$ has a presentation as a set of generators and relations. For example, this abelian group:
$$G = leftlangle a, b mid aba^{-1}b^{-1}, a^3, b^7 rightrangle$$
Now this group presentation is really shorthand for $G$ being the quotient of the free group $langle a,brangle$ by the normal subgroup generated by the elements $langle aba^{-1}b^{-1}, a^3, b^7 rangle$. If you want to get fancy and use more categorical language, that presentation of $G$ will be the cokernel of the map $langle x,y,z rangle to langle a, b rangle$, such that $x mapsto aba^{-1}b^{-1}$, $y mapsto a^3$, and $z mapsto b^7$.
answered Jan 30 at 5:57
Mike PierceMike Pierce
11.7k103585
$endgroup$
$begingroup$
May I please know what you mean by $G$ is the quotient of a group $F$? Quotient of which two groups is $G$? I know that any finitely generated abelian group can be written as the product of a torsion group and a free abelian group. Thanks
$endgroup$
– manifolded
Jan 30 at 6:02
add a comment |
$begingroup$
Every group $G$ is the quotient of a free group $F$. To say that $G$ is finitely generated means that $F$ has a finite set of generators. Saying basis and linearly independent is the right idea, but isn't quite proper, since we reserve those words for vector spaces. A finitely generated group has a finite generating set and possibly some relations among those generators.
To address your comment and expand a bit, every group $G$ has a presentation as a set of generators and relations. For example, this abelian group:
$$G = leftlangle a, b mid aba^{-1}b^{-1}, a^3, b^7 rightrangle$$
Now this group presentation is really shorthand for $G$ being the quotient of the free group $langle a,brangle$ by the normal subgroup generated by the elements $langle aba^{-1}b^{-1}, a^3, b^7 rangle$. If you want to get fancy and use more categorical language, that presentation of $G$ will be the cokernel of the map $langle x,y,z rangle to langle a, b rangle$, such that $x mapsto aba^{-1}b^{-1}$, $y mapsto a^3$, and $z mapsto b^7$.
answered Jan 30 at 5:57
Mike PierceMike Pierce
11.7k103585
$endgroup$
$begingroup$
May I please know what you mean by $G$ is the quotient of a group $F$? Quotient of which two groups is $G$? I know that any finitely generated abelian group can be written as the product of a torsion group and a free abelian group. Thanks
$endgroup$
– manifolded
Jan 30 at 6:02
add a comment |
$begingroup$
Every group $G$ is the quotient of a free group $F$. To say that $G$ is finitely generated means that $F$ has a finite set of generators. Saying basis and linearly independent is the right idea, but isn't quite proper, since we reserve those words for vector spaces. A finitely generated group has a finite generating set and possibly some relations among those generators.
To address your comment and expand a bit, every group $G$ has a presentation as a set of generators and relations. For example, this abelian group:
$$G = leftlangle a, b mid aba^{-1}b^{-1}, a^3, b^7 rightrangle$$
Now this group presentation is really shorthand for $G$ being the quotient of the free group $langle a,brangle$ by the normal subgroup generated by the elements $langle aba^{-1}b^{-1}, a^3, b^7 rangle$. If you want to get fancy and use more categorical language, that presentation of $G$ will be the cokernel of the map $langle x,y,z rangle to langle a, b rangle$, such that $x mapsto aba^{-1}b^{-1}$, $y mapsto a^3$, and $z mapsto b^7$.
answered Jan 30 at 5:57
Mike PierceMike Pierce
11.7k103585
$endgroup$
Every group $G$ is the quotient of a free group $F$. To say that $G$ is finitely generated means that $F$ has a finite set of generators. Saying basis and linearly independent is the right idea, but isn't quite proper, since we reserve those words for vector spaces. A finitely generated group has a finite generating set and possibly some relations among those generators.
To address your comment and expand a bit, every group $G$ has a presentation as a set of generators and relations. For example, this abelian group:
$$G = leftlangle a, b mid aba^{-1}b^{-1}, a^3, b^7 rightrangle$$
Now this group presentation is really shorthand for $G$ being the quotient of the free group $langle a,brangle$ by the normal subgroup generated by the elements $langle aba^{-1}b^{-1}, a^3, b^7 rangle$. If you want to get fancy and use more categorical language, that presentation of $G$ will be the cokernel of the map $langle x,y,z rangle to langle a, b rangle$, such that $x mapsto aba^{-1}b^{-1}$, $y mapsto a^3$, and $z mapsto b^7$.
answered Jan 30 at 5:57
Mike PierceMike Pierce
11.7k103585
edited Jan 31 at 17:45
answered Jan 30 at 5:57
Mike PierceMike Pierce
11.7k103585
answered Jan 30 at 5:57
Mike PierceMike Pierce
11.7k103585
answered Jan 30 at 5:57
Mike PierceMike Pierce
11.7k103585
11.7k103585
$begingroup$
May I please know what you mean by $G$ is the quotient of a group $F$? Quotient of which two groups is $G$? I know that any finitely generated abelian group can be written as the product of a torsion group and a free abelian group. Thanks
$endgroup$
– manifolded
Jan 30 at 6:02
add a comment |
$begingroup$
May I please know what you mean by $G$ is the quotient of a group $F$? Quotient of which two groups is $G$? I know that any finitely generated abelian group can be written as the product of a torsion group and a free abelian group. Thanks
$endgroup$
– manifolded
Jan 30 at 6:02
$begingroup$
May I please know what you mean by $G$ is the quotient of a group $F$? Quotient of which two groups is $G$? I know that any finitely generated abelian group can be written as the product of a torsion group and a free abelian group. Thanks
$endgroup$
– manifolded
Jan 30 at 6:02
$begingroup$
May I please know what you mean by $G$ is the quotient of a group $F$? Quotient of which two groups is $G$? I know that any finitely generated abelian group can be written as the product of a torsion group and a free abelian group. Thanks
$endgroup$
– manifolded
Jan 30 at 6:02
add a comment |
$begingroup$
I personally find the following perspective useful. This also generalizes nicely to other situations.
An abelian group $G$ is free whenever there is an isomorphism
$$
alphacolonmathbf{Z}^I to G
$$
for some index set $I$.
The usual generators of $mathbf{Z}^I$ are then mapped to generators of $G$ by $alpha$.
An abelian group $H$ is finitely generated whenever there is a surjective homomorphism
$$
betacolonmathbf{Z}^n to H.
$$
for some natural number $ninmathbf{N}$.
Again, $beta$ maps generators to generators, but now they may be subject to relations. In other words, there is some subgroup $R$ of $mathbf{Z}^n$ such that $H$ is isomorphic to $mathbf{Z} ^n / R$.
In the situations above, if $I$ happens to be a finite set, or if $beta$ happens to be injective (or equivalently, $R=0$), then $G$ or $H$ respectively are both free and finitely generated.
answered Jan 30 at 7:00
ffffforallffffforall
36028
$endgroup$
add a comment |
$begingroup$
I personally find the following perspective useful. This also generalizes nicely to other situations.
An abelian group $G$ is free whenever there is an isomorphism
$$
alphacolonmathbf{Z}^I to G
$$
for some index set $I$.
The usual generators of $mathbf{Z}^I$ are then mapped to generators of $G$ by $alpha$.
An abelian group $H$ is finitely generated whenever there is a surjective homomorphism
$$
betacolonmathbf{Z}^n to H.
$$
for some natural number $ninmathbf{N}$.
Again, $beta$ maps generators to generators, but now they may be subject to relations. In other words, there is some subgroup $R$ of $mathbf{Z}^n$ such that $H$ is isomorphic to $mathbf{Z} ^n / R$.
In the situations above, if $I$ happens to be a finite set, or if $beta$ happens to be injective (or equivalently, $R=0$), then $G$ or $H$ respectively are both free and finitely generated.
answered Jan 30 at 7:00
ffffforallffffforall
36028
$endgroup$
add a comment |
$begingroup$
I personally find the following perspective useful. This also generalizes nicely to other situations.
An abelian group $G$ is free whenever there is an isomorphism
$$
alphacolonmathbf{Z}^I to G
$$
for some index set $I$.
The usual generators of $mathbf{Z}^I$ are then mapped to generators of $G$ by $alpha$.
An abelian group $H$ is finitely generated whenever there is a surjective homomorphism
$$
betacolonmathbf{Z}^n to H.
$$
for some natural number $ninmathbf{N}$.
Again, $beta$ maps generators to generators, but now they may be subject to relations. In other words, there is some subgroup $R$ of $mathbf{Z}^n$ such that $H$ is isomorphic to $mathbf{Z} ^n / R$.
In the situations above, if $I$ happens to be a finite set, or if $beta$ happens to be injective (or equivalently, $R=0$), then $G$ or $H$ respectively are both free and finitely generated.
answered Jan 30 at 7:00
ffffforallffffforall
36028
$endgroup$
I personally find the following perspective useful. This also generalizes nicely to other situations.
An abelian group $G$ is free whenever there is an isomorphism
$$
alphacolonmathbf{Z}^I to G
$$
for some index set $I$.
The usual generators of $mathbf{Z}^I$ are then mapped to generators of $G$ by $alpha$.
An abelian group $H$ is finitely generated whenever there is a surjective homomorphism
$$
betacolonmathbf{Z}^n to H.
$$
for some natural number $ninmathbf{N}$.
Again, $beta$ maps generators to generators, but now they may be subject to relations. In other words, there is some subgroup $R$ of $mathbf{Z}^n$ such that $H$ is isomorphic to $mathbf{Z} ^n / R$.
In the situations above, if $I$ happens to be a finite set, or if $beta$ happens to be injective (or equivalently, $R=0$), then $G$ or $H$ respectively are both free and finitely generated.
answered Jan 30 at 7:00
ffffforallffffforall
36028
answered Jan 30 at 7:00
ffffforallffffforall
36028
answered Jan 30 at 7:00
ffffforallffffforall
36028
answered Jan 30 at 7:00
ffffforallffffforall
36028
36028
add a comment |
add a comment |
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$begingroup$
Every finite Abelian group is finitely generated, but unless it has order $1$, it is not free Abelian.
$endgroup$
– Lord Shark the Unknown
Jan 30 at 5:56
$begingroup$
I see, that is a good point.
$endgroup$
– manifolded
Jan 30 at 6:05