I think these groups don't exists
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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
142
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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
142
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
142
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
142
asked Nov 17 at 10:09
I_wil_break_wall
142
edited Nov 17 at 10:30
asked Nov 17 at 10:09
I_wil_break_wall
142
asked Nov 17 at 10:09
I_wil_break_wall
142
asked Nov 17 at 10:09
I_wil_break_wall
142
142
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
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6
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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
51.5k53466
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
6
down vote
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
51.5k53466
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
6
down vote
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
51.5k53466
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
6
down vote
up vote
6
down vote
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
51.5k53466
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
51.5k53466
edited Nov 17 at 12:34
answered Nov 17 at 11:18
Derek Holt
51.5k53466
answered Nov 17 at 11:18
Derek Holt
51.5k53466
answered Nov 17 at 11:18
Derek Holt
51.5k53466
51.5k53466
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 |
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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