Mixing
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Structure theorem for modules over PIDs and abelian groups.
Clash Royale CLAN TAG#URR8PPP
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$textbfProblem.$ Let $G$ be the free abelian group $mathbfZ^5$. Show that G has only finitely many subgroups of index 17.
I was looking at an old qualifying exam and this problem was in the "Modules and Rings" section. I am wondering if there is a module-theoretic solution to this problem that doesn't simply use the fact that any finitely generated group has only finitely many subgroups of a given index. I currently don't see how this problem is related to the structure of modules over PIDs.
abstract-algebra group-theory modules principal-ideal-domains
edited Jul 21 at 22:21
Shaun
7,34592972
asked Jul 21 at 21:57
SihOASHoihd
18112
add a comment |Â
up vote
0
down vote
favorite
$textbfProblem.$ Let $G$ be the free abelian group $mathbfZ^5$. Show that G has only finitely many subgroups of index 17.
I was looking at an old qualifying exam and this problem was in the "Modules and Rings" section. I am wondering if there is a module-theoretic solution to this problem that doesn't simply use the fact that any finitely generated group has only finitely many subgroups of a given index. I currently don't see how this problem is related to the structure of modules over PIDs.
abstract-algebra group-theory modules principal-ideal-domains
edited Jul 21 at 22:21
Shaun
7,34592972
asked Jul 21 at 21:57
SihOASHoihd
18112
You can generalize this in a different direction. For any $n>0$ and any finitely generated group $G$, $G$ has only finitely many subgroups of index $n$.
– Derek Holt
Jul 22 at 14:58
I know, I mentioned that.
– SihOASHoihd
Jul 22 at 15:06
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$textbfProblem.$ Let $G$ be the free abelian group $mathbfZ^5$. Show that G has only finitely many subgroups of index 17.
I was looking at an old qualifying exam and this problem was in the "Modules and Rings" section. I am wondering if there is a module-theoretic solution to this problem that doesn't simply use the fact that any finitely generated group has only finitely many subgroups of a given index. I currently don't see how this problem is related to the structure of modules over PIDs.
abstract-algebra group-theory modules principal-ideal-domains
edited Jul 21 at 22:21
Shaun
7,34592972
asked Jul 21 at 21:57
SihOASHoihd
18112
$textbfProblem.$ Let $G$ be the free abelian group $mathbfZ^5$. Show that G has only finitely many subgroups of index 17.
I was looking at an old qualifying exam and this problem was in the "Modules and Rings" section. I am wondering if there is a module-theoretic solution to this problem that doesn't simply use the fact that any finitely generated group has only finitely many subgroups of a given index. I currently don't see how this problem is related to the structure of modules over PIDs.
abstract-algebra group-theory modules principal-ideal-domains
edited Jul 21 at 22:21
Shaun
7,34592972
asked Jul 21 at 21:57
SihOASHoihd
18112
edited Jul 21 at 22:21
Shaun
7,34592972
edited Jul 21 at 22:21
Shaun
7,34592972
edited Jul 21 at 22:21
Shaun
7,34592972
7,34592972
asked Jul 21 at 21:57
SihOASHoihd
18112
asked Jul 21 at 21:57
SihOASHoihd
18112
asked Jul 21 at 21:57
SihOASHoihd
18112
18112
You can generalize this in a different direction. For any $n>0$ and any finitely generated group $G$, $G$ has only finitely many subgroups of index $n$.
– Derek Holt
Jul 22 at 14:58
I know, I mentioned that.
– SihOASHoihd
Jul 22 at 15:06
add a comment |Â
You can generalize this in a different direction. For any $n>0$ and any finitely generated group $G$, $G$ has only finitely many subgroups of index $n$.
– Derek Holt
Jul 22 at 14:58
I know, I mentioned that.
– SihOASHoihd
Jul 22 at 15:06
You can generalize this in a different direction. For any $n>0$ and any finitely generated group $G$, $G$ has only finitely many subgroups of index $n$.
– Derek Holt
Jul 22 at 14:58
You can generalize this in a different direction. For any $n>0$ and any finitely generated group $G$, $G$ has only finitely many subgroups of index $n$.
– Derek Holt
Jul 22 at 14:58
I know, I mentioned that.
– SihOASHoihd
Jul 22 at 15:06
I know, I mentioned that.
– SihOASHoihd
Jul 22 at 15:06
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Abelian groups = $mathbbZ$-modules, so it is a special case of module over PID.
To answer your second concern in the comments, there is one thing that the proof of the theorem applies to submodules. As an immediate corollary, you obtain that those are also finitely generated, and in fact by at most as many elements as the original. The main idea of the proof is that every submodule has a very special basis. Namely, if $Nleq M$ is a submodule, then it is possible to pick a basis $b_1, ldots, b_n$ in $M$ and elements $r_1, ldots, r_n$ in the PID such that $r_1b_1, ldots, r_nb_b$ is a basis of $N$, and $r_1mid r_2midcdotsmid r_n$.
Clearly if the index is 17, then $r_1=cdots=r_4=1$ and $r_5=17$. If you show that four elements of the original basis must be generated, and $17$ times the fifth also, then you obtain that there are five such subgroups.
answered Jul 21 at 22:10
A. Pongrácz
2,269221
I know, but "I currently don't see how this problem is related to the structure of modules over PIDs"
– SihOASHoihd
Jul 21 at 22:57
Here is one problem I am having. The structure theorem classifies finitely generated modules up to isomorphism. But in this problem we want to count all subgroups, not just up to isomorphism.
– SihOASHoihd
Jul 21 at 23:01
I am not even aware of anything interesting that the structure theorem implies about submodules of finitely generated modules. (There are no exercises in the modules over PIDs section of Dummit and Foote that look anything like these problems.)
– SihOASHoihd
Jul 21 at 23:07
A third thing: the structure theorem applied to free modules over PIDs seems trivial. It just says $R^n=R^n$.
– SihOASHoihd
Jul 21 at 23:16
I edit my answer according to your further inquiries.
– A. Pongrácz
Jul 21 at 23:29
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Abelian groups = $mathbbZ$-modules, so it is a special case of module over PID.
To answer your second concern in the comments, there is one thing that the proof of the theorem applies to submodules. As an immediate corollary, you obtain that those are also finitely generated, and in fact by at most as many elements as the original. The main idea of the proof is that every submodule has a very special basis. Namely, if $Nleq M$ is a submodule, then it is possible to pick a basis $b_1, ldots, b_n$ in $M$ and elements $r_1, ldots, r_n$ in the PID such that $r_1b_1, ldots, r_nb_b$ is a basis of $N$, and $r_1mid r_2midcdotsmid r_n$.
Clearly if the index is 17, then $r_1=cdots=r_4=1$ and $r_5=17$. If you show that four elements of the original basis must be generated, and $17$ times the fifth also, then you obtain that there are five such subgroups.
answered Jul 21 at 22:10
A. Pongrácz
2,269221
I know, but "I currently don't see how this problem is related to the structure of modules over PIDs"
– SihOASHoihd
Jul 21 at 22:57
Here is one problem I am having. The structure theorem classifies finitely generated modules up to isomorphism. But in this problem we want to count all subgroups, not just up to isomorphism.
– SihOASHoihd
Jul 21 at 23:01
I am not even aware of anything interesting that the structure theorem implies about submodules of finitely generated modules. (There are no exercises in the modules over PIDs section of Dummit and Foote that look anything like these problems.)
– SihOASHoihd
Jul 21 at 23:07
A third thing: the structure theorem applied to free modules over PIDs seems trivial. It just says $R^n=R^n$.
– SihOASHoihd
Jul 21 at 23:16
I edit my answer according to your further inquiries.
– A. Pongrácz
Jul 21 at 23:29
 |Â
show 1 more comment
up vote
1
down vote
accepted
Abelian groups = $mathbbZ$-modules, so it is a special case of module over PID.
To answer your second concern in the comments, there is one thing that the proof of the theorem applies to submodules. As an immediate corollary, you obtain that those are also finitely generated, and in fact by at most as many elements as the original. The main idea of the proof is that every submodule has a very special basis. Namely, if $Nleq M$ is a submodule, then it is possible to pick a basis $b_1, ldots, b_n$ in $M$ and elements $r_1, ldots, r_n$ in the PID such that $r_1b_1, ldots, r_nb_b$ is a basis of $N$, and $r_1mid r_2midcdotsmid r_n$.
Clearly if the index is 17, then $r_1=cdots=r_4=1$ and $r_5=17$. If you show that four elements of the original basis must be generated, and $17$ times the fifth also, then you obtain that there are five such subgroups.
answered Jul 21 at 22:10
A. Pongrácz
2,269221
I know, but "I currently don't see how this problem is related to the structure of modules over PIDs"
– SihOASHoihd
Jul 21 at 22:57
Here is one problem I am having. The structure theorem classifies finitely generated modules up to isomorphism. But in this problem we want to count all subgroups, not just up to isomorphism.
– SihOASHoihd
Jul 21 at 23:01
I am not even aware of anything interesting that the structure theorem implies about submodules of finitely generated modules. (There are no exercises in the modules over PIDs section of Dummit and Foote that look anything like these problems.)
– SihOASHoihd
Jul 21 at 23:07
A third thing: the structure theorem applied to free modules over PIDs seems trivial. It just says $R^n=R^n$.
– SihOASHoihd
Jul 21 at 23:16
I edit my answer according to your further inquiries.
– A. Pongrácz
Jul 21 at 23:29
 |Â
show 1 more comment
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Abelian groups = $mathbbZ$-modules, so it is a special case of module over PID.
To answer your second concern in the comments, there is one thing that the proof of the theorem applies to submodules. As an immediate corollary, you obtain that those are also finitely generated, and in fact by at most as many elements as the original. The main idea of the proof is that every submodule has a very special basis. Namely, if $Nleq M$ is a submodule, then it is possible to pick a basis $b_1, ldots, b_n$ in $M$ and elements $r_1, ldots, r_n$ in the PID such that $r_1b_1, ldots, r_nb_b$ is a basis of $N$, and $r_1mid r_2midcdotsmid r_n$.
Clearly if the index is 17, then $r_1=cdots=r_4=1$ and $r_5=17$. If you show that four elements of the original basis must be generated, and $17$ times the fifth also, then you obtain that there are five such subgroups.
answered Jul 21 at 22:10
A. Pongrácz
2,269221
Abelian groups = $mathbbZ$-modules, so it is a special case of module over PID.
To answer your second concern in the comments, there is one thing that the proof of the theorem applies to submodules. As an immediate corollary, you obtain that those are also finitely generated, and in fact by at most as many elements as the original. The main idea of the proof is that every submodule has a very special basis. Namely, if $Nleq M$ is a submodule, then it is possible to pick a basis $b_1, ldots, b_n$ in $M$ and elements $r_1, ldots, r_n$ in the PID such that $r_1b_1, ldots, r_nb_b$ is a basis of $N$, and $r_1mid r_2midcdotsmid r_n$.
Clearly if the index is 17, then $r_1=cdots=r_4=1$ and $r_5=17$. If you show that four elements of the original basis must be generated, and $17$ times the fifth also, then you obtain that there are five such subgroups.
answered Jul 21 at 22:10
A. Pongrácz
2,269221
edited Jul 21 at 23:37
answered Jul 21 at 22:10
A. Pongrácz
2,269221
answered Jul 21 at 22:10
A. Pongrácz
2,269221
answered Jul 21 at 22:10
A. Pongrácz
2,269221
2,269221
I know, but "I currently don't see how this problem is related to the structure of modules over PIDs"
– SihOASHoihd
Jul 21 at 22:57
Here is one problem I am having. The structure theorem classifies finitely generated modules up to isomorphism. But in this problem we want to count all subgroups, not just up to isomorphism.
– SihOASHoihd
Jul 21 at 23:01
I am not even aware of anything interesting that the structure theorem implies about submodules of finitely generated modules. (There are no exercises in the modules over PIDs section of Dummit and Foote that look anything like these problems.)
– SihOASHoihd
Jul 21 at 23:07
A third thing: the structure theorem applied to free modules over PIDs seems trivial. It just says $R^n=R^n$.
– SihOASHoihd
Jul 21 at 23:16
I edit my answer according to your further inquiries.
– A. Pongrácz
Jul 21 at 23:29
 |Â
show 1 more comment
I know, but "I currently don't see how this problem is related to the structure of modules over PIDs"
– SihOASHoihd
Jul 21 at 22:57
Here is one problem I am having. The structure theorem classifies finitely generated modules up to isomorphism. But in this problem we want to count all subgroups, not just up to isomorphism.
– SihOASHoihd
Jul 21 at 23:01
I am not even aware of anything interesting that the structure theorem implies about submodules of finitely generated modules. (There are no exercises in the modules over PIDs section of Dummit and Foote that look anything like these problems.)
– SihOASHoihd
Jul 21 at 23:07
A third thing: the structure theorem applied to free modules over PIDs seems trivial. It just says $R^n=R^n$.
– SihOASHoihd
Jul 21 at 23:16
I edit my answer according to your further inquiries.
– A. Pongrácz
Jul 21 at 23:29
I know, but "I currently don't see how this problem is related to the structure of modules over PIDs"
– SihOASHoihd
Jul 21 at 22:57
I know, but "I currently don't see how this problem is related to the structure of modules over PIDs"
– SihOASHoihd
Jul 21 at 22:57
Here is one problem I am having. The structure theorem classifies finitely generated modules up to isomorphism. But in this problem we want to count all subgroups, not just up to isomorphism.
– SihOASHoihd
Jul 21 at 23:01
Here is one problem I am having. The structure theorem classifies finitely generated modules up to isomorphism. But in this problem we want to count all subgroups, not just up to isomorphism.
– SihOASHoihd
Jul 21 at 23:01
I am not even aware of anything interesting that the structure theorem implies about submodules of finitely generated modules. (There are no exercises in the modules over PIDs section of Dummit and Foote that look anything like these problems.)
– SihOASHoihd
Jul 21 at 23:07
I am not even aware of anything interesting that the structure theorem implies about submodules of finitely generated modules. (There are no exercises in the modules over PIDs section of Dummit and Foote that look anything like these problems.)
– SihOASHoihd
Jul 21 at 23:07
A third thing: the structure theorem applied to free modules over PIDs seems trivial. It just says $R^n=R^n$.
– SihOASHoihd
Jul 21 at 23:16
A third thing: the structure theorem applied to free modules over PIDs seems trivial. It just says $R^n=R^n$.
– SihOASHoihd
Jul 21 at 23:16
I edit my answer according to your further inquiries.
– A. Pongrácz
Jul 21 at 23:29
I edit my answer according to your further inquiries.
– A. Pongrácz
Jul 21 at 23:29
 |Â
show 1 more comment
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You can generalize this in a different direction. For any $n>0$ and any finitely generated group $G$, $G$ has only finitely many subgroups of index $n$.
– Derek Holt
Jul 22 at 14:58
I know, I mentioned that.
– SihOASHoihd
Jul 22 at 15:06