I think these groups don't exists
up vote
0
down vote
favorite
A group $G$ is said to be decomposable if $G= A times B$ ( direct product ). I am looking for the example of indecomposable groups (non-abelian) with very large center relative to the order of group. Are there groups (non-abelian ) $G$ such that $G$ is indecomposable and $Z(G) = c|G|$ where $c$ is very small relative to $|G|$
group-theory
asked Nov 17 at 10:09
I_wil_break_wall
153
|
show 1 more comment
up vote
0
down vote
favorite
A group $G$ is said to be decomposable if $G= A times B$ ( direct product ). I am looking for the example of indecomposable groups (non-abelian) with very large center relative to the order of group. Are there groups (non-abelian ) $G$ such that $G$ is indecomposable and $Z(G) = c|G|$ where $c$ is very small relative to $|G|$
group-theory
asked Nov 17 at 10:09
I_wil_break_wall
153
What is $c$ here? Obviously it cannot be greater then $1$. On the other hand if it is arbitrary then this is trivially true: every group has such constant, namely $|Z(G)|/|G|$.
– freakish
Nov 17 at 10:29
Like $1/|G|$? Every simple group has trivial center and is indecomposable. You can't have smaller center.
– freakish
Nov 17 at 10:33
@ freakish but i want the opposite, group with very large center and indecomposable
– I_wil_break_wall
Nov 17 at 10:34
So you meant $c$ as big as possible in $(0,1)$ range.
– freakish
Nov 17 at 10:34
@ freakish yes. or in other words a indecomposable group with largest possible center.
– I_wil_break_wall
Nov 17 at 10:36
|
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
A group $G$ is said to be decomposable if $G= A times B$ ( direct product ). I am looking for the example of indecomposable groups (non-abelian) with very large center relative to the order of group. Are there groups (non-abelian ) $G$ such that $G$ is indecomposable and $Z(G) = c|G|$ where $c$ is very small relative to $|G|$
group-theory
asked Nov 17 at 10:09
I_wil_break_wall
153
A group $G$ is said to be decomposable if $G= A times B$ ( direct product ). I am looking for the example of indecomposable groups (non-abelian) with very large center relative to the order of group. Are there groups (non-abelian ) $G$ such that $G$ is indecomposable and $Z(G) = c|G|$ where $c$ is very small relative to $|G|$
group-theory
group-theory
asked Nov 17 at 10:09
I_wil_break_wall
153
asked Nov 17 at 10:09
I_wil_break_wall
153
edited Nov 17 at 10:30
asked Nov 17 at 10:09
I_wil_break_wall
153
asked Nov 17 at 10:09
I_wil_break_wall
153
asked Nov 17 at 10:09
I_wil_break_wall
153
153
What is $c$ here? Obviously it cannot be greater then $1$. On the other hand if it is arbitrary then this is trivially true: every group has such constant, namely $|Z(G)|/|G|$.
– freakish
Nov 17 at 10:29
Like $1/|G|$? Every simple group has trivial center and is indecomposable. You can't have smaller center.
– freakish
Nov 17 at 10:33
@ freakish but i want the opposite, group with very large center and indecomposable
– I_wil_break_wall
Nov 17 at 10:34
So you meant $c$ as big as possible in $(0,1)$ range.
– freakish
Nov 17 at 10:34
@ freakish yes. or in other words a indecomposable group with largest possible center.
– I_wil_break_wall
Nov 17 at 10:36
|
show 1 more comment
What is $c$ here? Obviously it cannot be greater then $1$. On the other hand if it is arbitrary then this is trivially true: every group has such constant, namely $|Z(G)|/|G|$.
– freakish
Nov 17 at 10:29
Like $1/|G|$? Every simple group has trivial center and is indecomposable. You can't have smaller center.
– freakish
Nov 17 at 10:33
@ freakish but i want the opposite, group with very large center and indecomposable
– I_wil_break_wall
Nov 17 at 10:34
So you meant $c$ as big as possible in $(0,1)$ range.
– freakish
Nov 17 at 10:34
@ freakish yes. or in other words a indecomposable group with largest possible center.
– I_wil_break_wall
Nov 17 at 10:36
What is $c$ here? Obviously it cannot be greater then $1$. On the other hand if it is arbitrary then this is trivially true: every group has such constant, namely $|Z(G)|/|G|$.
– freakish
Nov 17 at 10:29
What is $c$ here? Obviously it cannot be greater then $1$. On the other hand if it is arbitrary then this is trivially true: every group has such constant, namely $|Z(G)|/|G|$.
– freakish
Nov 17 at 10:29
Like $1/|G|$? Every simple group has trivial center and is indecomposable. You can't have smaller center.
– freakish
Nov 17 at 10:33
Like $1/|G|$? Every simple group has trivial center and is indecomposable. You can't have smaller center.
– freakish
Nov 17 at 10:33
@ freakish but i want the opposite, group with very large center and indecomposable
– I_wil_break_wall
Nov 17 at 10:34
@ freakish but i want the opposite, group with very large center and indecomposable
– I_wil_break_wall
Nov 17 at 10:34
So you meant $c$ as big as possible in $(0,1)$ range.
– freakish
Nov 17 at 10:34
So you meant $c$ as big as possible in $(0,1)$ range.
– freakish
Nov 17 at 10:34
@ freakish yes. or in other words a indecomposable group with largest possible center.
– I_wil_break_wall
Nov 17 at 10:36
@ freakish yes. or in other words a indecomposable group with largest possible center.
– I_wil_break_wall
Nov 17 at 10:36
|
show 1 more comment
1 Answer
1
active
oldest
votes
up vote
8
down vote
accepted
It is well-known that $G/Z(G)$ cannot be nontrivial and cyclic, so we must have $|G/Z(G)| ge 4$.
For all $nge 2$, the group of order $2^{n+1}$ defined by the presentation
$$langle x,y mid x^{2^n}=y^2=1,y^{-1}xy=x^{2^{n-1}+1} rangle$$
has centre $langle x^2 rangle$ of index $4$, and it is easily shown to be indecomposable. So there are arbitrarily large finite groups with $|G/Z(G)|=4$.
These groups are sometimes called modular groups. They are one of the four families of nonabelian $2$-groups that have a maximal cyclic subgroup, which are discussed here.
answered Nov 17 at 11:18
Derek Holt
52k53468
Thanks for the answer. Please name the group you have given above in the answer. Are there well known groups which satisfy requirements I have mentioned in my question?
– I_wil_break_wall
Nov 17 at 12:01
See my edit. I think you could get further examples by allowing $y$ to have order a larger power of $2$, but apart from that I do not know of any other indecomposable groups with $|G/Z(G)|=4$.
– Derek Holt
Nov 17 at 12:33
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
accepted
It is well-known that $G/Z(G)$ cannot be nontrivial and cyclic, so we must have $|G/Z(G)| ge 4$.
For all $nge 2$, the group of order $2^{n+1}$ defined by the presentation
$$langle x,y mid x^{2^n}=y^2=1,y^{-1}xy=x^{2^{n-1}+1} rangle$$
has centre $langle x^2 rangle$ of index $4$, and it is easily shown to be indecomposable. So there are arbitrarily large finite groups with $|G/Z(G)|=4$.
These groups are sometimes called modular groups. They are one of the four families of nonabelian $2$-groups that have a maximal cyclic subgroup, which are discussed here.
answered Nov 17 at 11:18
Derek Holt
52k53468
Thanks for the answer. Please name the group you have given above in the answer. Are there well known groups which satisfy requirements I have mentioned in my question?
– I_wil_break_wall
Nov 17 at 12:01
See my edit. I think you could get further examples by allowing $y$ to have order a larger power of $2$, but apart from that I do not know of any other indecomposable groups with $|G/Z(G)|=4$.
– Derek Holt
Nov 17 at 12:33
add a comment |
up vote
8
down vote
accepted
It is well-known that $G/Z(G)$ cannot be nontrivial and cyclic, so we must have $|G/Z(G)| ge 4$.
For all $nge 2$, the group of order $2^{n+1}$ defined by the presentation
$$langle x,y mid x^{2^n}=y^2=1,y^{-1}xy=x^{2^{n-1}+1} rangle$$
has centre $langle x^2 rangle$ of index $4$, and it is easily shown to be indecomposable. So there are arbitrarily large finite groups with $|G/Z(G)|=4$.
These groups are sometimes called modular groups. They are one of the four families of nonabelian $2$-groups that have a maximal cyclic subgroup, which are discussed here.
answered Nov 17 at 11:18
Derek Holt
52k53468
Thanks for the answer. Please name the group you have given above in the answer. Are there well known groups which satisfy requirements I have mentioned in my question?
– I_wil_break_wall
Nov 17 at 12:01
See my edit. I think you could get further examples by allowing $y$ to have order a larger power of $2$, but apart from that I do not know of any other indecomposable groups with $|G/Z(G)|=4$.
– Derek Holt
Nov 17 at 12:33
add a comment |
up vote
8
down vote
accepted
up vote
8
down vote
accepted
It is well-known that $G/Z(G)$ cannot be nontrivial and cyclic, so we must have $|G/Z(G)| ge 4$.
For all $nge 2$, the group of order $2^{n+1}$ defined by the presentation
$$langle x,y mid x^{2^n}=y^2=1,y^{-1}xy=x^{2^{n-1}+1} rangle$$
has centre $langle x^2 rangle$ of index $4$, and it is easily shown to be indecomposable. So there are arbitrarily large finite groups with $|G/Z(G)|=4$.
These groups are sometimes called modular groups. They are one of the four families of nonabelian $2$-groups that have a maximal cyclic subgroup, which are discussed here.
answered Nov 17 at 11:18
Derek Holt
52k53468
It is well-known that $G/Z(G)$ cannot be nontrivial and cyclic, so we must have $|G/Z(G)| ge 4$.
For all $nge 2$, the group of order $2^{n+1}$ defined by the presentation
$$langle x,y mid x^{2^n}=y^2=1,y^{-1}xy=x^{2^{n-1}+1} rangle$$
has centre $langle x^2 rangle$ of index $4$, and it is easily shown to be indecomposable. So there are arbitrarily large finite groups with $|G/Z(G)|=4$.
These groups are sometimes called modular groups. They are one of the four families of nonabelian $2$-groups that have a maximal cyclic subgroup, which are discussed here.
answered Nov 17 at 11:18
Derek Holt
52k53468
edited Nov 17 at 12:34
answered Nov 17 at 11:18
Derek Holt
52k53468
answered Nov 17 at 11:18
Derek Holt
52k53468
answered Nov 17 at 11:18
Derek Holt
52k53468
52k53468
Thanks for the answer. Please name the group you have given above in the answer. Are there well known groups which satisfy requirements I have mentioned in my question?
– I_wil_break_wall
Nov 17 at 12:01
See my edit. I think you could get further examples by allowing $y$ to have order a larger power of $2$, but apart from that I do not know of any other indecomposable groups with $|G/Z(G)|=4$.
– Derek Holt
Nov 17 at 12:33
add a comment |
Thanks for the answer. Please name the group you have given above in the answer. Are there well known groups which satisfy requirements I have mentioned in my question?
– I_wil_break_wall
Nov 17 at 12:01
See my edit. I think you could get further examples by allowing $y$ to have order a larger power of $2$, but apart from that I do not know of any other indecomposable groups with $|G/Z(G)|=4$.
– Derek Holt
Nov 17 at 12:33
Thanks for the answer. Please name the group you have given above in the answer. Are there well known groups which satisfy requirements I have mentioned in my question?
– I_wil_break_wall
Nov 17 at 12:01
Thanks for the answer. Please name the group you have given above in the answer. Are there well known groups which satisfy requirements I have mentioned in my question?
– I_wil_break_wall
Nov 17 at 12:01
See my edit. I think you could get further examples by allowing $y$ to have order a larger power of $2$, but apart from that I do not know of any other indecomposable groups with $|G/Z(G)|=4$.
– Derek Holt
Nov 17 at 12:33
See my edit. I think you could get further examples by allowing $y$ to have order a larger power of $2$, but apart from that I do not know of any other indecomposable groups with $|G/Z(G)|=4$.
– Derek Holt
Nov 17 at 12:33
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3002163%2fi-think-these-groups-dont-exists%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
What is $c$ here? Obviously it cannot be greater then $1$. On the other hand if it is arbitrary then this is trivially true: every group has such constant, namely $|Z(G)|/|G|$.
– freakish
Nov 17 at 10:29
Like $1/|G|$? Every simple group has trivial center and is indecomposable. You can't have smaller center.
– freakish
Nov 17 at 10:33
@ freakish but i want the opposite, group with very large center and indecomposable
– I_wil_break_wall
Nov 17 at 10:34
So you meant $c$ as big as possible in $(0,1)$ range.
– freakish
Nov 17 at 10:34
@ freakish yes. or in other words a indecomposable group with largest possible center.
– I_wil_break_wall
Nov 17 at 10:36