If $G$ is linear over $mathbb{Z}$ then $operatorname{Aut}(G)$ is linear over $mathbb{Z}$ as well?
If $G$ is linear over $mathbb{Z}$ then $operatorname{Aut}(G)$ is linear over
$mathbb{Z}$ as well?
We have $G cong H leq GL(n, mathbb{Z})$ for some $n$ (this is what I mean when I say that $G$ is linear over $mathbb{Z}$)
Then is $operatorname{Aut}(G) cong N leq GL(n^2, mathbb{Z})$?
I need this in context of:
$G$ polycyclic, $Gamma leq operatorname{Aut}(G)$ solvable $implies Gamma$
polycyclic
If the topic assertion is correct then this can proven as:
Every polycyclic group is linear of $mathbb{Z}$, and every solvable linear group of $mathbb{Z}$ is polycyclic.
abstract-algebra group-theory solvable-groups
asked Nov 24 at 15:19
Mariah
1,282518
add a comment |
If $G$ is linear over $mathbb{Z}$ then $operatorname{Aut}(G)$ is linear over
$mathbb{Z}$ as well?
We have $G cong H leq GL(n, mathbb{Z})$ for some $n$ (this is what I mean when I say that $G$ is linear over $mathbb{Z}$)
Then is $operatorname{Aut}(G) cong N leq GL(n^2, mathbb{Z})$?
I need this in context of:
$G$ polycyclic, $Gamma leq operatorname{Aut}(G)$ solvable $implies Gamma$
polycyclic
If the topic assertion is correct then this can proven as:
Every polycyclic group is linear of $mathbb{Z}$, and every solvable linear group of $mathbb{Z}$ is polycyclic.
abstract-algebra group-theory solvable-groups
asked Nov 24 at 15:19
Mariah
1,282518
2
What does " being linear over $;Bbb Z;$" mean, anyway?
– DonAntonio
Nov 24 at 15:24
2
@DonAntonio We're using it as saying that $G$ is isomorphic to a subgroup of $GL(n, mathbb{Z})$ for some $n$
– Mariah
Nov 24 at 15:25
add a comment |
If $G$ is linear over $mathbb{Z}$ then $operatorname{Aut}(G)$ is linear over
$mathbb{Z}$ as well?
We have $G cong H leq GL(n, mathbb{Z})$ for some $n$ (this is what I mean when I say that $G$ is linear over $mathbb{Z}$)
Then is $operatorname{Aut}(G) cong N leq GL(n^2, mathbb{Z})$?
I need this in context of:
$G$ polycyclic, $Gamma leq operatorname{Aut}(G)$ solvable $implies Gamma$
polycyclic
If the topic assertion is correct then this can proven as:
Every polycyclic group is linear of $mathbb{Z}$, and every solvable linear group of $mathbb{Z}$ is polycyclic.
abstract-algebra group-theory solvable-groups
asked Nov 24 at 15:19
Mariah
1,282518
If $G$ is linear over $mathbb{Z}$ then $operatorname{Aut}(G)$ is linear over
$mathbb{Z}$ as well?
We have $G cong H leq GL(n, mathbb{Z})$ for some $n$ (this is what I mean when I say that $G$ is linear over $mathbb{Z}$)
Then is $operatorname{Aut}(G) cong N leq GL(n^2, mathbb{Z})$?
I need this in context of:
$G$ polycyclic, $Gamma leq operatorname{Aut}(G)$ solvable $implies Gamma$
polycyclic
If the topic assertion is correct then this can proven as:
Every polycyclic group is linear of $mathbb{Z}$, and every solvable linear group of $mathbb{Z}$ is polycyclic.
abstract-algebra group-theory solvable-groups
abstract-algebra group-theory solvable-groups
asked Nov 24 at 15:19
Mariah
1,282518
asked Nov 24 at 15:19
Mariah
1,282518
edited Nov 24 at 15:46
asked Nov 24 at 15:19
Mariah
1,282518
asked Nov 24 at 15:19
Mariah
1,282518
asked Nov 24 at 15:19
Mariah
1,282518
1,282518
2
What does " being linear over $;Bbb Z;$" mean, anyway?
– DonAntonio
Nov 24 at 15:24
2
@DonAntonio We're using it as saying that $G$ is isomorphic to a subgroup of $GL(n, mathbb{Z})$ for some $n$
– Mariah
Nov 24 at 15:25
add a comment |
2
What does " being linear over $;Bbb Z;$" mean, anyway?
– DonAntonio
Nov 24 at 15:24
2
@DonAntonio We're using it as saying that $G$ is isomorphic to a subgroup of $GL(n, mathbb{Z})$ for some $n$
– Mariah
Nov 24 at 15:25
2
2
What does " being linear over $;Bbb Z;$" mean, anyway?
– DonAntonio
Nov 24 at 15:24
What does " being linear over $;Bbb Z;$" mean, anyway?
– DonAntonio
Nov 24 at 15:24
2
2
@DonAntonio We're using it as saying that $G$ is isomorphic to a subgroup of $GL(n, mathbb{Z})$ for some $n$
– Mariah
Nov 24 at 15:25
@DonAntonio We're using it as saying that $G$ is isomorphic to a subgroup of $GL(n, mathbb{Z})$ for some $n$
– Mariah
Nov 24 at 15:25
add a comment |
1 Answer
1
active
oldest
votes
The answer to your question is no, because all free groups $F_n$ of finite rank are linear over ${mathbb Z}$, but it is proved here that ${rm Aut}(F_n)$ is not linear (over any field) for $n ge 3$.
It is proved in Chapter 2 of Dan Segal's book on polycyclic groups that solvable subgroups of ${rm GL}(n,{mathbb Z})$ are polycyclic.
answered Nov 24 at 15:52
Derek Holt
52.3k53570
would you mind checking out my new question; its related to the one here and I will appreciate your guidance there: math.stackexchange.com/questions/3012808/…
– Mariah
Nov 25 at 13:47
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The answer to your question is no, because all free groups $F_n$ of finite rank are linear over ${mathbb Z}$, but it is proved here that ${rm Aut}(F_n)$ is not linear (over any field) for $n ge 3$.
It is proved in Chapter 2 of Dan Segal's book on polycyclic groups that solvable subgroups of ${rm GL}(n,{mathbb Z})$ are polycyclic.
answered Nov 24 at 15:52
Derek Holt
52.3k53570
would you mind checking out my new question; its related to the one here and I will appreciate your guidance there: math.stackexchange.com/questions/3012808/…
– Mariah
Nov 25 at 13:47
add a comment |
The answer to your question is no, because all free groups $F_n$ of finite rank are linear over ${mathbb Z}$, but it is proved here that ${rm Aut}(F_n)$ is not linear (over any field) for $n ge 3$.
It is proved in Chapter 2 of Dan Segal's book on polycyclic groups that solvable subgroups of ${rm GL}(n,{mathbb Z})$ are polycyclic.
answered Nov 24 at 15:52
Derek Holt
52.3k53570
would you mind checking out my new question; its related to the one here and I will appreciate your guidance there: math.stackexchange.com/questions/3012808/…
– Mariah
Nov 25 at 13:47
add a comment |
The answer to your question is no, because all free groups $F_n$ of finite rank are linear over ${mathbb Z}$, but it is proved here that ${rm Aut}(F_n)$ is not linear (over any field) for $n ge 3$.
It is proved in Chapter 2 of Dan Segal's book on polycyclic groups that solvable subgroups of ${rm GL}(n,{mathbb Z})$ are polycyclic.
answered Nov 24 at 15:52
Derek Holt
52.3k53570
The answer to your question is no, because all free groups $F_n$ of finite rank are linear over ${mathbb Z}$, but it is proved here that ${rm Aut}(F_n)$ is not linear (over any field) for $n ge 3$.
It is proved in Chapter 2 of Dan Segal's book on polycyclic groups that solvable subgroups of ${rm GL}(n,{mathbb Z})$ are polycyclic.
answered Nov 24 at 15:52
Derek Holt
52.3k53570
answered Nov 24 at 15:52
Derek Holt
52.3k53570
answered Nov 24 at 15:52
Derek Holt
52.3k53570
answered Nov 24 at 15:52
Derek Holt
52.3k53570
52.3k53570
would you mind checking out my new question; its related to the one here and I will appreciate your guidance there: math.stackexchange.com/questions/3012808/…
– Mariah
Nov 25 at 13:47
add a comment |
would you mind checking out my new question; its related to the one here and I will appreciate your guidance there: math.stackexchange.com/questions/3012808/…
– Mariah
Nov 25 at 13:47
would you mind checking out my new question; its related to the one here and I will appreciate your guidance there: math.stackexchange.com/questions/3012808/…
– Mariah
Nov 25 at 13:47
would you mind checking out my new question; its related to the one here and I will appreciate your guidance there: math.stackexchange.com/questions/3012808/…
– Mariah
Nov 25 at 13:47
add a comment |
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2
What does " being linear over $;Bbb Z;$" mean, anyway?
– DonAntonio
Nov 24 at 15:24
2
@DonAntonio We're using it as saying that $G$ is isomorphic to a subgroup of $GL(n, mathbb{Z})$ for some $n$
– Mariah
Nov 24 at 15:25