Thinking of a cube as $mathbb Z_6timesmathbb Z_4$

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I was wondering if it makes sense to think of a cube as an additive group in this way. For example $(4,1)$ corresponds to face $4$ edge $1$.



I am a beginner in group theory and I hope this makes sense!



If I added together face $4$ edge $1$ to face $3$ edge $0$: $(4,1)+(3,0)=(1,1)$ would correspond to face $1$ edge $1$?







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  • 1




    Have you heard of the dihedral group? This isn't a "wrong" way to view the cube, but it's for sure less common that ones based on the dihedral group.
    – Mark
    Aug 23 at 5:49






  • 2




    That is a clear and easy to visualise way of describing the $24$ different orientations of the cube, but the operation, to me, doesn't seem to be as natural and easy to visualise.
    – Arthur
    Aug 23 at 6:43















up vote
1
down vote

favorite












I was wondering if it makes sense to think of a cube as an additive group in this way. For example $(4,1)$ corresponds to face $4$ edge $1$.



I am a beginner in group theory and I hope this makes sense!



If I added together face $4$ edge $1$ to face $3$ edge $0$: $(4,1)+(3,0)=(1,1)$ would correspond to face $1$ edge $1$?







share|cite|improve this question


















  • 1




    Have you heard of the dihedral group? This isn't a "wrong" way to view the cube, but it's for sure less common that ones based on the dihedral group.
    – Mark
    Aug 23 at 5:49






  • 2




    That is a clear and easy to visualise way of describing the $24$ different orientations of the cube, but the operation, to me, doesn't seem to be as natural and easy to visualise.
    – Arthur
    Aug 23 at 6:43













up vote
1
down vote

favorite









up vote
1
down vote

favorite











I was wondering if it makes sense to think of a cube as an additive group in this way. For example $(4,1)$ corresponds to face $4$ edge $1$.



I am a beginner in group theory and I hope this makes sense!



If I added together face $4$ edge $1$ to face $3$ edge $0$: $(4,1)+(3,0)=(1,1)$ would correspond to face $1$ edge $1$?







share|cite|improve this question














I was wondering if it makes sense to think of a cube as an additive group in this way. For example $(4,1)$ corresponds to face $4$ edge $1$.



I am a beginner in group theory and I hope this makes sense!



If I added together face $4$ edge $1$ to face $3$ edge $0$: $(4,1)+(3,0)=(1,1)$ would correspond to face $1$ edge $1$?









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 23 at 5:56









martini

69.2k45788




69.2k45788










asked Aug 23 at 5:14









Jungleshrimp

1369




1369







  • 1




    Have you heard of the dihedral group? This isn't a "wrong" way to view the cube, but it's for sure less common that ones based on the dihedral group.
    – Mark
    Aug 23 at 5:49






  • 2




    That is a clear and easy to visualise way of describing the $24$ different orientations of the cube, but the operation, to me, doesn't seem to be as natural and easy to visualise.
    – Arthur
    Aug 23 at 6:43













  • 1




    Have you heard of the dihedral group? This isn't a "wrong" way to view the cube, but it's for sure less common that ones based on the dihedral group.
    – Mark
    Aug 23 at 5:49






  • 2




    That is a clear and easy to visualise way of describing the $24$ different orientations of the cube, but the operation, to me, doesn't seem to be as natural and easy to visualise.
    – Arthur
    Aug 23 at 6:43








1




1




Have you heard of the dihedral group? This isn't a "wrong" way to view the cube, but it's for sure less common that ones based on the dihedral group.
– Mark
Aug 23 at 5:49




Have you heard of the dihedral group? This isn't a "wrong" way to view the cube, but it's for sure less common that ones based on the dihedral group.
– Mark
Aug 23 at 5:49




2




2




That is a clear and easy to visualise way of describing the $24$ different orientations of the cube, but the operation, to me, doesn't seem to be as natural and easy to visualise.
– Arthur
Aug 23 at 6:43





That is a clear and easy to visualise way of describing the $24$ different orientations of the cube, but the operation, to me, doesn't seem to be as natural and easy to visualise.
– Arthur
Aug 23 at 6:43











3 Answers
3






active

oldest

votes

















up vote
2
down vote



accepted










While your approach isn't invalid, I want you to ask back. What can be achieved via your definition of cube group? Why is it useful? Do not take this as an offense because I didn't mean it. They are questions needed for mathematics.



You defined how to match an element of a cube to an element of $mathbbZ_6 times mathbbZ_4$. There's basically no problem. Also, you said $(4,1) + (3, 0) = (1, 1)$ on $mathbbZ_6 times mathbbZ_4$. That's correct. Then you linked it to an operation on cube: adding face $4$ edge $1$ to face $3$ edge $0$, to get face $1$ edge $1$. But what's meaning on that? You can address meaningful explanation of that operation? Maybe a geometrical meaning?



If you can present such a meaning: that will be very nice. Maybe the group will be named after you. Or unfortunately somebody else already did it.



For example describing structure of cube, is a symmetry group of the cube. This is a group about cube, or more strictly, rotations and reflections of the cube(and octahedron). An element of the group means one exact rotation and reflection(that means, kinda preserving faces', edges' and vertices' locations) of the cube. They are total 24 ways to rotate, and 2 ways to reflect. Also, group operation(multiplying two elements of group) means composition of such rotations and reflections. So it addresses something, namely, symmetric structure of a cube.



So you defined a cube group structure of your own. That's great. The next thing is, to find it useful. Maybe you cannot find it, but it's OK. That's how we do mathematics. Just keep trying.






share|cite|improve this answer




















  • I dislike the line "Maybe the group will be named after you. Or unfortunately somebody else already did it." This is for a variety of reasons, but mostly: It is far too optimistic, and is not a good motivation for doing something.
    – user1729
    Aug 24 at 8:59











  • @user1729 good point. Maybe I'll think about it myself.
    – Jeong Jinmyeong
    Aug 24 at 23:19

















up vote
4
down vote













A possible problem with treating the cube as $mathbbZ_6 times mathbbZ_4$ is making sense of what that allows you to do. In your question, at the end, you note that you could add "face $4$, edge $1$" to "face $3$, edge $0$" and get "face $1$, edge $1$". But what does that actually mean?



If we label our cube so that each face has a number like a traditional die then opposite pairs of faces add to 7 (so 1 is opposite 6, 2 is opposite 5 and 3 is opposite 4). So your representation allows you to "add" edges of faces that aren't connected to each other. Is that useful? (It might be, by the way, but I can't think of a use at the moment.)



So: either we need to add some more structure to your representation in order to "forbid" additions that don't make sense, or we need to change the representation in order to ensure that the operations we can perform all make sense. Both are valid approaches: we could say for example that we can only "add" elements of the group together if they share a common edge or a common face. You'll come across this in your course when you start talking about generators and relations which essentially say: here are specific relationships, and these are the permissible operations on them.



The dihedral group $mathbbD_n$ was mentioned by a commentor above: this takes the symmetries of the square as the essential elements: rotations and reflections, and then specifies how they fit together. If $a$ is rotation by $90^circ$ about the $y$-axis and $b$ is reflection in the $y$-axis then you can state that $a^4 = 1$ since four rotations take you back to where you start, and $b^2 =1$ since two reflections take you back to where you started.



As @user1729 notes in the comments, the symmetry group of the cube is more complicated, and you'll find it easier to start with the dihedral groups since you mention being new to group theory. They also note that your group $mathbbZ_6 times mathbbZ_4$ is abelian (i.e. given any two elements $x,y$ in your group you have $xy = yx$). This means that the order in which you apply your operations doesn't matter. In the dihedral group, and most symmetry groups, it does matter, and the groups are non-abelian. You can check this easily yourself: draw a square and label the corners $1,2,3$ and $4$. Then reflect it about the $y$-axis and rotate it $90^circ$ clockwise and note where the labels end up. Then start from the first square and rotate then reflect and see that the labels end up in different places.






share|cite|improve this answer


















  • 1




    The symmetries of a cube is not given by a dihedral group, but by something more complicated. See here for more details (a cube is dual to an octahedron so has the same set of symmetries).
    – user1729
    Aug 23 at 9:12






  • 1




    Also, its worth pointing out that the group in the question, $mathbbZ_6timesmathbbZ_4$, is abelian while the symmetry groups of $n$-gons, $ngeq3$, and of the cube are all non-abelian.
    – user1729
    Aug 23 at 9:14










  • @user1729 thank-you, you are quite right! I will update the answer to note that it's the square, not the cube that has the dihedral group. Plus I hadn't thought about Abelian-ness, but that's a very good point too
    – postmortes
    Aug 23 at 10:29

















up vote
1
down vote













I don't see how faces and edges are related here. I makes sense, but it doesn't seem to have anything to do with the cube. For instance, you could take four types of pens in six different colors. Then define addition of pen i with color j to be like the group $mathbbZ_6 times mathbbZ_4$. ... If it helps you think about the definition of the group, then that is fine, but often when one defines a group based on a geometric object like a cube, one is thinking about some sort of geometry related to it, but sometimes it helps to visualize things better.



It might be better to think of the group as a bike lock with two numbers to pick from, the first number to pick from is from 0 to 5 and the second number to pick from is from 0 to 3 and how you add two combinations of numbers is to rotate them the number of times in each slot. This naturally captures the idea of the group and the definition of the group is deeply tied into the geometry/movement of the object used to explain it.



For example, this Bike lock could be used to $mathbbZ_10 times mathbbZ_10 times mathbbZ_10 times mathbbZ_10$.






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  • Yes, that is correct. ... I will change it now.
    – 4-ier
    Aug 23 at 6:48










  • Thank you @Arthur
    – 4-ier
    Aug 23 at 6:48










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3 Answers
3






active

oldest

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3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
2
down vote



accepted










While your approach isn't invalid, I want you to ask back. What can be achieved via your definition of cube group? Why is it useful? Do not take this as an offense because I didn't mean it. They are questions needed for mathematics.



You defined how to match an element of a cube to an element of $mathbbZ_6 times mathbbZ_4$. There's basically no problem. Also, you said $(4,1) + (3, 0) = (1, 1)$ on $mathbbZ_6 times mathbbZ_4$. That's correct. Then you linked it to an operation on cube: adding face $4$ edge $1$ to face $3$ edge $0$, to get face $1$ edge $1$. But what's meaning on that? You can address meaningful explanation of that operation? Maybe a geometrical meaning?



If you can present such a meaning: that will be very nice. Maybe the group will be named after you. Or unfortunately somebody else already did it.



For example describing structure of cube, is a symmetry group of the cube. This is a group about cube, or more strictly, rotations and reflections of the cube(and octahedron). An element of the group means one exact rotation and reflection(that means, kinda preserving faces', edges' and vertices' locations) of the cube. They are total 24 ways to rotate, and 2 ways to reflect. Also, group operation(multiplying two elements of group) means composition of such rotations and reflections. So it addresses something, namely, symmetric structure of a cube.



So you defined a cube group structure of your own. That's great. The next thing is, to find it useful. Maybe you cannot find it, but it's OK. That's how we do mathematics. Just keep trying.






share|cite|improve this answer




















  • I dislike the line "Maybe the group will be named after you. Or unfortunately somebody else already did it." This is for a variety of reasons, but mostly: It is far too optimistic, and is not a good motivation for doing something.
    – user1729
    Aug 24 at 8:59











  • @user1729 good point. Maybe I'll think about it myself.
    – Jeong Jinmyeong
    Aug 24 at 23:19














up vote
2
down vote



accepted










While your approach isn't invalid, I want you to ask back. What can be achieved via your definition of cube group? Why is it useful? Do not take this as an offense because I didn't mean it. They are questions needed for mathematics.



You defined how to match an element of a cube to an element of $mathbbZ_6 times mathbbZ_4$. There's basically no problem. Also, you said $(4,1) + (3, 0) = (1, 1)$ on $mathbbZ_6 times mathbbZ_4$. That's correct. Then you linked it to an operation on cube: adding face $4$ edge $1$ to face $3$ edge $0$, to get face $1$ edge $1$. But what's meaning on that? You can address meaningful explanation of that operation? Maybe a geometrical meaning?



If you can present such a meaning: that will be very nice. Maybe the group will be named after you. Or unfortunately somebody else already did it.



For example describing structure of cube, is a symmetry group of the cube. This is a group about cube, or more strictly, rotations and reflections of the cube(and octahedron). An element of the group means one exact rotation and reflection(that means, kinda preserving faces', edges' and vertices' locations) of the cube. They are total 24 ways to rotate, and 2 ways to reflect. Also, group operation(multiplying two elements of group) means composition of such rotations and reflections. So it addresses something, namely, symmetric structure of a cube.



So you defined a cube group structure of your own. That's great. The next thing is, to find it useful. Maybe you cannot find it, but it's OK. That's how we do mathematics. Just keep trying.






share|cite|improve this answer




















  • I dislike the line "Maybe the group will be named after you. Or unfortunately somebody else already did it." This is for a variety of reasons, but mostly: It is far too optimistic, and is not a good motivation for doing something.
    – user1729
    Aug 24 at 8:59











  • @user1729 good point. Maybe I'll think about it myself.
    – Jeong Jinmyeong
    Aug 24 at 23:19












up vote
2
down vote



accepted







up vote
2
down vote



accepted






While your approach isn't invalid, I want you to ask back. What can be achieved via your definition of cube group? Why is it useful? Do not take this as an offense because I didn't mean it. They are questions needed for mathematics.



You defined how to match an element of a cube to an element of $mathbbZ_6 times mathbbZ_4$. There's basically no problem. Also, you said $(4,1) + (3, 0) = (1, 1)$ on $mathbbZ_6 times mathbbZ_4$. That's correct. Then you linked it to an operation on cube: adding face $4$ edge $1$ to face $3$ edge $0$, to get face $1$ edge $1$. But what's meaning on that? You can address meaningful explanation of that operation? Maybe a geometrical meaning?



If you can present such a meaning: that will be very nice. Maybe the group will be named after you. Or unfortunately somebody else already did it.



For example describing structure of cube, is a symmetry group of the cube. This is a group about cube, or more strictly, rotations and reflections of the cube(and octahedron). An element of the group means one exact rotation and reflection(that means, kinda preserving faces', edges' and vertices' locations) of the cube. They are total 24 ways to rotate, and 2 ways to reflect. Also, group operation(multiplying two elements of group) means composition of such rotations and reflections. So it addresses something, namely, symmetric structure of a cube.



So you defined a cube group structure of your own. That's great. The next thing is, to find it useful. Maybe you cannot find it, but it's OK. That's how we do mathematics. Just keep trying.






share|cite|improve this answer












While your approach isn't invalid, I want you to ask back. What can be achieved via your definition of cube group? Why is it useful? Do not take this as an offense because I didn't mean it. They are questions needed for mathematics.



You defined how to match an element of a cube to an element of $mathbbZ_6 times mathbbZ_4$. There's basically no problem. Also, you said $(4,1) + (3, 0) = (1, 1)$ on $mathbbZ_6 times mathbbZ_4$. That's correct. Then you linked it to an operation on cube: adding face $4$ edge $1$ to face $3$ edge $0$, to get face $1$ edge $1$. But what's meaning on that? You can address meaningful explanation of that operation? Maybe a geometrical meaning?



If you can present such a meaning: that will be very nice. Maybe the group will be named after you. Or unfortunately somebody else already did it.



For example describing structure of cube, is a symmetry group of the cube. This is a group about cube, or more strictly, rotations and reflections of the cube(and octahedron). An element of the group means one exact rotation and reflection(that means, kinda preserving faces', edges' and vertices' locations) of the cube. They are total 24 ways to rotate, and 2 ways to reflect. Also, group operation(multiplying two elements of group) means composition of such rotations and reflections. So it addresses something, namely, symmetric structure of a cube.



So you defined a cube group structure of your own. That's great. The next thing is, to find it useful. Maybe you cannot find it, but it's OK. That's how we do mathematics. Just keep trying.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Aug 23 at 6:42









Jeong Jinmyeong

34617




34617











  • I dislike the line "Maybe the group will be named after you. Or unfortunately somebody else already did it." This is for a variety of reasons, but mostly: It is far too optimistic, and is not a good motivation for doing something.
    – user1729
    Aug 24 at 8:59











  • @user1729 good point. Maybe I'll think about it myself.
    – Jeong Jinmyeong
    Aug 24 at 23:19
















  • I dislike the line "Maybe the group will be named after you. Or unfortunately somebody else already did it." This is for a variety of reasons, but mostly: It is far too optimistic, and is not a good motivation for doing something.
    – user1729
    Aug 24 at 8:59











  • @user1729 good point. Maybe I'll think about it myself.
    – Jeong Jinmyeong
    Aug 24 at 23:19















I dislike the line "Maybe the group will be named after you. Or unfortunately somebody else already did it." This is for a variety of reasons, but mostly: It is far too optimistic, and is not a good motivation for doing something.
– user1729
Aug 24 at 8:59





I dislike the line "Maybe the group will be named after you. Or unfortunately somebody else already did it." This is for a variety of reasons, but mostly: It is far too optimistic, and is not a good motivation for doing something.
– user1729
Aug 24 at 8:59













@user1729 good point. Maybe I'll think about it myself.
– Jeong Jinmyeong
Aug 24 at 23:19




@user1729 good point. Maybe I'll think about it myself.
– Jeong Jinmyeong
Aug 24 at 23:19










up vote
4
down vote













A possible problem with treating the cube as $mathbbZ_6 times mathbbZ_4$ is making sense of what that allows you to do. In your question, at the end, you note that you could add "face $4$, edge $1$" to "face $3$, edge $0$" and get "face $1$, edge $1$". But what does that actually mean?



If we label our cube so that each face has a number like a traditional die then opposite pairs of faces add to 7 (so 1 is opposite 6, 2 is opposite 5 and 3 is opposite 4). So your representation allows you to "add" edges of faces that aren't connected to each other. Is that useful? (It might be, by the way, but I can't think of a use at the moment.)



So: either we need to add some more structure to your representation in order to "forbid" additions that don't make sense, or we need to change the representation in order to ensure that the operations we can perform all make sense. Both are valid approaches: we could say for example that we can only "add" elements of the group together if they share a common edge or a common face. You'll come across this in your course when you start talking about generators and relations which essentially say: here are specific relationships, and these are the permissible operations on them.



The dihedral group $mathbbD_n$ was mentioned by a commentor above: this takes the symmetries of the square as the essential elements: rotations and reflections, and then specifies how they fit together. If $a$ is rotation by $90^circ$ about the $y$-axis and $b$ is reflection in the $y$-axis then you can state that $a^4 = 1$ since four rotations take you back to where you start, and $b^2 =1$ since two reflections take you back to where you started.



As @user1729 notes in the comments, the symmetry group of the cube is more complicated, and you'll find it easier to start with the dihedral groups since you mention being new to group theory. They also note that your group $mathbbZ_6 times mathbbZ_4$ is abelian (i.e. given any two elements $x,y$ in your group you have $xy = yx$). This means that the order in which you apply your operations doesn't matter. In the dihedral group, and most symmetry groups, it does matter, and the groups are non-abelian. You can check this easily yourself: draw a square and label the corners $1,2,3$ and $4$. Then reflect it about the $y$-axis and rotate it $90^circ$ clockwise and note where the labels end up. Then start from the first square and rotate then reflect and see that the labels end up in different places.






share|cite|improve this answer


















  • 1




    The symmetries of a cube is not given by a dihedral group, but by something more complicated. See here for more details (a cube is dual to an octahedron so has the same set of symmetries).
    – user1729
    Aug 23 at 9:12






  • 1




    Also, its worth pointing out that the group in the question, $mathbbZ_6timesmathbbZ_4$, is abelian while the symmetry groups of $n$-gons, $ngeq3$, and of the cube are all non-abelian.
    – user1729
    Aug 23 at 9:14










  • @user1729 thank-you, you are quite right! I will update the answer to note that it's the square, not the cube that has the dihedral group. Plus I hadn't thought about Abelian-ness, but that's a very good point too
    – postmortes
    Aug 23 at 10:29














up vote
4
down vote













A possible problem with treating the cube as $mathbbZ_6 times mathbbZ_4$ is making sense of what that allows you to do. In your question, at the end, you note that you could add "face $4$, edge $1$" to "face $3$, edge $0$" and get "face $1$, edge $1$". But what does that actually mean?



If we label our cube so that each face has a number like a traditional die then opposite pairs of faces add to 7 (so 1 is opposite 6, 2 is opposite 5 and 3 is opposite 4). So your representation allows you to "add" edges of faces that aren't connected to each other. Is that useful? (It might be, by the way, but I can't think of a use at the moment.)



So: either we need to add some more structure to your representation in order to "forbid" additions that don't make sense, or we need to change the representation in order to ensure that the operations we can perform all make sense. Both are valid approaches: we could say for example that we can only "add" elements of the group together if they share a common edge or a common face. You'll come across this in your course when you start talking about generators and relations which essentially say: here are specific relationships, and these are the permissible operations on them.



The dihedral group $mathbbD_n$ was mentioned by a commentor above: this takes the symmetries of the square as the essential elements: rotations and reflections, and then specifies how they fit together. If $a$ is rotation by $90^circ$ about the $y$-axis and $b$ is reflection in the $y$-axis then you can state that $a^4 = 1$ since four rotations take you back to where you start, and $b^2 =1$ since two reflections take you back to where you started.



As @user1729 notes in the comments, the symmetry group of the cube is more complicated, and you'll find it easier to start with the dihedral groups since you mention being new to group theory. They also note that your group $mathbbZ_6 times mathbbZ_4$ is abelian (i.e. given any two elements $x,y$ in your group you have $xy = yx$). This means that the order in which you apply your operations doesn't matter. In the dihedral group, and most symmetry groups, it does matter, and the groups are non-abelian. You can check this easily yourself: draw a square and label the corners $1,2,3$ and $4$. Then reflect it about the $y$-axis and rotate it $90^circ$ clockwise and note where the labels end up. Then start from the first square and rotate then reflect and see that the labels end up in different places.






share|cite|improve this answer


















  • 1




    The symmetries of a cube is not given by a dihedral group, but by something more complicated. See here for more details (a cube is dual to an octahedron so has the same set of symmetries).
    – user1729
    Aug 23 at 9:12






  • 1




    Also, its worth pointing out that the group in the question, $mathbbZ_6timesmathbbZ_4$, is abelian while the symmetry groups of $n$-gons, $ngeq3$, and of the cube are all non-abelian.
    – user1729
    Aug 23 at 9:14










  • @user1729 thank-you, you are quite right! I will update the answer to note that it's the square, not the cube that has the dihedral group. Plus I hadn't thought about Abelian-ness, but that's a very good point too
    – postmortes
    Aug 23 at 10:29












up vote
4
down vote










up vote
4
down vote









A possible problem with treating the cube as $mathbbZ_6 times mathbbZ_4$ is making sense of what that allows you to do. In your question, at the end, you note that you could add "face $4$, edge $1$" to "face $3$, edge $0$" and get "face $1$, edge $1$". But what does that actually mean?



If we label our cube so that each face has a number like a traditional die then opposite pairs of faces add to 7 (so 1 is opposite 6, 2 is opposite 5 and 3 is opposite 4). So your representation allows you to "add" edges of faces that aren't connected to each other. Is that useful? (It might be, by the way, but I can't think of a use at the moment.)



So: either we need to add some more structure to your representation in order to "forbid" additions that don't make sense, or we need to change the representation in order to ensure that the operations we can perform all make sense. Both are valid approaches: we could say for example that we can only "add" elements of the group together if they share a common edge or a common face. You'll come across this in your course when you start talking about generators and relations which essentially say: here are specific relationships, and these are the permissible operations on them.



The dihedral group $mathbbD_n$ was mentioned by a commentor above: this takes the symmetries of the square as the essential elements: rotations and reflections, and then specifies how they fit together. If $a$ is rotation by $90^circ$ about the $y$-axis and $b$ is reflection in the $y$-axis then you can state that $a^4 = 1$ since four rotations take you back to where you start, and $b^2 =1$ since two reflections take you back to where you started.



As @user1729 notes in the comments, the symmetry group of the cube is more complicated, and you'll find it easier to start with the dihedral groups since you mention being new to group theory. They also note that your group $mathbbZ_6 times mathbbZ_4$ is abelian (i.e. given any two elements $x,y$ in your group you have $xy = yx$). This means that the order in which you apply your operations doesn't matter. In the dihedral group, and most symmetry groups, it does matter, and the groups are non-abelian. You can check this easily yourself: draw a square and label the corners $1,2,3$ and $4$. Then reflect it about the $y$-axis and rotate it $90^circ$ clockwise and note where the labels end up. Then start from the first square and rotate then reflect and see that the labels end up in different places.






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A possible problem with treating the cube as $mathbbZ_6 times mathbbZ_4$ is making sense of what that allows you to do. In your question, at the end, you note that you could add "face $4$, edge $1$" to "face $3$, edge $0$" and get "face $1$, edge $1$". But what does that actually mean?



If we label our cube so that each face has a number like a traditional die then opposite pairs of faces add to 7 (so 1 is opposite 6, 2 is opposite 5 and 3 is opposite 4). So your representation allows you to "add" edges of faces that aren't connected to each other. Is that useful? (It might be, by the way, but I can't think of a use at the moment.)



So: either we need to add some more structure to your representation in order to "forbid" additions that don't make sense, or we need to change the representation in order to ensure that the operations we can perform all make sense. Both are valid approaches: we could say for example that we can only "add" elements of the group together if they share a common edge or a common face. You'll come across this in your course when you start talking about generators and relations which essentially say: here are specific relationships, and these are the permissible operations on them.



The dihedral group $mathbbD_n$ was mentioned by a commentor above: this takes the symmetries of the square as the essential elements: rotations and reflections, and then specifies how they fit together. If $a$ is rotation by $90^circ$ about the $y$-axis and $b$ is reflection in the $y$-axis then you can state that $a^4 = 1$ since four rotations take you back to where you start, and $b^2 =1$ since two reflections take you back to where you started.



As @user1729 notes in the comments, the symmetry group of the cube is more complicated, and you'll find it easier to start with the dihedral groups since you mention being new to group theory. They also note that your group $mathbbZ_6 times mathbbZ_4$ is abelian (i.e. given any two elements $x,y$ in your group you have $xy = yx$). This means that the order in which you apply your operations doesn't matter. In the dihedral group, and most symmetry groups, it does matter, and the groups are non-abelian. You can check this easily yourself: draw a square and label the corners $1,2,3$ and $4$. Then reflect it about the $y$-axis and rotate it $90^circ$ clockwise and note where the labels end up. Then start from the first square and rotate then reflect and see that the labels end up in different places.







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share|cite|improve this answer



share|cite|improve this answer








edited Aug 23 at 12:32

























answered Aug 23 at 6:51









postmortes

1,56311016




1,56311016







  • 1




    The symmetries of a cube is not given by a dihedral group, but by something more complicated. See here for more details (a cube is dual to an octahedron so has the same set of symmetries).
    – user1729
    Aug 23 at 9:12






  • 1




    Also, its worth pointing out that the group in the question, $mathbbZ_6timesmathbbZ_4$, is abelian while the symmetry groups of $n$-gons, $ngeq3$, and of the cube are all non-abelian.
    – user1729
    Aug 23 at 9:14










  • @user1729 thank-you, you are quite right! I will update the answer to note that it's the square, not the cube that has the dihedral group. Plus I hadn't thought about Abelian-ness, but that's a very good point too
    – postmortes
    Aug 23 at 10:29












  • 1




    The symmetries of a cube is not given by a dihedral group, but by something more complicated. See here for more details (a cube is dual to an octahedron so has the same set of symmetries).
    – user1729
    Aug 23 at 9:12






  • 1




    Also, its worth pointing out that the group in the question, $mathbbZ_6timesmathbbZ_4$, is abelian while the symmetry groups of $n$-gons, $ngeq3$, and of the cube are all non-abelian.
    – user1729
    Aug 23 at 9:14










  • @user1729 thank-you, you are quite right! I will update the answer to note that it's the square, not the cube that has the dihedral group. Plus I hadn't thought about Abelian-ness, but that's a very good point too
    – postmortes
    Aug 23 at 10:29







1




1




The symmetries of a cube is not given by a dihedral group, but by something more complicated. See here for more details (a cube is dual to an octahedron so has the same set of symmetries).
– user1729
Aug 23 at 9:12




The symmetries of a cube is not given by a dihedral group, but by something more complicated. See here for more details (a cube is dual to an octahedron so has the same set of symmetries).
– user1729
Aug 23 at 9:12




1




1




Also, its worth pointing out that the group in the question, $mathbbZ_6timesmathbbZ_4$, is abelian while the symmetry groups of $n$-gons, $ngeq3$, and of the cube are all non-abelian.
– user1729
Aug 23 at 9:14




Also, its worth pointing out that the group in the question, $mathbbZ_6timesmathbbZ_4$, is abelian while the symmetry groups of $n$-gons, $ngeq3$, and of the cube are all non-abelian.
– user1729
Aug 23 at 9:14












@user1729 thank-you, you are quite right! I will update the answer to note that it's the square, not the cube that has the dihedral group. Plus I hadn't thought about Abelian-ness, but that's a very good point too
– postmortes
Aug 23 at 10:29




@user1729 thank-you, you are quite right! I will update the answer to note that it's the square, not the cube that has the dihedral group. Plus I hadn't thought about Abelian-ness, but that's a very good point too
– postmortes
Aug 23 at 10:29










up vote
1
down vote













I don't see how faces and edges are related here. I makes sense, but it doesn't seem to have anything to do with the cube. For instance, you could take four types of pens in six different colors. Then define addition of pen i with color j to be like the group $mathbbZ_6 times mathbbZ_4$. ... If it helps you think about the definition of the group, then that is fine, but often when one defines a group based on a geometric object like a cube, one is thinking about some sort of geometry related to it, but sometimes it helps to visualize things better.



It might be better to think of the group as a bike lock with two numbers to pick from, the first number to pick from is from 0 to 5 and the second number to pick from is from 0 to 3 and how you add two combinations of numbers is to rotate them the number of times in each slot. This naturally captures the idea of the group and the definition of the group is deeply tied into the geometry/movement of the object used to explain it.



For example, this Bike lock could be used to $mathbbZ_10 times mathbbZ_10 times mathbbZ_10 times mathbbZ_10$.






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  • Yes, that is correct. ... I will change it now.
    – 4-ier
    Aug 23 at 6:48










  • Thank you @Arthur
    – 4-ier
    Aug 23 at 6:48














up vote
1
down vote













I don't see how faces and edges are related here. I makes sense, but it doesn't seem to have anything to do with the cube. For instance, you could take four types of pens in six different colors. Then define addition of pen i with color j to be like the group $mathbbZ_6 times mathbbZ_4$. ... If it helps you think about the definition of the group, then that is fine, but often when one defines a group based on a geometric object like a cube, one is thinking about some sort of geometry related to it, but sometimes it helps to visualize things better.



It might be better to think of the group as a bike lock with two numbers to pick from, the first number to pick from is from 0 to 5 and the second number to pick from is from 0 to 3 and how you add two combinations of numbers is to rotate them the number of times in each slot. This naturally captures the idea of the group and the definition of the group is deeply tied into the geometry/movement of the object used to explain it.



For example, this Bike lock could be used to $mathbbZ_10 times mathbbZ_10 times mathbbZ_10 times mathbbZ_10$.






share|cite|improve this answer






















  • Yes, that is correct. ... I will change it now.
    – 4-ier
    Aug 23 at 6:48










  • Thank you @Arthur
    – 4-ier
    Aug 23 at 6:48












up vote
1
down vote










up vote
1
down vote









I don't see how faces and edges are related here. I makes sense, but it doesn't seem to have anything to do with the cube. For instance, you could take four types of pens in six different colors. Then define addition of pen i with color j to be like the group $mathbbZ_6 times mathbbZ_4$. ... If it helps you think about the definition of the group, then that is fine, but often when one defines a group based on a geometric object like a cube, one is thinking about some sort of geometry related to it, but sometimes it helps to visualize things better.



It might be better to think of the group as a bike lock with two numbers to pick from, the first number to pick from is from 0 to 5 and the second number to pick from is from 0 to 3 and how you add two combinations of numbers is to rotate them the number of times in each slot. This naturally captures the idea of the group and the definition of the group is deeply tied into the geometry/movement of the object used to explain it.



For example, this Bike lock could be used to $mathbbZ_10 times mathbbZ_10 times mathbbZ_10 times mathbbZ_10$.






share|cite|improve this answer














I don't see how faces and edges are related here. I makes sense, but it doesn't seem to have anything to do with the cube. For instance, you could take four types of pens in six different colors. Then define addition of pen i with color j to be like the group $mathbbZ_6 times mathbbZ_4$. ... If it helps you think about the definition of the group, then that is fine, but often when one defines a group based on a geometric object like a cube, one is thinking about some sort of geometry related to it, but sometimes it helps to visualize things better.



It might be better to think of the group as a bike lock with two numbers to pick from, the first number to pick from is from 0 to 5 and the second number to pick from is from 0 to 3 and how you add two combinations of numbers is to rotate them the number of times in each slot. This naturally captures the idea of the group and the definition of the group is deeply tied into the geometry/movement of the object used to explain it.



For example, this Bike lock could be used to $mathbbZ_10 times mathbbZ_10 times mathbbZ_10 times mathbbZ_10$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Aug 23 at 6:48

























answered Aug 23 at 6:28









4-ier

5989




5989











  • Yes, that is correct. ... I will change it now.
    – 4-ier
    Aug 23 at 6:48










  • Thank you @Arthur
    – 4-ier
    Aug 23 at 6:48
















  • Yes, that is correct. ... I will change it now.
    – 4-ier
    Aug 23 at 6:48










  • Thank you @Arthur
    – 4-ier
    Aug 23 at 6:48















Yes, that is correct. ... I will change it now.
– 4-ier
Aug 23 at 6:48




Yes, that is correct. ... I will change it now.
– 4-ier
Aug 23 at 6:48












Thank you @Arthur
– 4-ier
Aug 23 at 6:48




Thank you @Arthur
– 4-ier
Aug 23 at 6:48

















 

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