If $G=langle{a_i}_{i=1}^rrangle$ is nilpotent with $|a_i|=m_i$, show that $|G|$ divides a power of...
use this notation for the following $textbf{Theorem}$ -
$textbf{notation}-$ $G^n$ is the subgroup generated by $n$th power of elements of $G$
$textbf{Theorem}$- In a finitely generated nilpotent group, $cap{G^p}$, for any infinite set of primes, is finite,(follows by a paper of HIGMAN) does it help me to prove this question above. I don't think so.
Is it true, that a group generated by finite elements of finite order is finite. If it is true then to prove $G$ is finite in the problem is easy, and don not require nilpotency. But it should not be true, but what can be an example?
group-theory finite-groups nilpotent-groups
edited Nov 28 '18 at 18:58
Shaun
8,805113680
asked Sep 25 '14 at 17:35
Bhaskar Vashishth
7,58712053
|
show 4 more comments
use this notation for the following $textbf{Theorem}$ -
$textbf{notation}-$ $G^n$ is the subgroup generated by $n$th power of elements of $G$
$textbf{Theorem}$- In a finitely generated nilpotent group, $cap{G^p}$, for any infinite set of primes, is finite,(follows by a paper of HIGMAN) does it help me to prove this question above. I don't think so.
Is it true, that a group generated by finite elements of finite order is finite. If it is true then to prove $G$ is finite in the problem is easy, and don not require nilpotency. But it should not be true, but what can be an example?
group-theory finite-groups nilpotent-groups
edited Nov 28 '18 at 18:58
Shaun
8,805113680
asked Sep 25 '14 at 17:35
Bhaskar Vashishth
7,58712053
1
Tarski monsters are examples of finitely generated groups whose generators have finite order.
– Teri
Sep 25 '14 at 17:41
so we need nilpotency to say $G$ is finite. How can i approach this problem?
– Bhaskar Vashishth
Sep 25 '14 at 17:43
Use induction on the nilpotence class, and consider the central quotient in the non-abelian case.
– James
Sep 25 '14 at 19:24
I can't use induction, until I show $G$ is finite
– Bhaskar Vashishth
Sep 25 '14 at 19:25
1
@BhaskarVashishth The induction is on nilpotence class, which is finite by assumption.
– zibadawa timmy
Sep 25 '14 at 19:33
|
show 4 more comments
use this notation for the following $textbf{Theorem}$ -
$textbf{notation}-$ $G^n$ is the subgroup generated by $n$th power of elements of $G$
$textbf{Theorem}$- In a finitely generated nilpotent group, $cap{G^p}$, for any infinite set of primes, is finite,(follows by a paper of HIGMAN) does it help me to prove this question above. I don't think so.
Is it true, that a group generated by finite elements of finite order is finite. If it is true then to prove $G$ is finite in the problem is easy, and don not require nilpotency. But it should not be true, but what can be an example?
group-theory finite-groups nilpotent-groups
edited Nov 28 '18 at 18:58
Shaun
8,805113680
asked Sep 25 '14 at 17:35
Bhaskar Vashishth
7,58712053
use this notation for the following $textbf{Theorem}$ -
$textbf{notation}-$ $G^n$ is the subgroup generated by $n$th power of elements of $G$
$textbf{Theorem}$- In a finitely generated nilpotent group, $cap{G^p}$, for any infinite set of primes, is finite,(follows by a paper of HIGMAN) does it help me to prove this question above. I don't think so.
Is it true, that a group generated by finite elements of finite order is finite. If it is true then to prove $G$ is finite in the problem is easy, and don not require nilpotency. But it should not be true, but what can be an example?
group-theory finite-groups nilpotent-groups
group-theory finite-groups nilpotent-groups
edited Nov 28 '18 at 18:58
Shaun
8,805113680
asked Sep 25 '14 at 17:35
Bhaskar Vashishth
7,58712053
edited Nov 28 '18 at 18:58
Shaun
8,805113680
asked Sep 25 '14 at 17:35
Bhaskar Vashishth
7,58712053
edited Nov 28 '18 at 18:58
Shaun
8,805113680
edited Nov 28 '18 at 18:58
Shaun
8,805113680
edited Nov 28 '18 at 18:58
Shaun
8,805113680
8,805113680
asked Sep 25 '14 at 17:35
Bhaskar Vashishth
7,58712053
asked Sep 25 '14 at 17:35
Bhaskar Vashishth
7,58712053
asked Sep 25 '14 at 17:35
Bhaskar Vashishth
7,58712053
7,58712053
1
Tarski monsters are examples of finitely generated groups whose generators have finite order.
– Teri
Sep 25 '14 at 17:41
so we need nilpotency to say $G$ is finite. How can i approach this problem?
– Bhaskar Vashishth
Sep 25 '14 at 17:43
Use induction on the nilpotence class, and consider the central quotient in the non-abelian case.
– James
Sep 25 '14 at 19:24
I can't use induction, until I show $G$ is finite
– Bhaskar Vashishth
Sep 25 '14 at 19:25
1
@BhaskarVashishth The induction is on nilpotence class, which is finite by assumption.
– zibadawa timmy
Sep 25 '14 at 19:33
|
show 4 more comments
1
Tarski monsters are examples of finitely generated groups whose generators have finite order.
– Teri
Sep 25 '14 at 17:41
so we need nilpotency to say $G$ is finite. How can i approach this problem?
– Bhaskar Vashishth
Sep 25 '14 at 17:43
Use induction on the nilpotence class, and consider the central quotient in the non-abelian case.
– James
Sep 25 '14 at 19:24
I can't use induction, until I show $G$ is finite
– Bhaskar Vashishth
Sep 25 '14 at 19:25
1
@BhaskarVashishth The induction is on nilpotence class, which is finite by assumption.
– zibadawa timmy
Sep 25 '14 at 19:33
1
1
Tarski monsters are examples of finitely generated groups whose generators have finite order.
– Teri
Sep 25 '14 at 17:41
Tarski monsters are examples of finitely generated groups whose generators have finite order.
– Teri
Sep 25 '14 at 17:41
so we need nilpotency to say $G$ is finite. How can i approach this problem?
– Bhaskar Vashishth
Sep 25 '14 at 17:43
so we need nilpotency to say $G$ is finite. How can i approach this problem?
– Bhaskar Vashishth
Sep 25 '14 at 17:43
Use induction on the nilpotence class, and consider the central quotient in the non-abelian case.
– James
Sep 25 '14 at 19:24
Use induction on the nilpotence class, and consider the central quotient in the non-abelian case.
– James
Sep 25 '14 at 19:24
I can't use induction, until I show $G$ is finite
– Bhaskar Vashishth
Sep 25 '14 at 19:25
I can't use induction, until I show $G$ is finite
– Bhaskar Vashishth
Sep 25 '14 at 19:25
1
1
@BhaskarVashishth The induction is on nilpotence class, which is finite by assumption.
– zibadawa timmy
Sep 25 '14 at 19:33
@BhaskarVashishth The induction is on nilpotence class, which is finite by assumption.
– zibadawa timmy
Sep 25 '14 at 19:33
|
show 4 more comments
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1
Tarski monsters are examples of finitely generated groups whose generators have finite order.
– Teri
Sep 25 '14 at 17:41
so we need nilpotency to say $G$ is finite. How can i approach this problem?
– Bhaskar Vashishth
Sep 25 '14 at 17:43
Use induction on the nilpotence class, and consider the central quotient in the non-abelian case.
– James
Sep 25 '14 at 19:24
I can't use induction, until I show $G$ is finite
– Bhaskar Vashishth
Sep 25 '14 at 19:25
1
@BhaskarVashishth The induction is on nilpotence class, which is finite by assumption.
– zibadawa timmy
Sep 25 '14 at 19:33