Find $G/Z(G)$ given the following information about the group?
$begingroup$
$G$ is a finite group generated by two elements $a$ and $b$, we are given the following data:
Order of a= $2$
Order of $b=2$
Order of $ab=8$.
If $Z(G)$ denotes the center then what is $G/Z(G)$ isomorphic to?
Attempt:
To be honest I don't know how to start this. I thought of taking $D_8$ as a concrete example of such a group but was not able to proceed much.
One thing I can see that this group is non abelian as if it was abelian then it would imply that $(ab)^2=e$ which contradicts the fact that order of $ab$ is 8.
What can I say more about this group?
group-theory finite-groups dihedral-groups
edited Dec 31 '18 at 16:33
Shaun
8,929113681
asked Dec 7 '18 at 18:28
StammeringMathematicianStammeringMathematician
2,3461322
$endgroup$
add a comment |
$begingroup$
$G$ is a finite group generated by two elements $a$ and $b$, we are given the following data:
Order of a= $2$
Order of $b=2$
Order of $ab=8$.
If $Z(G)$ denotes the center then what is $G/Z(G)$ isomorphic to?
Attempt:
To be honest I don't know how to start this. I thought of taking $D_8$ as a concrete example of such a group but was not able to proceed much.
One thing I can see that this group is non abelian as if it was abelian then it would imply that $(ab)^2=e$ which contradicts the fact that order of $ab$ is 8.
What can I say more about this group?
group-theory finite-groups dihedral-groups
edited Dec 31 '18 at 16:33
Shaun
8,929113681
asked Dec 7 '18 at 18:28
StammeringMathematicianStammeringMathematician
2,3461322
$endgroup$
add a comment |
$begingroup$
$G$ is a finite group generated by two elements $a$ and $b$, we are given the following data:
Order of a= $2$
Order of $b=2$
Order of $ab=8$.
If $Z(G)$ denotes the center then what is $G/Z(G)$ isomorphic to?
Attempt:
To be honest I don't know how to start this. I thought of taking $D_8$ as a concrete example of such a group but was not able to proceed much.
One thing I can see that this group is non abelian as if it was abelian then it would imply that $(ab)^2=e$ which contradicts the fact that order of $ab$ is 8.
What can I say more about this group?
group-theory finite-groups dihedral-groups
edited Dec 31 '18 at 16:33
Shaun
8,929113681
asked Dec 7 '18 at 18:28
StammeringMathematicianStammeringMathematician
2,3461322
$endgroup$
$G$ is a finite group generated by two elements $a$ and $b$, we are given the following data:
Order of a= $2$
Order of $b=2$
Order of $ab=8$.
If $Z(G)$ denotes the center then what is $G/Z(G)$ isomorphic to?
Attempt:
To be honest I don't know how to start this. I thought of taking $D_8$ as a concrete example of such a group but was not able to proceed much.
One thing I can see that this group is non abelian as if it was abelian then it would imply that $(ab)^2=e$ which contradicts the fact that order of $ab$ is 8.
What can I say more about this group?
group-theory finite-groups dihedral-groups
group-theory finite-groups dihedral-groups
edited Dec 31 '18 at 16:33
Shaun
8,929113681
asked Dec 7 '18 at 18:28
StammeringMathematicianStammeringMathematician
2,3461322
edited Dec 31 '18 at 16:33
Shaun
8,929113681
asked Dec 7 '18 at 18:28
StammeringMathematicianStammeringMathematician
2,3461322
edited Dec 31 '18 at 16:33
Shaun
8,929113681
edited Dec 31 '18 at 16:33
Shaun
8,929113681
edited Dec 31 '18 at 16:33
Shaun
8,929113681
8,929113681
asked Dec 7 '18 at 18:28
StammeringMathematicianStammeringMathematician
2,3461322
asked Dec 7 '18 at 18:28
StammeringMathematicianStammeringMathematician
2,3461322
asked Dec 7 '18 at 18:28
StammeringMathematicianStammeringMathematician
2,3461322
2,3461322
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
$G$ is indeed the dihedral group of order $16$.
A useful fact: The center of dihedral group $D_{2n}$ (notation: $D_{2n}$ is the dihedral group of order $2n$) is trivial if $n$ is odd and is $pm 1$ if $n$ even. This is easily seen from the relation $rbar{r}=bar{r}r^{-1}$.
For $n>2$, the quotient $D_{2n}/Z$ is also generated by two elements of order $2$, so is dihedral of the appropriate order.
answered Dec 7 '18 at 18:40
user10354138user10354138
7,3772925
$endgroup$
$begingroup$
The quotient doesn't appear to be dihedral.
$endgroup$
– Chris Custer
Jan 3 at 3:37
$begingroup$
@ChrisCuster Huh? It is dihedral from the relations (which automatically hold for quotients) $bar{r}^2=(rbar{r})^2=1$ and order.
$endgroup$
– user10354138
Jan 4 at 22:47
$begingroup$
Oops. My mistake.
$endgroup$
– Chris Custer
Jan 5 at 0:48
add a comment |
$begingroup$
In general, $G/Z(G)congoperatorname{Inn}G$, the group of inner automorphisms of $G$.
Since in this case $G=D_{16}$ (see this), and $Z(D_{16})={1,r^4}$, we have $operatorname{Inn}Gcong D_{16}/mathbb Z_2cong D_8$.
answered Dec 31 '18 at 17:20
Chris CusterChris Custer
11.5k3824
$endgroup$
add a comment |
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2 Answers
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active
oldest
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2 Answers
2
active
oldest
votes
active
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oldest
votes
$begingroup$
$G$ is indeed the dihedral group of order $16$.
A useful fact: The center of dihedral group $D_{2n}$ (notation: $D_{2n}$ is the dihedral group of order $2n$) is trivial if $n$ is odd and is $pm 1$ if $n$ even. This is easily seen from the relation $rbar{r}=bar{r}r^{-1}$.
For $n>2$, the quotient $D_{2n}/Z$ is also generated by two elements of order $2$, so is dihedral of the appropriate order.
answered Dec 7 '18 at 18:40
user10354138user10354138
7,3772925
$endgroup$
$begingroup$
The quotient doesn't appear to be dihedral.
$endgroup$
– Chris Custer
Jan 3 at 3:37
$begingroup$
@ChrisCuster Huh? It is dihedral from the relations (which automatically hold for quotients) $bar{r}^2=(rbar{r})^2=1$ and order.
$endgroup$
– user10354138
Jan 4 at 22:47
$begingroup$
Oops. My mistake.
$endgroup$
– Chris Custer
Jan 5 at 0:48
add a comment |
$begingroup$
$G$ is indeed the dihedral group of order $16$.
A useful fact: The center of dihedral group $D_{2n}$ (notation: $D_{2n}$ is the dihedral group of order $2n$) is trivial if $n$ is odd and is $pm 1$ if $n$ even. This is easily seen from the relation $rbar{r}=bar{r}r^{-1}$.
For $n>2$, the quotient $D_{2n}/Z$ is also generated by two elements of order $2$, so is dihedral of the appropriate order.
answered Dec 7 '18 at 18:40
user10354138user10354138
7,3772925
$endgroup$
$begingroup$
The quotient doesn't appear to be dihedral.
$endgroup$
– Chris Custer
Jan 3 at 3:37
$begingroup$
@ChrisCuster Huh? It is dihedral from the relations (which automatically hold for quotients) $bar{r}^2=(rbar{r})^2=1$ and order.
$endgroup$
– user10354138
Jan 4 at 22:47
$begingroup$
Oops. My mistake.
$endgroup$
– Chris Custer
Jan 5 at 0:48
add a comment |
$begingroup$
$G$ is indeed the dihedral group of order $16$.
A useful fact: The center of dihedral group $D_{2n}$ (notation: $D_{2n}$ is the dihedral group of order $2n$) is trivial if $n$ is odd and is $pm 1$ if $n$ even. This is easily seen from the relation $rbar{r}=bar{r}r^{-1}$.
For $n>2$, the quotient $D_{2n}/Z$ is also generated by two elements of order $2$, so is dihedral of the appropriate order.
answered Dec 7 '18 at 18:40
user10354138user10354138
7,3772925
$endgroup$
$G$ is indeed the dihedral group of order $16$.
A useful fact: The center of dihedral group $D_{2n}$ (notation: $D_{2n}$ is the dihedral group of order $2n$) is trivial if $n$ is odd and is $pm 1$ if $n$ even. This is easily seen from the relation $rbar{r}=bar{r}r^{-1}$.
For $n>2$, the quotient $D_{2n}/Z$ is also generated by two elements of order $2$, so is dihedral of the appropriate order.
answered Dec 7 '18 at 18:40
user10354138user10354138
7,3772925
answered Dec 7 '18 at 18:40
user10354138user10354138
7,3772925
answered Dec 7 '18 at 18:40
user10354138user10354138
7,3772925
answered Dec 7 '18 at 18:40
user10354138user10354138
7,3772925
7,3772925
$begingroup$
The quotient doesn't appear to be dihedral.
$endgroup$
– Chris Custer
Jan 3 at 3:37
$begingroup$
@ChrisCuster Huh? It is dihedral from the relations (which automatically hold for quotients) $bar{r}^2=(rbar{r})^2=1$ and order.
$endgroup$
– user10354138
Jan 4 at 22:47
$begingroup$
Oops. My mistake.
$endgroup$
– Chris Custer
Jan 5 at 0:48
add a comment |
$begingroup$
The quotient doesn't appear to be dihedral.
$endgroup$
– Chris Custer
Jan 3 at 3:37
$begingroup$
@ChrisCuster Huh? It is dihedral from the relations (which automatically hold for quotients) $bar{r}^2=(rbar{r})^2=1$ and order.
$endgroup$
– user10354138
Jan 4 at 22:47
$begingroup$
Oops. My mistake.
$endgroup$
– Chris Custer
Jan 5 at 0:48
$begingroup$
The quotient doesn't appear to be dihedral.
$endgroup$
– Chris Custer
Jan 3 at 3:37
$begingroup$
The quotient doesn't appear to be dihedral.
$endgroup$
– Chris Custer
Jan 3 at 3:37
$begingroup$
@ChrisCuster Huh? It is dihedral from the relations (which automatically hold for quotients) $bar{r}^2=(rbar{r})^2=1$ and order.
$endgroup$
– user10354138
Jan 4 at 22:47
$begingroup$
@ChrisCuster Huh? It is dihedral from the relations (which automatically hold for quotients) $bar{r}^2=(rbar{r})^2=1$ and order.
$endgroup$
– user10354138
Jan 4 at 22:47
$begingroup$
Oops. My mistake.
$endgroup$
– Chris Custer
Jan 5 at 0:48
$begingroup$
Oops. My mistake.
$endgroup$
– Chris Custer
Jan 5 at 0:48
add a comment |
$begingroup$
In general, $G/Z(G)congoperatorname{Inn}G$, the group of inner automorphisms of $G$.
Since in this case $G=D_{16}$ (see this), and $Z(D_{16})={1,r^4}$, we have $operatorname{Inn}Gcong D_{16}/mathbb Z_2cong D_8$.
answered Dec 31 '18 at 17:20
Chris CusterChris Custer
11.5k3824
$endgroup$
add a comment |
$begingroup$
In general, $G/Z(G)congoperatorname{Inn}G$, the group of inner automorphisms of $G$.
Since in this case $G=D_{16}$ (see this), and $Z(D_{16})={1,r^4}$, we have $operatorname{Inn}Gcong D_{16}/mathbb Z_2cong D_8$.
answered Dec 31 '18 at 17:20
Chris CusterChris Custer
11.5k3824
$endgroup$
add a comment |
$begingroup$
In general, $G/Z(G)congoperatorname{Inn}G$, the group of inner automorphisms of $G$.
Since in this case $G=D_{16}$ (see this), and $Z(D_{16})={1,r^4}$, we have $operatorname{Inn}Gcong D_{16}/mathbb Z_2cong D_8$.
answered Dec 31 '18 at 17:20
Chris CusterChris Custer
11.5k3824
$endgroup$
In general, $G/Z(G)congoperatorname{Inn}G$, the group of inner automorphisms of $G$.
Since in this case $G=D_{16}$ (see this), and $Z(D_{16})={1,r^4}$, we have $operatorname{Inn}Gcong D_{16}/mathbb Z_2cong D_8$.
answered Dec 31 '18 at 17:20
Chris CusterChris Custer
11.5k3824
edited Jan 5 at 0:50
answered Dec 31 '18 at 17:20
Chris CusterChris Custer
11.5k3824
answered Dec 31 '18 at 17:20
Chris CusterChris Custer
11.5k3824
answered Dec 31 '18 at 17:20
Chris CusterChris Custer
11.5k3824
11.5k3824
add a comment |
add a comment |
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