Mixing
The Dihedral Group D_12 is a Non-Abelian Group
The question comes from page 29 of Evan Chan's Napkin (https://usamo.files.wordpress.com/2018/08/napkin-2018-08-22.pdf).
I'm having a hard time understanding his trivia. From what I see, D_12 appears to be commutative but maybe I'm missing something.
P.S. SOLVED: I was reading compositions the wrong direction and misunderstanding the type of reflections shown in his diagrams. Thank you all for the help and for reminding me why I love this site!
group-theory
asked Nov 21 '18 at 0:54
Connor James
1188
|
show 2 more comments
The question comes from page 29 of Evan Chan's Napkin (https://usamo.files.wordpress.com/2018/08/napkin-2018-08-22.pdf).
I'm having a hard time understanding his trivia. From what I see, D_12 appears to be commutative but maybe I'm missing something.
P.S. SOLVED: I was reading compositions the wrong direction and misunderstanding the type of reflections shown in his diagrams. Thank you all for the help and for reminding me why I love this site!
group-theory
asked Nov 21 '18 at 0:54
Connor James
1188
@JohnNash isn't that for D_10 though? Did I draw the rotations on D_{12} correctly?
– Connor James
Nov 21 '18 at 0:59
1
@JohnNash OP is asking about $D_{12}$
– amWhy
Nov 21 '18 at 1:00
360/6=60 so it rotates 60 degree. Now Write all elements then check commutative property.
– John Nash
Nov 21 '18 at 1:05
1
Take R₆₀ and H . Now you can see They do not commute. Where H is horizontal reflection
– John Nash
Nov 21 '18 at 1:15
@JohnNash The original author's reflections were not relative to the horizontal but relative to the vertices themselves. Did he mess up?
– Connor James
Nov 21 '18 at 1:21
|
show 2 more comments
The question comes from page 29 of Evan Chan's Napkin (https://usamo.files.wordpress.com/2018/08/napkin-2018-08-22.pdf).
I'm having a hard time understanding his trivia. From what I see, D_12 appears to be commutative but maybe I'm missing something.
P.S. SOLVED: I was reading compositions the wrong direction and misunderstanding the type of reflections shown in his diagrams. Thank you all for the help and for reminding me why I love this site!
group-theory
asked Nov 21 '18 at 0:54
Connor James
1188
The question comes from page 29 of Evan Chan's Napkin (https://usamo.files.wordpress.com/2018/08/napkin-2018-08-22.pdf).
I'm having a hard time understanding his trivia. From what I see, D_12 appears to be commutative but maybe I'm missing something.
P.S. SOLVED: I was reading compositions the wrong direction and misunderstanding the type of reflections shown in his diagrams. Thank you all for the help and for reminding me why I love this site!
group-theory
group-theory
asked Nov 21 '18 at 0:54
Connor James
1188
asked Nov 21 '18 at 0:54
Connor James
1188
edited Nov 21 '18 at 1:44
asked Nov 21 '18 at 0:54
Connor James
1188
asked Nov 21 '18 at 0:54
Connor James
1188
asked Nov 21 '18 at 0:54
Connor James
1188
1188
@JohnNash isn't that for D_10 though? Did I draw the rotations on D_{12} correctly?
– Connor James
Nov 21 '18 at 0:59
1
@JohnNash OP is asking about $D_{12}$
– amWhy
Nov 21 '18 at 1:00
360/6=60 so it rotates 60 degree. Now Write all elements then check commutative property.
– John Nash
Nov 21 '18 at 1:05
1
Take R₆₀ and H . Now you can see They do not commute. Where H is horizontal reflection
– John Nash
Nov 21 '18 at 1:15
@JohnNash The original author's reflections were not relative to the horizontal but relative to the vertices themselves. Did he mess up?
– Connor James
Nov 21 '18 at 1:21
|
show 2 more comments
@JohnNash isn't that for D_10 though? Did I draw the rotations on D_{12} correctly?
– Connor James
Nov 21 '18 at 0:59
1
@JohnNash OP is asking about $D_{12}$
– amWhy
Nov 21 '18 at 1:00
360/6=60 so it rotates 60 degree. Now Write all elements then check commutative property.
– John Nash
Nov 21 '18 at 1:05
1
Take R₆₀ and H . Now you can see They do not commute. Where H is horizontal reflection
– John Nash
Nov 21 '18 at 1:15
@JohnNash The original author's reflections were not relative to the horizontal but relative to the vertices themselves. Did he mess up?
– Connor James
Nov 21 '18 at 1:21
@JohnNash isn't that for D_10 though? Did I draw the rotations on D_{12} correctly?
– Connor James
Nov 21 '18 at 0:59
@JohnNash isn't that for D_10 though? Did I draw the rotations on D_{12} correctly?
– Connor James
Nov 21 '18 at 0:59
1
1
@JohnNash OP is asking about $D_{12}$
– amWhy
Nov 21 '18 at 1:00
@JohnNash OP is asking about $D_{12}$
– amWhy
Nov 21 '18 at 1:00
360/6=60 so it rotates 60 degree. Now Write all elements then check commutative property.
– John Nash
Nov 21 '18 at 1:05
360/6=60 so it rotates 60 degree. Now Write all elements then check commutative property.
– John Nash
Nov 21 '18 at 1:05
1
1
Take R₆₀ and H . Now you can see They do not commute. Where H is horizontal reflection
– John Nash
Nov 21 '18 at 1:15
Take R₆₀ and H . Now you can see They do not commute. Where H is horizontal reflection
– John Nash
Nov 21 '18 at 1:15
@JohnNash The original author's reflections were not relative to the horizontal but relative to the vertices themselves. Did he mess up?
– Connor James
Nov 21 '18 at 1:21
@JohnNash The original author's reflections were not relative to the horizontal but relative to the vertices themselves. Did he mess up?
– Connor James
Nov 21 '18 at 1:21
|
show 2 more comments
2 Answers
2
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Hint: the dihedral group with 6 elements, i.e., the group of isometries of an equilateral triangle is non-abelian and is a subgroup of the group of isometries of a regular hexagon (the dihedral group with 12 elements). (Different authors have different conventions about the notation for the isometry groups of regular $n$-gons a.k.a. dihedral groups: some write $D_n$ where $n$ is the number of elements in the group and some write $D_n$ where $n$ is the number of vertices in the $n$-gon.) Rotations and reflections do not commute in general in any dihedral group.
answered Nov 21 '18 at 1:18
Rob Arthan
29.1k42866
add a comment |
Let $;r;$ be rotation in $;60^circ;$ (counterclockwise, say) and $;s;$ refletion through the diagonal $;1-4;$ (the vertical one in your first diagram), then in your upper diagrams, after you apply $;s;$ (the application of $;s;$ is correct), you actually get $;5-6-1-2-3-4;$ , beginning at the top and counterclockwise, not what you wrote there, and the second diagram is correct...so not commutative.
answered Nov 21 '18 at 1:10
DonAntonio
177k1492225
So, if I understand correctly when the reflection $;s;$ is applied, I should always do it through the vertical, not relative to where 1 and 4 are after rotation, right? If that's the case, why didn't he do that in his diagram of D_10?
– Connor James
Nov 21 '18 at 1:18
@ConnorJames You can choose any diagonal of the $;n,-$ gon going through the center. I chose the vertical diagonal because it seems to me the easiest to "see", but you can choose any other diagonal. In $;D_{10};$ , the "diagonal" is in fact a segment from a vertex to the middle point of the opposite side...
– DonAntonio
Nov 21 '18 at 1:54
thank you. My problem is I read the compositions backwards and though he was moving his reflection axis to stay with his first vertex. Now I see it doesn't matter where you reflect, you just have to be consistent.
– Connor James
Nov 21 '18 at 1:56
@ConnorJames Exactly: you chose one diagonal to reflect through it, stay with it all the time. Rotations are usually easier to deal with.
– DonAntonio
Nov 21 '18 at 9:24
add a comment |
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Hint: the dihedral group with 6 elements, i.e., the group of isometries of an equilateral triangle is non-abelian and is a subgroup of the group of isometries of a regular hexagon (the dihedral group with 12 elements). (Different authors have different conventions about the notation for the isometry groups of regular $n$-gons a.k.a. dihedral groups: some write $D_n$ where $n$ is the number of elements in the group and some write $D_n$ where $n$ is the number of vertices in the $n$-gon.) Rotations and reflections do not commute in general in any dihedral group.
answered Nov 21 '18 at 1:18
Rob Arthan
29.1k42866
add a comment |
Hint: the dihedral group with 6 elements, i.e., the group of isometries of an equilateral triangle is non-abelian and is a subgroup of the group of isometries of a regular hexagon (the dihedral group with 12 elements). (Different authors have different conventions about the notation for the isometry groups of regular $n$-gons a.k.a. dihedral groups: some write $D_n$ where $n$ is the number of elements in the group and some write $D_n$ where $n$ is the number of vertices in the $n$-gon.) Rotations and reflections do not commute in general in any dihedral group.
answered Nov 21 '18 at 1:18
Rob Arthan
29.1k42866
add a comment |
Hint: the dihedral group with 6 elements, i.e., the group of isometries of an equilateral triangle is non-abelian and is a subgroup of the group of isometries of a regular hexagon (the dihedral group with 12 elements). (Different authors have different conventions about the notation for the isometry groups of regular $n$-gons a.k.a. dihedral groups: some write $D_n$ where $n$ is the number of elements in the group and some write $D_n$ where $n$ is the number of vertices in the $n$-gon.) Rotations and reflections do not commute in general in any dihedral group.
answered Nov 21 '18 at 1:18
Rob Arthan
29.1k42866
Hint: the dihedral group with 6 elements, i.e., the group of isometries of an equilateral triangle is non-abelian and is a subgroup of the group of isometries of a regular hexagon (the dihedral group with 12 elements). (Different authors have different conventions about the notation for the isometry groups of regular $n$-gons a.k.a. dihedral groups: some write $D_n$ where $n$ is the number of elements in the group and some write $D_n$ where $n$ is the number of vertices in the $n$-gon.) Rotations and reflections do not commute in general in any dihedral group.
answered Nov 21 '18 at 1:18
Rob Arthan
29.1k42866
answered Nov 21 '18 at 1:18
Rob Arthan
29.1k42866
answered Nov 21 '18 at 1:18
Rob Arthan
29.1k42866
answered Nov 21 '18 at 1:18
Rob Arthan
29.1k42866
29.1k42866
add a comment |
add a comment |
Let $;r;$ be rotation in $;60^circ;$ (counterclockwise, say) and $;s;$ refletion through the diagonal $;1-4;$ (the vertical one in your first diagram), then in your upper diagrams, after you apply $;s;$ (the application of $;s;$ is correct), you actually get $;5-6-1-2-3-4;$ , beginning at the top and counterclockwise, not what you wrote there, and the second diagram is correct...so not commutative.
answered Nov 21 '18 at 1:10
DonAntonio
177k1492225
So, if I understand correctly when the reflection $;s;$ is applied, I should always do it through the vertical, not relative to where 1 and 4 are after rotation, right? If that's the case, why didn't he do that in his diagram of D_10?
– Connor James
Nov 21 '18 at 1:18
@ConnorJames You can choose any diagonal of the $;n,-$ gon going through the center. I chose the vertical diagonal because it seems to me the easiest to "see", but you can choose any other diagonal. In $;D_{10};$ , the "diagonal" is in fact a segment from a vertex to the middle point of the opposite side...
– DonAntonio
Nov 21 '18 at 1:54
thank you. My problem is I read the compositions backwards and though he was moving his reflection axis to stay with his first vertex. Now I see it doesn't matter where you reflect, you just have to be consistent.
– Connor James
Nov 21 '18 at 1:56
@ConnorJames Exactly: you chose one diagonal to reflect through it, stay with it all the time. Rotations are usually easier to deal with.
– DonAntonio
Nov 21 '18 at 9:24
add a comment |
Let $;r;$ be rotation in $;60^circ;$ (counterclockwise, say) and $;s;$ refletion through the diagonal $;1-4;$ (the vertical one in your first diagram), then in your upper diagrams, after you apply $;s;$ (the application of $;s;$ is correct), you actually get $;5-6-1-2-3-4;$ , beginning at the top and counterclockwise, not what you wrote there, and the second diagram is correct...so not commutative.
answered Nov 21 '18 at 1:10
DonAntonio
177k1492225
So, if I understand correctly when the reflection $;s;$ is applied, I should always do it through the vertical, not relative to where 1 and 4 are after rotation, right? If that's the case, why didn't he do that in his diagram of D_10?
– Connor James
Nov 21 '18 at 1:18
@ConnorJames You can choose any diagonal of the $;n,-$ gon going through the center. I chose the vertical diagonal because it seems to me the easiest to "see", but you can choose any other diagonal. In $;D_{10};$ , the "diagonal" is in fact a segment from a vertex to the middle point of the opposite side...
– DonAntonio
Nov 21 '18 at 1:54
thank you. My problem is I read the compositions backwards and though he was moving his reflection axis to stay with his first vertex. Now I see it doesn't matter where you reflect, you just have to be consistent.
– Connor James
Nov 21 '18 at 1:56
@ConnorJames Exactly: you chose one diagonal to reflect through it, stay with it all the time. Rotations are usually easier to deal with.
– DonAntonio
Nov 21 '18 at 9:24
add a comment |
Let $;r;$ be rotation in $;60^circ;$ (counterclockwise, say) and $;s;$ refletion through the diagonal $;1-4;$ (the vertical one in your first diagram), then in your upper diagrams, after you apply $;s;$ (the application of $;s;$ is correct), you actually get $;5-6-1-2-3-4;$ , beginning at the top and counterclockwise, not what you wrote there, and the second diagram is correct...so not commutative.
answered Nov 21 '18 at 1:10
DonAntonio
177k1492225
Let $;r;$ be rotation in $;60^circ;$ (counterclockwise, say) and $;s;$ refletion through the diagonal $;1-4;$ (the vertical one in your first diagram), then in your upper diagrams, after you apply $;s;$ (the application of $;s;$ is correct), you actually get $;5-6-1-2-3-4;$ , beginning at the top and counterclockwise, not what you wrote there, and the second diagram is correct...so not commutative.
answered Nov 21 '18 at 1:10
DonAntonio
177k1492225
answered Nov 21 '18 at 1:10
DonAntonio
177k1492225
answered Nov 21 '18 at 1:10
DonAntonio
177k1492225
answered Nov 21 '18 at 1:10
DonAntonio
177k1492225
177k1492225
So, if I understand correctly when the reflection $;s;$ is applied, I should always do it through the vertical, not relative to where 1 and 4 are after rotation, right? If that's the case, why didn't he do that in his diagram of D_10?
– Connor James
Nov 21 '18 at 1:18
@ConnorJames You can choose any diagonal of the $;n,-$ gon going through the center. I chose the vertical diagonal because it seems to me the easiest to "see", but you can choose any other diagonal. In $;D_{10};$ , the "diagonal" is in fact a segment from a vertex to the middle point of the opposite side...
– DonAntonio
Nov 21 '18 at 1:54
thank you. My problem is I read the compositions backwards and though he was moving his reflection axis to stay with his first vertex. Now I see it doesn't matter where you reflect, you just have to be consistent.
– Connor James
Nov 21 '18 at 1:56
@ConnorJames Exactly: you chose one diagonal to reflect through it, stay with it all the time. Rotations are usually easier to deal with.
– DonAntonio
Nov 21 '18 at 9:24
add a comment |
So, if I understand correctly when the reflection $;s;$ is applied, I should always do it through the vertical, not relative to where 1 and 4 are after rotation, right? If that's the case, why didn't he do that in his diagram of D_10?
– Connor James
Nov 21 '18 at 1:18
@ConnorJames You can choose any diagonal of the $;n,-$ gon going through the center. I chose the vertical diagonal because it seems to me the easiest to "see", but you can choose any other diagonal. In $;D_{10};$ , the "diagonal" is in fact a segment from a vertex to the middle point of the opposite side...
– DonAntonio
Nov 21 '18 at 1:54
thank you. My problem is I read the compositions backwards and though he was moving his reflection axis to stay with his first vertex. Now I see it doesn't matter where you reflect, you just have to be consistent.
– Connor James
Nov 21 '18 at 1:56
@ConnorJames Exactly: you chose one diagonal to reflect through it, stay with it all the time. Rotations are usually easier to deal with.
– DonAntonio
Nov 21 '18 at 9:24
So, if I understand correctly when the reflection $;s;$ is applied, I should always do it through the vertical, not relative to where 1 and 4 are after rotation, right? If that's the case, why didn't he do that in his diagram of D_10?
– Connor James
Nov 21 '18 at 1:18
So, if I understand correctly when the reflection $;s;$ is applied, I should always do it through the vertical, not relative to where 1 and 4 are after rotation, right? If that's the case, why didn't he do that in his diagram of D_10?
– Connor James
Nov 21 '18 at 1:18
@ConnorJames You can choose any diagonal of the $;n,-$ gon going through the center. I chose the vertical diagonal because it seems to me the easiest to "see", but you can choose any other diagonal. In $;D_{10};$ , the "diagonal" is in fact a segment from a vertex to the middle point of the opposite side...
– DonAntonio
Nov 21 '18 at 1:54
@ConnorJames You can choose any diagonal of the $;n,-$ gon going through the center. I chose the vertical diagonal because it seems to me the easiest to "see", but you can choose any other diagonal. In $;D_{10};$ , the "diagonal" is in fact a segment from a vertex to the middle point of the opposite side...
– DonAntonio
Nov 21 '18 at 1:54
thank you. My problem is I read the compositions backwards and though he was moving his reflection axis to stay with his first vertex. Now I see it doesn't matter where you reflect, you just have to be consistent.
– Connor James
Nov 21 '18 at 1:56
thank you. My problem is I read the compositions backwards and though he was moving his reflection axis to stay with his first vertex. Now I see it doesn't matter where you reflect, you just have to be consistent.
– Connor James
Nov 21 '18 at 1:56
@ConnorJames Exactly: you chose one diagonal to reflect through it, stay with it all the time. Rotations are usually easier to deal with.
– DonAntonio
Nov 21 '18 at 9:24
@ConnorJames Exactly: you chose one diagonal to reflect through it, stay with it all the time. Rotations are usually easier to deal with.
– DonAntonio
Nov 21 '18 at 9:24
add a comment |
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@JohnNash isn't that for D_10 though? Did I draw the rotations on D_{12} correctly?
– Connor James
Nov 21 '18 at 0:59
1
@JohnNash OP is asking about $D_{12}$
– amWhy
Nov 21 '18 at 1:00
360/6=60 so it rotates 60 degree. Now Write all elements then check commutative property.
– John Nash
Nov 21 '18 at 1:05
1
Take R₆₀ and H . Now you can see They do not commute. Where H is horizontal reflection
– John Nash
Nov 21 '18 at 1:15
@JohnNash The original author's reflections were not relative to the horizontal but relative to the vertices themselves. Did he mess up?
– Connor James
Nov 21 '18 at 1:21