Mixing
Normal subgroups of dihedral groups
$begingroup$
In relation to my previous question, I am curious about what exactly are the normal subgroups of a dihedral group $D_n$ of order $2n$.
It is easy to see that cyclic subgroups of $D_n$ is normal. But I suspect that case analysis is needed to decide whether dihedral subgroups of $D_n$ is normal.
A little bit of Internet search suggests the use of semidirect product $(mathbb Z/nmathbb Z) rtimes (mathbb Z/2mathbb Z) cong D_n$, but I do not know the condition for subgroups of a semidirect product to be normal.
I would be grateful if you could suggest a way to enumerate the normal subgroups of $D_n$ that does not resort to too much of case analysis.
group-theory finite-groups
edited Apr 13 '17 at 12:20
Community♦
1
asked Sep 5 '13 at 12:07
PteromysPteromys
2,46121745
$endgroup$
add a comment |
$begingroup$
In relation to my previous question, I am curious about what exactly are the normal subgroups of a dihedral group $D_n$ of order $2n$.
It is easy to see that cyclic subgroups of $D_n$ is normal. But I suspect that case analysis is needed to decide whether dihedral subgroups of $D_n$ is normal.
A little bit of Internet search suggests the use of semidirect product $(mathbb Z/nmathbb Z) rtimes (mathbb Z/2mathbb Z) cong D_n$, but I do not know the condition for subgroups of a semidirect product to be normal.
I would be grateful if you could suggest a way to enumerate the normal subgroups of $D_n$ that does not resort to too much of case analysis.
group-theory finite-groups
edited Apr 13 '17 at 12:20
Community♦
1
asked Sep 5 '13 at 12:07
PteromysPteromys
2,46121745
$endgroup$
add a comment |
$begingroup$
In relation to my previous question, I am curious about what exactly are the normal subgroups of a dihedral group $D_n$ of order $2n$.
It is easy to see that cyclic subgroups of $D_n$ is normal. But I suspect that case analysis is needed to decide whether dihedral subgroups of $D_n$ is normal.
A little bit of Internet search suggests the use of semidirect product $(mathbb Z/nmathbb Z) rtimes (mathbb Z/2mathbb Z) cong D_n$, but I do not know the condition for subgroups of a semidirect product to be normal.
I would be grateful if you could suggest a way to enumerate the normal subgroups of $D_n$ that does not resort to too much of case analysis.
group-theory finite-groups
edited Apr 13 '17 at 12:20
Community♦
1
asked Sep 5 '13 at 12:07
PteromysPteromys
2,46121745
$endgroup$
In relation to my previous question, I am curious about what exactly are the normal subgroups of a dihedral group $D_n$ of order $2n$.
It is easy to see that cyclic subgroups of $D_n$ is normal. But I suspect that case analysis is needed to decide whether dihedral subgroups of $D_n$ is normal.
A little bit of Internet search suggests the use of semidirect product $(mathbb Z/nmathbb Z) rtimes (mathbb Z/2mathbb Z) cong D_n$, but I do not know the condition for subgroups of a semidirect product to be normal.
I would be grateful if you could suggest a way to enumerate the normal subgroups of $D_n$ that does not resort to too much of case analysis.
group-theory finite-groups
group-theory finite-groups
edited Apr 13 '17 at 12:20
Community♦
1
asked Sep 5 '13 at 12:07
PteromysPteromys
2,46121745
edited Apr 13 '17 at 12:20
Community♦
1
asked Sep 5 '13 at 12:07
PteromysPteromys
2,46121745
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
edited Apr 13 '17 at 12:20
Community♦
1
1
asked Sep 5 '13 at 12:07
PteromysPteromys
2,46121745
asked Sep 5 '13 at 12:07
PteromysPteromys
2,46121745
asked Sep 5 '13 at 12:07
PteromysPteromys
2,46121745
2,46121745
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here is a nice answer: the dihedral group is generated by a rotation $R$ and a reflection $F$ subject to the relations $R^n=F^2=1$ and $(RF)^2=1$.
For $n$ odd the normal subgroups are given by $D_n$ and $langle R^d rangle$ for all divisors $dmid n$. If $n$ is even, there are two more normal subgroups, i.e., $langle R^2,F rangle$ and $langle R^2,RF rangle$.
answered Sep 5 '13 at 12:13
Dietrich BurdeDietrich Burde
82k649107
$endgroup$
1
$begingroup$
You have lost track of the group $D_n$ itself.
$endgroup$
– Alex M.
Jun 12 '16 at 10:37
$begingroup$
Yes, you are right. Of course, the group itself should be included.
$endgroup$
– Dietrich Burde
Jun 12 '16 at 11:39
2
$begingroup$
By saying $langle{R^d}rangle$ is a normal subgroup for all divisors $d mid {n}$, you've actually already included {1}, because $n mid {n}$.
$endgroup$
– Rasputin
Oct 31 '16 at 20:22
$begingroup$
@ Dietrich Burde : Is there any other normal subgroups of $D_{2n}$ besides these normal subgroups
$endgroup$
– user120386
May 22 '17 at 10:08
$begingroup$
@user120386 No, this is a complete classification of normal subgroups, see Keith Conrad's article on dihedral groups.
$endgroup$
– Dietrich Burde
May 22 '17 at 13:51
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Here is a nice answer: the dihedral group is generated by a rotation $R$ and a reflection $F$ subject to the relations $R^n=F^2=1$ and $(RF)^2=1$.
For $n$ odd the normal subgroups are given by $D_n$ and $langle R^d rangle$ for all divisors $dmid n$. If $n$ is even, there are two more normal subgroups, i.e., $langle R^2,F rangle$ and $langle R^2,RF rangle$.
answered Sep 5 '13 at 12:13
Dietrich BurdeDietrich Burde
82k649107
$endgroup$
1
$begingroup$
You have lost track of the group $D_n$ itself.
$endgroup$
– Alex M.
Jun 12 '16 at 10:37
$begingroup$
Yes, you are right. Of course, the group itself should be included.
$endgroup$
– Dietrich Burde
Jun 12 '16 at 11:39
2
$begingroup$
By saying $langle{R^d}rangle$ is a normal subgroup for all divisors $d mid {n}$, you've actually already included {1}, because $n mid {n}$.
$endgroup$
– Rasputin
Oct 31 '16 at 20:22
$begingroup$
@ Dietrich Burde : Is there any other normal subgroups of $D_{2n}$ besides these normal subgroups
$endgroup$
– user120386
May 22 '17 at 10:08
$begingroup$
@user120386 No, this is a complete classification of normal subgroups, see Keith Conrad's article on dihedral groups.
$endgroup$
– Dietrich Burde
May 22 '17 at 13:51
add a comment |
$begingroup$
Here is a nice answer: the dihedral group is generated by a rotation $R$ and a reflection $F$ subject to the relations $R^n=F^2=1$ and $(RF)^2=1$.
For $n$ odd the normal subgroups are given by $D_n$ and $langle R^d rangle$ for all divisors $dmid n$. If $n$ is even, there are two more normal subgroups, i.e., $langle R^2,F rangle$ and $langle R^2,RF rangle$.
answered Sep 5 '13 at 12:13
Dietrich BurdeDietrich Burde
82k649107
$endgroup$
1
$begingroup$
You have lost track of the group $D_n$ itself.
$endgroup$
– Alex M.
Jun 12 '16 at 10:37
$begingroup$
Yes, you are right. Of course, the group itself should be included.
$endgroup$
– Dietrich Burde
Jun 12 '16 at 11:39
2
$begingroup$
By saying $langle{R^d}rangle$ is a normal subgroup for all divisors $d mid {n}$, you've actually already included {1}, because $n mid {n}$.
$endgroup$
– Rasputin
Oct 31 '16 at 20:22
$begingroup$
@ Dietrich Burde : Is there any other normal subgroups of $D_{2n}$ besides these normal subgroups
$endgroup$
– user120386
May 22 '17 at 10:08
$begingroup$
@user120386 No, this is a complete classification of normal subgroups, see Keith Conrad's article on dihedral groups.
$endgroup$
– Dietrich Burde
May 22 '17 at 13:51
add a comment |
$begingroup$
Here is a nice answer: the dihedral group is generated by a rotation $R$ and a reflection $F$ subject to the relations $R^n=F^2=1$ and $(RF)^2=1$.
For $n$ odd the normal subgroups are given by $D_n$ and $langle R^d rangle$ for all divisors $dmid n$. If $n$ is even, there are two more normal subgroups, i.e., $langle R^2,F rangle$ and $langle R^2,RF rangle$.
answered Sep 5 '13 at 12:13
Dietrich BurdeDietrich Burde
82k649107
$endgroup$
Here is a nice answer: the dihedral group is generated by a rotation $R$ and a reflection $F$ subject to the relations $R^n=F^2=1$ and $(RF)^2=1$.
For $n$ odd the normal subgroups are given by $D_n$ and $langle R^d rangle$ for all divisors $dmid n$. If $n$ is even, there are two more normal subgroups, i.e., $langle R^2,F rangle$ and $langle R^2,RF rangle$.
answered Sep 5 '13 at 12:13
Dietrich BurdeDietrich Burde
82k649107
edited Feb 1 at 15:33
answered Sep 5 '13 at 12:13
Dietrich BurdeDietrich Burde
82k649107
answered Sep 5 '13 at 12:13
Dietrich BurdeDietrich Burde
82k649107
answered Sep 5 '13 at 12:13
Dietrich BurdeDietrich Burde
82k649107
82k649107
1
$begingroup$
You have lost track of the group $D_n$ itself.
$endgroup$
– Alex M.
Jun 12 '16 at 10:37
$begingroup$
Yes, you are right. Of course, the group itself should be included.
$endgroup$
– Dietrich Burde
Jun 12 '16 at 11:39
2
$begingroup$
By saying $langle{R^d}rangle$ is a normal subgroup for all divisors $d mid {n}$, you've actually already included {1}, because $n mid {n}$.
$endgroup$
– Rasputin
Oct 31 '16 at 20:22
$begingroup$
@ Dietrich Burde : Is there any other normal subgroups of $D_{2n}$ besides these normal subgroups
$endgroup$
– user120386
May 22 '17 at 10:08
$begingroup$
@user120386 No, this is a complete classification of normal subgroups, see Keith Conrad's article on dihedral groups.
$endgroup$
– Dietrich Burde
May 22 '17 at 13:51
add a comment |
1
$begingroup$
You have lost track of the group $D_n$ itself.
$endgroup$
– Alex M.
Jun 12 '16 at 10:37
$begingroup$
Yes, you are right. Of course, the group itself should be included.
$endgroup$
– Dietrich Burde
Jun 12 '16 at 11:39
2
$begingroup$
By saying $langle{R^d}rangle$ is a normal subgroup for all divisors $d mid {n}$, you've actually already included {1}, because $n mid {n}$.
$endgroup$
– Rasputin
Oct 31 '16 at 20:22
$begingroup$
@ Dietrich Burde : Is there any other normal subgroups of $D_{2n}$ besides these normal subgroups
$endgroup$
– user120386
May 22 '17 at 10:08
$begingroup$
@user120386 No, this is a complete classification of normal subgroups, see Keith Conrad's article on dihedral groups.
$endgroup$
– Dietrich Burde
May 22 '17 at 13:51
1
1
$begingroup$
You have lost track of the group $D_n$ itself.
$endgroup$
– Alex M.
Jun 12 '16 at 10:37
$begingroup$
You have lost track of the group $D_n$ itself.
$endgroup$
– Alex M.
Jun 12 '16 at 10:37
$begingroup$
Yes, you are right. Of course, the group itself should be included.
$endgroup$
– Dietrich Burde
Jun 12 '16 at 11:39
$begingroup$
Yes, you are right. Of course, the group itself should be included.
$endgroup$
– Dietrich Burde
Jun 12 '16 at 11:39
2
2
$begingroup$
By saying $langle{R^d}rangle$ is a normal subgroup for all divisors $d mid {n}$, you've actually already included {1}, because $n mid {n}$.
$endgroup$
– Rasputin
Oct 31 '16 at 20:22
$begingroup$
By saying $langle{R^d}rangle$ is a normal subgroup for all divisors $d mid {n}$, you've actually already included {1}, because $n mid {n}$.
$endgroup$
– Rasputin
Oct 31 '16 at 20:22
$begingroup$
@ Dietrich Burde : Is there any other normal subgroups of $D_{2n}$ besides these normal subgroups
$endgroup$
– user120386
May 22 '17 at 10:08
$begingroup$
@ Dietrich Burde : Is there any other normal subgroups of $D_{2n}$ besides these normal subgroups
$endgroup$
– user120386
May 22 '17 at 10:08
$begingroup$
@user120386 No, this is a complete classification of normal subgroups, see Keith Conrad's article on dihedral groups.
$endgroup$
– Dietrich Burde
May 22 '17 at 13:51
$begingroup$
@user120386 No, this is a complete classification of normal subgroups, see Keith Conrad's article on dihedral groups.
$endgroup$
– Dietrich Burde
May 22 '17 at 13:51
add a comment |
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