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A polycyclic group is Solvable and Noetherian
Clash Royale CLAN TAG#URR8PPP
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I know how to prove this problem, but I just need a small check towards the end of the proof on how to use induction on the length of chains to show that $G_1$ is Noetherian.(below) I would appreciate your feedback!
Question:Every Polycyclic group is Noetherian and Solvable.
Proof: Since $G$ is polycyclic, there is a finite series of subgroups $$e=G_kle G_k-1leq…le G_1le G_o=G$$ such that $G_j+1$ is normal in $G_j$ (for $0le jle k-1$) and $G_j/G_j+1$ is cyclic(which implies abelian) and so, $G$ is Solvable by definition. It remains to show that $G$ is Noetherian. Since $G_1$ is normal in $G_0=G$, we only need to show that $G_1$ and $G_o/G_1$ are Noetherian. Now, since $G_o/G_1$ is cyclic, it is Noetherian. We use induction on the length of $k$ to show that $G_1$ is Noetherian. If $k=1$, then we have $e=G_1$ is the identity which is Noetherian. At this point here on, this is where I am asking for a check(help).
Here's my argument: Suppose that $G_1$ is Noetherian for a chain of lenth $k=n$ and let $k=n+1$. Then we have the chain $e=G_n+1le G_nle...le G_2le G_1le G_0=G$. By the inductive hypothesis, $G_2$ is Noetherian. Since then $G_2$ is normal in $G_1$, and $G_1/G_2$ is Noetherian(because it is cyclic), we conclude that $G_1$ is Noetherian.
group-theory
asked Aug 1 at 16:05
Djinola
647
add a comment |Â
up vote
1
down vote
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I know how to prove this problem, but I just need a small check towards the end of the proof on how to use induction on the length of chains to show that $G_1$ is Noetherian.(below) I would appreciate your feedback!
Question:Every Polycyclic group is Noetherian and Solvable.
Proof: Since $G$ is polycyclic, there is a finite series of subgroups $$e=G_kle G_k-1leq…le G_1le G_o=G$$ such that $G_j+1$ is normal in $G_j$ (for $0le jle k-1$) and $G_j/G_j+1$ is cyclic(which implies abelian) and so, $G$ is Solvable by definition. It remains to show that $G$ is Noetherian. Since $G_1$ is normal in $G_0=G$, we only need to show that $G_1$ and $G_o/G_1$ are Noetherian. Now, since $G_o/G_1$ is cyclic, it is Noetherian. We use induction on the length of $k$ to show that $G_1$ is Noetherian. If $k=1$, then we have $e=G_1$ is the identity which is Noetherian. At this point here on, this is where I am asking for a check(help).
Here's my argument: Suppose that $G_1$ is Noetherian for a chain of lenth $k=n$ and let $k=n+1$. Then we have the chain $e=G_n+1le G_nle...le G_2le G_1le G_0=G$. By the inductive hypothesis, $G_2$ is Noetherian. Since then $G_2$ is normal in $G_1$, and $G_1/G_2$ is Noetherian(because it is cyclic), we conclude that $G_1$ is Noetherian.
group-theory
asked Aug 1 at 16:05
Djinola
647
You have to prove that if $N$ is normal in $G$ and both $N$ and $G/N$ are Noetherian, then $G$ is Noetherian.
– egreg
Aug 1 at 16:43
1
This is fine, but you've phrased it weirdly. Use the induction at the beginning: "assuming all polycyclic groups of series length $<k$ are Noetherian, consider $G$ of length $k$." Now you get $G_1$ is Noetherian by induction right away.
– Steve D
Aug 1 at 17:54
Thanks @Steve D. I go it!
– Djinola
Aug 1 at 19:56
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I know how to prove this problem, but I just need a small check towards the end of the proof on how to use induction on the length of chains to show that $G_1$ is Noetherian.(below) I would appreciate your feedback!
Question:Every Polycyclic group is Noetherian and Solvable.
Proof: Since $G$ is polycyclic, there is a finite series of subgroups $$e=G_kle G_k-1leq…le G_1le G_o=G$$ such that $G_j+1$ is normal in $G_j$ (for $0le jle k-1$) and $G_j/G_j+1$ is cyclic(which implies abelian) and so, $G$ is Solvable by definition. It remains to show that $G$ is Noetherian. Since $G_1$ is normal in $G_0=G$, we only need to show that $G_1$ and $G_o/G_1$ are Noetherian. Now, since $G_o/G_1$ is cyclic, it is Noetherian. We use induction on the length of $k$ to show that $G_1$ is Noetherian. If $k=1$, then we have $e=G_1$ is the identity which is Noetherian. At this point here on, this is where I am asking for a check(help).
Here's my argument: Suppose that $G_1$ is Noetherian for a chain of lenth $k=n$ and let $k=n+1$. Then we have the chain $e=G_n+1le G_nle...le G_2le G_1le G_0=G$. By the inductive hypothesis, $G_2$ is Noetherian. Since then $G_2$ is normal in $G_1$, and $G_1/G_2$ is Noetherian(because it is cyclic), we conclude that $G_1$ is Noetherian.
group-theory
asked Aug 1 at 16:05
Djinola
647
I know how to prove this problem, but I just need a small check towards the end of the proof on how to use induction on the length of chains to show that $G_1$ is Noetherian.(below) I would appreciate your feedback!
Question:Every Polycyclic group is Noetherian and Solvable.
Proof: Since $G$ is polycyclic, there is a finite series of subgroups $$e=G_kle G_k-1leq…le G_1le G_o=G$$ such that $G_j+1$ is normal in $G_j$ (for $0le jle k-1$) and $G_j/G_j+1$ is cyclic(which implies abelian) and so, $G$ is Solvable by definition. It remains to show that $G$ is Noetherian. Since $G_1$ is normal in $G_0=G$, we only need to show that $G_1$ and $G_o/G_1$ are Noetherian. Now, since $G_o/G_1$ is cyclic, it is Noetherian. We use induction on the length of $k$ to show that $G_1$ is Noetherian. If $k=1$, then we have $e=G_1$ is the identity which is Noetherian. At this point here on, this is where I am asking for a check(help).
Here's my argument: Suppose that $G_1$ is Noetherian for a chain of lenth $k=n$ and let $k=n+1$. Then we have the chain $e=G_n+1le G_nle...le G_2le G_1le G_0=G$. By the inductive hypothesis, $G_2$ is Noetherian. Since then $G_2$ is normal in $G_1$, and $G_1/G_2$ is Noetherian(because it is cyclic), we conclude that $G_1$ is Noetherian.
group-theory
asked Aug 1 at 16:05
Djinola
647
asked Aug 1 at 16:05
Djinola
647
asked Aug 1 at 16:05
Djinola
647
asked Aug 1 at 16:05
Djinola
647
647
You have to prove that if $N$ is normal in $G$ and both $N$ and $G/N$ are Noetherian, then $G$ is Noetherian.
– egreg
Aug 1 at 16:43
1
This is fine, but you've phrased it weirdly. Use the induction at the beginning: "assuming all polycyclic groups of series length $<k$ are Noetherian, consider $G$ of length $k$." Now you get $G_1$ is Noetherian by induction right away.
– Steve D
Aug 1 at 17:54
Thanks @Steve D. I go it!
– Djinola
Aug 1 at 19:56
add a comment |Â
You have to prove that if $N$ is normal in $G$ and both $N$ and $G/N$ are Noetherian, then $G$ is Noetherian.
– egreg
Aug 1 at 16:43
1
This is fine, but you've phrased it weirdly. Use the induction at the beginning: "assuming all polycyclic groups of series length $<k$ are Noetherian, consider $G$ of length $k$." Now you get $G_1$ is Noetherian by induction right away.
– Steve D
Aug 1 at 17:54
Thanks @Steve D. I go it!
– Djinola
Aug 1 at 19:56
You have to prove that if $N$ is normal in $G$ and both $N$ and $G/N$ are Noetherian, then $G$ is Noetherian.
– egreg
Aug 1 at 16:43
You have to prove that if $N$ is normal in $G$ and both $N$ and $G/N$ are Noetherian, then $G$ is Noetherian.
– egreg
Aug 1 at 16:43
1
1
This is fine, but you've phrased it weirdly. Use the induction at the beginning: "assuming all polycyclic groups of series length $<k$ are Noetherian, consider $G$ of length $k$." Now you get $G_1$ is Noetherian by induction right away.
– Steve D
Aug 1 at 17:54
This is fine, but you've phrased it weirdly. Use the induction at the beginning: "assuming all polycyclic groups of series length $<k$ are Noetherian, consider $G$ of length $k$." Now you get $G_1$ is Noetherian by induction right away.
– Steve D
Aug 1 at 17:54
Thanks @Steve D. I go it!
– Djinola
Aug 1 at 19:56
Thanks @Steve D. I go it!
– Djinola
Aug 1 at 19:56
add a comment |Â
1 Answer
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More is true: A solvable group is Noetherian if and only if it is polycyclic, see Lemma 2 of the paper On linear Noetherian groups by Zassenhaus.
answered Aug 1 at 17:00
Dietrich Burde
74.5k64184
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
More is true: A solvable group is Noetherian if and only if it is polycyclic, see Lemma 2 of the paper On linear Noetherian groups by Zassenhaus.
answered Aug 1 at 17:00
Dietrich Burde
74.5k64184
add a comment |Â
up vote
1
down vote
More is true: A solvable group is Noetherian if and only if it is polycyclic, see Lemma 2 of the paper On linear Noetherian groups by Zassenhaus.
answered Aug 1 at 17:00
Dietrich Burde
74.5k64184
add a comment |Â
up vote
1
down vote
up vote
1
down vote
More is true: A solvable group is Noetherian if and only if it is polycyclic, see Lemma 2 of the paper On linear Noetherian groups by Zassenhaus.
answered Aug 1 at 17:00
Dietrich Burde
74.5k64184
More is true: A solvable group is Noetherian if and only if it is polycyclic, see Lemma 2 of the paper On linear Noetherian groups by Zassenhaus.
answered Aug 1 at 17:00
Dietrich Burde
74.5k64184
answered Aug 1 at 17:00
Dietrich Burde
74.5k64184
answered Aug 1 at 17:00
Dietrich Burde
74.5k64184
answered Aug 1 at 17:00
Dietrich Burde
74.5k64184
74.5k64184
add a comment |Â
add a comment |Â
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You have to prove that if $N$ is normal in $G$ and both $N$ and $G/N$ are Noetherian, then $G$ is Noetherian.
– egreg
Aug 1 at 16:43
1
This is fine, but you've phrased it weirdly. Use the induction at the beginning: "assuming all polycyclic groups of series length $<k$ are Noetherian, consider $G$ of length $k$." Now you get $G_1$ is Noetherian by induction right away.
– Steve D
Aug 1 at 17:54
Thanks @Steve D. I go it!
– Djinola
Aug 1 at 19:56