Mixing
Burnsides Theorem Application - Necklace
Having a lot of trouble understanding this question: How many different necklaces can be made from eight black beads and four white beads?
What I know:
Burnsides Theorem is given by: $$frac{1}{|G|} sum_{g in G} |operatorname{fix}_{Omega} (g)|$$
8+4=12 and this corresponds to the Dihedral group of order 12 (I use the notation $D_{2n})$ so I will note this as $D_{24}$, also known as the symmetric 12-gon.
So we know that $|G|=24$ and the total number of necklace configurations is given by $binom{12}{4}=495$ so I'm guessing we have to $Fix(e)=495$? But I'm unsure how to apply the rest of Burnsides theorem from here and work out the configurations of rotations and reflections.
Any help would be great, thanks.
abstract-algebra group-theory finite-groups
asked Jan 1 at 7:14
ReetyReety
13911
add a comment |
Having a lot of trouble understanding this question: How many different necklaces can be made from eight black beads and four white beads?
What I know:
Burnsides Theorem is given by: $$frac{1}{|G|} sum_{g in G} |operatorname{fix}_{Omega} (g)|$$
8+4=12 and this corresponds to the Dihedral group of order 12 (I use the notation $D_{2n})$ so I will note this as $D_{24}$, also known as the symmetric 12-gon.
So we know that $|G|=24$ and the total number of necklace configurations is given by $binom{12}{4}=495$ so I'm guessing we have to $Fix(e)=495$? But I'm unsure how to apply the rest of Burnsides theorem from here and work out the configurations of rotations and reflections.
Any help would be great, thanks.
abstract-algebra group-theory finite-groups
asked Jan 1 at 7:14
ReetyReety
13911
Think about the rotations and reflections of the dodecagon, and consider how each acts on its vertices.
– Lord Shark the Unknown
Jan 1 at 7:17
That's what I'm having trouble with visualising, this is known as Fix(g) right? @LordSharktheUnknown
– Reety
Jan 1 at 7:25
Make/buy a small dodecahedron. I did for, e.g., my module on the group theory of virology when I was an undergraduate; it helped considerably.
– Shaun
Jan 1 at 11:41
add a comment |
Having a lot of trouble understanding this question: How many different necklaces can be made from eight black beads and four white beads?
What I know:
Burnsides Theorem is given by: $$frac{1}{|G|} sum_{g in G} |operatorname{fix}_{Omega} (g)|$$
8+4=12 and this corresponds to the Dihedral group of order 12 (I use the notation $D_{2n})$ so I will note this as $D_{24}$, also known as the symmetric 12-gon.
So we know that $|G|=24$ and the total number of necklace configurations is given by $binom{12}{4}=495$ so I'm guessing we have to $Fix(e)=495$? But I'm unsure how to apply the rest of Burnsides theorem from here and work out the configurations of rotations and reflections.
Any help would be great, thanks.
abstract-algebra group-theory finite-groups
asked Jan 1 at 7:14
ReetyReety
13911
Having a lot of trouble understanding this question: How many different necklaces can be made from eight black beads and four white beads?
What I know:
Burnsides Theorem is given by: $$frac{1}{|G|} sum_{g in G} |operatorname{fix}_{Omega} (g)|$$
8+4=12 and this corresponds to the Dihedral group of order 12 (I use the notation $D_{2n})$ so I will note this as $D_{24}$, also known as the symmetric 12-gon.
So we know that $|G|=24$ and the total number of necklace configurations is given by $binom{12}{4}=495$ so I'm guessing we have to $Fix(e)=495$? But I'm unsure how to apply the rest of Burnsides theorem from here and work out the configurations of rotations and reflections.
Any help would be great, thanks.
abstract-algebra group-theory finite-groups
abstract-algebra group-theory finite-groups
asked Jan 1 at 7:14
ReetyReety
13911
asked Jan 1 at 7:14
ReetyReety
13911
asked Jan 1 at 7:14
ReetyReety
13911
asked Jan 1 at 7:14
ReetyReety
13911
asked Jan 1 at 7:14
ReetyReety
13911
13911
Think about the rotations and reflections of the dodecagon, and consider how each acts on its vertices.
– Lord Shark the Unknown
Jan 1 at 7:17
That's what I'm having trouble with visualising, this is known as Fix(g) right? @LordSharktheUnknown
– Reety
Jan 1 at 7:25
Make/buy a small dodecahedron. I did for, e.g., my module on the group theory of virology when I was an undergraduate; it helped considerably.
– Shaun
Jan 1 at 11:41
add a comment |
Think about the rotations and reflections of the dodecagon, and consider how each acts on its vertices.
– Lord Shark the Unknown
Jan 1 at 7:17
That's what I'm having trouble with visualising, this is known as Fix(g) right? @LordSharktheUnknown
– Reety
Jan 1 at 7:25
Make/buy a small dodecahedron. I did for, e.g., my module on the group theory of virology when I was an undergraduate; it helped considerably.
– Shaun
Jan 1 at 11:41
Think about the rotations and reflections of the dodecagon, and consider how each acts on its vertices.
– Lord Shark the Unknown
Jan 1 at 7:17
Think about the rotations and reflections of the dodecagon, and consider how each acts on its vertices.
– Lord Shark the Unknown
Jan 1 at 7:17
That's what I'm having trouble with visualising, this is known as Fix(g) right? @LordSharktheUnknown
– Reety
Jan 1 at 7:25
That's what I'm having trouble with visualising, this is known as Fix(g) right? @LordSharktheUnknown
– Reety
Jan 1 at 7:25
Make/buy a small dodecahedron. I did for, e.g., my module on the group theory of virology when I was an undergraduate; it helped considerably.
– Shaun
Jan 1 at 11:41
Make/buy a small dodecahedron. I did for, e.g., my module on the group theory of virology when I was an undergraduate; it helped considerably.
– Shaun
Jan 1 at 11:41
add a comment |
1 Answer
1
active
oldest
votes
Given that we have dihedral symmetry we suppose we are working with
bracelets (naming convention by OEIS). This requires the cycle index
$Z(D_{12})$ of the dihedral group $D_{12}.$ We have for the cyclic
group that
$$Z(C_{12}) = sum_{d|12} varphi(d) a_d^{12/d}
= frac{1}{12}
(a_1^{12} + a_2^6 + 2 a_3^4 + 2 a_4^3 + 2 a_6^2 + 4 a_{12}).$$
We get twelve more permutations corresponding to flips about an axis
passing through opposite slots or opposite edges, for a result of
$$Z(D_{12}) = frac{1}{24}
(a_1^{12} + a_2^6 + 2 a_3^4 + 2 a_4^3 + 2 a_6^2 + 4 a_{12})
+ frac{1}{24} ( 6 a_1^2 a_2^5 + 6 a_2^6).$$
By the Polya Enumeration Theorem (PET) we are interested in the
quantity
$$[B^8 W^4] Z(D_{12}; B+W).$$
Working through the terms we find
- $[B^8 W^4] (B+W)^{12} = {12choose 8},$
- $[B^8 W^4] (B^2+W^2)^{6} = [B^4 W^2] (B+W)^{6}
= {6choose 4},$ - $[B^8 W^4] 2 (B^3+W^3)^{4} = 0$
- $[B^8 W^4] 2 (B^4+W^4)^{3} = [B^2 W] 2 (B+W)^{3}
= 2 {3choose 2}$ - $[B^8 W^4] 2 (B^6+W^6)^{2} = 0$
- $[B^8 W^4] 4 (B^{12}+W^{12}) = 0$
We get from the reflections
- $[B^8 W^4] 6 (B+W)^2 (B^2+W^2)^5
\ = [B^6 W^4] 6 (B^2+W^2)^5 + [B^7 W^3] 12 (B^2+W^2)^5 +
[B^8 W^2] 6 (B^2+W^2)^5
\ = [B^3 W^2] 6 (B+W)^5 + [B^4 W] 6 (B+W)^5
\ = 6 {5choose 3} + 6 {5choose 4}$ - $[B^8 W^4] 6 (B^2+W^2)^{6} = [B^4 W^2] 6 (B+W)^{6}
= 6 {6choose 4}.$
Collecting everything we find for our result that it is
$$frac{1}{24} left({12choose 8} + {6choose 4} + 2 {3choose 2}
+ 6{5choose 3} + 6{5choose 4} + 6 {6choose 4}right).$$
which yields
$$bbox[5px,border:2px solid #00A000]{29.}$$
answered Jan 1 at 15:11
Marko RiedelMarko Riedel
39.4k339107
Of topic, is this one also doable by PET or Burnside: math.stackexchange.com/questions/314788/…
– greedoid
Jan 1 at 15:23
It appears that the answer is yes, consult e.g. MSE 2894653.
– Marko Riedel
Jan 1 at 15:30
@MarkoRiedel I haven't been introduced to this theorem/notation before would you mind explaining what the $a_{n}$ to the power, variables are. Thanks.
– Reety
Jan 2 at 2:58
Wikipedia on cycle indices is quite useful here.
– Marko Riedel
Jan 2 at 14:03
add a comment |
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Given that we have dihedral symmetry we suppose we are working with
bracelets (naming convention by OEIS). This requires the cycle index
$Z(D_{12})$ of the dihedral group $D_{12}.$ We have for the cyclic
group that
$$Z(C_{12}) = sum_{d|12} varphi(d) a_d^{12/d}
= frac{1}{12}
(a_1^{12} + a_2^6 + 2 a_3^4 + 2 a_4^3 + 2 a_6^2 + 4 a_{12}).$$
We get twelve more permutations corresponding to flips about an axis
passing through opposite slots or opposite edges, for a result of
$$Z(D_{12}) = frac{1}{24}
(a_1^{12} + a_2^6 + 2 a_3^4 + 2 a_4^3 + 2 a_6^2 + 4 a_{12})
+ frac{1}{24} ( 6 a_1^2 a_2^5 + 6 a_2^6).$$
By the Polya Enumeration Theorem (PET) we are interested in the
quantity
$$[B^8 W^4] Z(D_{12}; B+W).$$
Working through the terms we find
- $[B^8 W^4] (B+W)^{12} = {12choose 8},$
- $[B^8 W^4] (B^2+W^2)^{6} = [B^4 W^2] (B+W)^{6}
= {6choose 4},$ - $[B^8 W^4] 2 (B^3+W^3)^{4} = 0$
- $[B^8 W^4] 2 (B^4+W^4)^{3} = [B^2 W] 2 (B+W)^{3}
= 2 {3choose 2}$ - $[B^8 W^4] 2 (B^6+W^6)^{2} = 0$
- $[B^8 W^4] 4 (B^{12}+W^{12}) = 0$
We get from the reflections
- $[B^8 W^4] 6 (B+W)^2 (B^2+W^2)^5
\ = [B^6 W^4] 6 (B^2+W^2)^5 + [B^7 W^3] 12 (B^2+W^2)^5 +
[B^8 W^2] 6 (B^2+W^2)^5
\ = [B^3 W^2] 6 (B+W)^5 + [B^4 W] 6 (B+W)^5
\ = 6 {5choose 3} + 6 {5choose 4}$ - $[B^8 W^4] 6 (B^2+W^2)^{6} = [B^4 W^2] 6 (B+W)^{6}
= 6 {6choose 4}.$
Collecting everything we find for our result that it is
$$frac{1}{24} left({12choose 8} + {6choose 4} + 2 {3choose 2}
+ 6{5choose 3} + 6{5choose 4} + 6 {6choose 4}right).$$
which yields
$$bbox[5px,border:2px solid #00A000]{29.}$$
answered Jan 1 at 15:11
Marko RiedelMarko Riedel
39.4k339107
Of topic, is this one also doable by PET or Burnside: math.stackexchange.com/questions/314788/…
– greedoid
Jan 1 at 15:23
It appears that the answer is yes, consult e.g. MSE 2894653.
– Marko Riedel
Jan 1 at 15:30
@MarkoRiedel I haven't been introduced to this theorem/notation before would you mind explaining what the $a_{n}$ to the power, variables are. Thanks.
– Reety
Jan 2 at 2:58
Wikipedia on cycle indices is quite useful here.
– Marko Riedel
Jan 2 at 14:03
add a comment |
Given that we have dihedral symmetry we suppose we are working with
bracelets (naming convention by OEIS). This requires the cycle index
$Z(D_{12})$ of the dihedral group $D_{12}.$ We have for the cyclic
group that
$$Z(C_{12}) = sum_{d|12} varphi(d) a_d^{12/d}
= frac{1}{12}
(a_1^{12} + a_2^6 + 2 a_3^4 + 2 a_4^3 + 2 a_6^2 + 4 a_{12}).$$
We get twelve more permutations corresponding to flips about an axis
passing through opposite slots or opposite edges, for a result of
$$Z(D_{12}) = frac{1}{24}
(a_1^{12} + a_2^6 + 2 a_3^4 + 2 a_4^3 + 2 a_6^2 + 4 a_{12})
+ frac{1}{24} ( 6 a_1^2 a_2^5 + 6 a_2^6).$$
By the Polya Enumeration Theorem (PET) we are interested in the
quantity
$$[B^8 W^4] Z(D_{12}; B+W).$$
Working through the terms we find
- $[B^8 W^4] (B+W)^{12} = {12choose 8},$
- $[B^8 W^4] (B^2+W^2)^{6} = [B^4 W^2] (B+W)^{6}
= {6choose 4},$ - $[B^8 W^4] 2 (B^3+W^3)^{4} = 0$
- $[B^8 W^4] 2 (B^4+W^4)^{3} = [B^2 W] 2 (B+W)^{3}
= 2 {3choose 2}$ - $[B^8 W^4] 2 (B^6+W^6)^{2} = 0$
- $[B^8 W^4] 4 (B^{12}+W^{12}) = 0$
We get from the reflections
- $[B^8 W^4] 6 (B+W)^2 (B^2+W^2)^5
\ = [B^6 W^4] 6 (B^2+W^2)^5 + [B^7 W^3] 12 (B^2+W^2)^5 +
[B^8 W^2] 6 (B^2+W^2)^5
\ = [B^3 W^2] 6 (B+W)^5 + [B^4 W] 6 (B+W)^5
\ = 6 {5choose 3} + 6 {5choose 4}$ - $[B^8 W^4] 6 (B^2+W^2)^{6} = [B^4 W^2] 6 (B+W)^{6}
= 6 {6choose 4}.$
Collecting everything we find for our result that it is
$$frac{1}{24} left({12choose 8} + {6choose 4} + 2 {3choose 2}
+ 6{5choose 3} + 6{5choose 4} + 6 {6choose 4}right).$$
which yields
$$bbox[5px,border:2px solid #00A000]{29.}$$
answered Jan 1 at 15:11
Marko RiedelMarko Riedel
39.4k339107
Of topic, is this one also doable by PET or Burnside: math.stackexchange.com/questions/314788/…
– greedoid
Jan 1 at 15:23
It appears that the answer is yes, consult e.g. MSE 2894653.
– Marko Riedel
Jan 1 at 15:30
@MarkoRiedel I haven't been introduced to this theorem/notation before would you mind explaining what the $a_{n}$ to the power, variables are. Thanks.
– Reety
Jan 2 at 2:58
Wikipedia on cycle indices is quite useful here.
– Marko Riedel
Jan 2 at 14:03
add a comment |
Given that we have dihedral symmetry we suppose we are working with
bracelets (naming convention by OEIS). This requires the cycle index
$Z(D_{12})$ of the dihedral group $D_{12}.$ We have for the cyclic
group that
$$Z(C_{12}) = sum_{d|12} varphi(d) a_d^{12/d}
= frac{1}{12}
(a_1^{12} + a_2^6 + 2 a_3^4 + 2 a_4^3 + 2 a_6^2 + 4 a_{12}).$$
We get twelve more permutations corresponding to flips about an axis
passing through opposite slots or opposite edges, for a result of
$$Z(D_{12}) = frac{1}{24}
(a_1^{12} + a_2^6 + 2 a_3^4 + 2 a_4^3 + 2 a_6^2 + 4 a_{12})
+ frac{1}{24} ( 6 a_1^2 a_2^5 + 6 a_2^6).$$
By the Polya Enumeration Theorem (PET) we are interested in the
quantity
$$[B^8 W^4] Z(D_{12}; B+W).$$
Working through the terms we find
- $[B^8 W^4] (B+W)^{12} = {12choose 8},$
- $[B^8 W^4] (B^2+W^2)^{6} = [B^4 W^2] (B+W)^{6}
= {6choose 4},$ - $[B^8 W^4] 2 (B^3+W^3)^{4} = 0$
- $[B^8 W^4] 2 (B^4+W^4)^{3} = [B^2 W] 2 (B+W)^{3}
= 2 {3choose 2}$ - $[B^8 W^4] 2 (B^6+W^6)^{2} = 0$
- $[B^8 W^4] 4 (B^{12}+W^{12}) = 0$
We get from the reflections
- $[B^8 W^4] 6 (B+W)^2 (B^2+W^2)^5
\ = [B^6 W^4] 6 (B^2+W^2)^5 + [B^7 W^3] 12 (B^2+W^2)^5 +
[B^8 W^2] 6 (B^2+W^2)^5
\ = [B^3 W^2] 6 (B+W)^5 + [B^4 W] 6 (B+W)^5
\ = 6 {5choose 3} + 6 {5choose 4}$ - $[B^8 W^4] 6 (B^2+W^2)^{6} = [B^4 W^2] 6 (B+W)^{6}
= 6 {6choose 4}.$
Collecting everything we find for our result that it is
$$frac{1}{24} left({12choose 8} + {6choose 4} + 2 {3choose 2}
+ 6{5choose 3} + 6{5choose 4} + 6 {6choose 4}right).$$
which yields
$$bbox[5px,border:2px solid #00A000]{29.}$$
answered Jan 1 at 15:11
Marko RiedelMarko Riedel
39.4k339107
Given that we have dihedral symmetry we suppose we are working with
bracelets (naming convention by OEIS). This requires the cycle index
$Z(D_{12})$ of the dihedral group $D_{12}.$ We have for the cyclic
group that
$$Z(C_{12}) = sum_{d|12} varphi(d) a_d^{12/d}
= frac{1}{12}
(a_1^{12} + a_2^6 + 2 a_3^4 + 2 a_4^3 + 2 a_6^2 + 4 a_{12}).$$
We get twelve more permutations corresponding to flips about an axis
passing through opposite slots or opposite edges, for a result of
$$Z(D_{12}) = frac{1}{24}
(a_1^{12} + a_2^6 + 2 a_3^4 + 2 a_4^3 + 2 a_6^2 + 4 a_{12})
+ frac{1}{24} ( 6 a_1^2 a_2^5 + 6 a_2^6).$$
By the Polya Enumeration Theorem (PET) we are interested in the
quantity
$$[B^8 W^4] Z(D_{12}; B+W).$$
Working through the terms we find
- $[B^8 W^4] (B+W)^{12} = {12choose 8},$
- $[B^8 W^4] (B^2+W^2)^{6} = [B^4 W^2] (B+W)^{6}
= {6choose 4},$ - $[B^8 W^4] 2 (B^3+W^3)^{4} = 0$
- $[B^8 W^4] 2 (B^4+W^4)^{3} = [B^2 W] 2 (B+W)^{3}
= 2 {3choose 2}$ - $[B^8 W^4] 2 (B^6+W^6)^{2} = 0$
- $[B^8 W^4] 4 (B^{12}+W^{12}) = 0$
We get from the reflections
- $[B^8 W^4] 6 (B+W)^2 (B^2+W^2)^5
\ = [B^6 W^4] 6 (B^2+W^2)^5 + [B^7 W^3] 12 (B^2+W^2)^5 +
[B^8 W^2] 6 (B^2+W^2)^5
\ = [B^3 W^2] 6 (B+W)^5 + [B^4 W] 6 (B+W)^5
\ = 6 {5choose 3} + 6 {5choose 4}$ - $[B^8 W^4] 6 (B^2+W^2)^{6} = [B^4 W^2] 6 (B+W)^{6}
= 6 {6choose 4}.$
Collecting everything we find for our result that it is
$$frac{1}{24} left({12choose 8} + {6choose 4} + 2 {3choose 2}
+ 6{5choose 3} + 6{5choose 4} + 6 {6choose 4}right).$$
which yields
$$bbox[5px,border:2px solid #00A000]{29.}$$
answered Jan 1 at 15:11
Marko RiedelMarko Riedel
39.4k339107
edited Jan 1 at 15:37
answered Jan 1 at 15:11
Marko RiedelMarko Riedel
39.4k339107
answered Jan 1 at 15:11
Marko RiedelMarko Riedel
39.4k339107
answered Jan 1 at 15:11
Marko RiedelMarko Riedel
39.4k339107
39.4k339107
Of topic, is this one also doable by PET or Burnside: math.stackexchange.com/questions/314788/…
– greedoid
Jan 1 at 15:23
It appears that the answer is yes, consult e.g. MSE 2894653.
– Marko Riedel
Jan 1 at 15:30
@MarkoRiedel I haven't been introduced to this theorem/notation before would you mind explaining what the $a_{n}$ to the power, variables are. Thanks.
– Reety
Jan 2 at 2:58
Wikipedia on cycle indices is quite useful here.
– Marko Riedel
Jan 2 at 14:03
add a comment |
Of topic, is this one also doable by PET or Burnside: math.stackexchange.com/questions/314788/…
– greedoid
Jan 1 at 15:23
It appears that the answer is yes, consult e.g. MSE 2894653.
– Marko Riedel
Jan 1 at 15:30
@MarkoRiedel I haven't been introduced to this theorem/notation before would you mind explaining what the $a_{n}$ to the power, variables are. Thanks.
– Reety
Jan 2 at 2:58
Wikipedia on cycle indices is quite useful here.
– Marko Riedel
Jan 2 at 14:03
Of topic, is this one also doable by PET or Burnside: math.stackexchange.com/questions/314788/…
– greedoid
Jan 1 at 15:23
Of topic, is this one also doable by PET or Burnside: math.stackexchange.com/questions/314788/…
– greedoid
Jan 1 at 15:23
It appears that the answer is yes, consult e.g. MSE 2894653.
– Marko Riedel
Jan 1 at 15:30
It appears that the answer is yes, consult e.g. MSE 2894653.
– Marko Riedel
Jan 1 at 15:30
@MarkoRiedel I haven't been introduced to this theorem/notation before would you mind explaining what the $a_{n}$ to the power, variables are. Thanks.
– Reety
Jan 2 at 2:58
@MarkoRiedel I haven't been introduced to this theorem/notation before would you mind explaining what the $a_{n}$ to the power, variables are. Thanks.
– Reety
Jan 2 at 2:58
Wikipedia on cycle indices is quite useful here.
– Marko Riedel
Jan 2 at 14:03
Wikipedia on cycle indices is quite useful here.
– Marko Riedel
Jan 2 at 14:03
add a comment |
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Think about the rotations and reflections of the dodecagon, and consider how each acts on its vertices.
– Lord Shark the Unknown
Jan 1 at 7:17
That's what I'm having trouble with visualising, this is known as Fix(g) right? @LordSharktheUnknown
– Reety
Jan 1 at 7:25
Make/buy a small dodecahedron. I did for, e.g., my module on the group theory of virology when I was an undergraduate; it helped considerably.
– Shaun
Jan 1 at 11:41