How to get generators for a finite point group?
How do I find generators of Octahedral group when viewing it as a subgroup of $S_8$?
Update: Alex Bartel's suggestion gives the following permutations and Cayley graph
Update 2: second suggestion by Alex gives following Cayley graph, which seems to be isomorphic to the Cayley graph of symmetric group given elsewhere, check!
group-theory finite-groups
edited Dec 29 '18 at 3:17
the_fox
2,47211431
asked Dec 21 '10 at 0:44
Yaroslav BulatovYaroslav Bulatov
1,86911526
add a comment |
How do I find generators of Octahedral group when viewing it as a subgroup of $S_8$?
Update: Alex Bartel's suggestion gives the following permutations and Cayley graph
Update 2: second suggestion by Alex gives following Cayley graph, which seems to be isomorphic to the Cayley graph of symmetric group given elsewhere, check!
group-theory finite-groups
edited Dec 29 '18 at 3:17
the_fox
2,47211431
asked Dec 21 '10 at 0:44
Yaroslav BulatovYaroslav Bulatov
1,86911526
I don't understand your question, but here is some random information: it is well known that the group of rotations of the cube is $S_4$, the full symmetry group on 4 letters. This can be seen concretely by letting the rotations permute the 4 long diagonals of the cube. Of course, by labelling the vertices and writing down what each rotation does to them, you can realise it as a subgroup of $S_8$, or by doing the same with the faces you get a subgroup of $S_6$, and so on.
– Alex B.
Dec 21 '10 at 1:54
I also don't understand your question. Can you say more precisely what you want?
– Qiaochu Yuan
Dec 21 '10 at 1:59
reworded for clarity
– Yaroslav Bulatov
Dec 21 '10 at 2:06
For the reworded question: just label the vertices, rotate and see how the vertices are permuted. It's enough to do it for a set of generators of $S_4$ (which is the group of rotations), so e.g. do it for a rotation of order 4 and one of order 3. What exactly are you having difficulties with?
– Alex B.
Dec 21 '10 at 2:17
Is it possible to do it without referring to geometry of the cube? IE, S4 has generators (1,2) and (1,2,3,4), what would be the corresponding generators when viewing it as a subgroup of S8?
– Yaroslav Bulatov
Dec 21 '10 at 2:22
add a comment |
How do I find generators of Octahedral group when viewing it as a subgroup of $S_8$?
Update: Alex Bartel's suggestion gives the following permutations and Cayley graph
Update 2: second suggestion by Alex gives following Cayley graph, which seems to be isomorphic to the Cayley graph of symmetric group given elsewhere, check!
group-theory finite-groups
edited Dec 29 '18 at 3:17
the_fox
2,47211431
asked Dec 21 '10 at 0:44
Yaroslav BulatovYaroslav Bulatov
1,86911526
How do I find generators of Octahedral group when viewing it as a subgroup of $S_8$?
Update: Alex Bartel's suggestion gives the following permutations and Cayley graph
Update 2: second suggestion by Alex gives following Cayley graph, which seems to be isomorphic to the Cayley graph of symmetric group given elsewhere, check!
group-theory finite-groups
group-theory finite-groups
edited Dec 29 '18 at 3:17
the_fox
2,47211431
asked Dec 21 '10 at 0:44
Yaroslav BulatovYaroslav Bulatov
1,86911526
edited Dec 29 '18 at 3:17
the_fox
2,47211431
asked Dec 21 '10 at 0:44
Yaroslav BulatovYaroslav Bulatov
1,86911526
edited Dec 29 '18 at 3:17
the_fox
2,47211431
edited Dec 29 '18 at 3:17
the_fox
2,47211431
edited Dec 29 '18 at 3:17
the_fox
2,47211431
2,47211431
asked Dec 21 '10 at 0:44
Yaroslav BulatovYaroslav Bulatov
1,86911526
asked Dec 21 '10 at 0:44
Yaroslav BulatovYaroslav Bulatov
1,86911526
asked Dec 21 '10 at 0:44
Yaroslav BulatovYaroslav Bulatov
1,86911526
1,86911526
I don't understand your question, but here is some random information: it is well known that the group of rotations of the cube is $S_4$, the full symmetry group on 4 letters. This can be seen concretely by letting the rotations permute the 4 long diagonals of the cube. Of course, by labelling the vertices and writing down what each rotation does to them, you can realise it as a subgroup of $S_8$, or by doing the same with the faces you get a subgroup of $S_6$, and so on.
– Alex B.
Dec 21 '10 at 1:54
I also don't understand your question. Can you say more precisely what you want?
– Qiaochu Yuan
Dec 21 '10 at 1:59
reworded for clarity
– Yaroslav Bulatov
Dec 21 '10 at 2:06
For the reworded question: just label the vertices, rotate and see how the vertices are permuted. It's enough to do it for a set of generators of $S_4$ (which is the group of rotations), so e.g. do it for a rotation of order 4 and one of order 3. What exactly are you having difficulties with?
– Alex B.
Dec 21 '10 at 2:17
Is it possible to do it without referring to geometry of the cube? IE, S4 has generators (1,2) and (1,2,3,4), what would be the corresponding generators when viewing it as a subgroup of S8?
– Yaroslav Bulatov
Dec 21 '10 at 2:22
add a comment |
I don't understand your question, but here is some random information: it is well known that the group of rotations of the cube is $S_4$, the full symmetry group on 4 letters. This can be seen concretely by letting the rotations permute the 4 long diagonals of the cube. Of course, by labelling the vertices and writing down what each rotation does to them, you can realise it as a subgroup of $S_8$, or by doing the same with the faces you get a subgroup of $S_6$, and so on.
– Alex B.
Dec 21 '10 at 1:54
I also don't understand your question. Can you say more precisely what you want?
– Qiaochu Yuan
Dec 21 '10 at 1:59
reworded for clarity
– Yaroslav Bulatov
Dec 21 '10 at 2:06
For the reworded question: just label the vertices, rotate and see how the vertices are permuted. It's enough to do it for a set of generators of $S_4$ (which is the group of rotations), so e.g. do it for a rotation of order 4 and one of order 3. What exactly are you having difficulties with?
– Alex B.
Dec 21 '10 at 2:17
Is it possible to do it without referring to geometry of the cube? IE, S4 has generators (1,2) and (1,2,3,4), what would be the corresponding generators when viewing it as a subgroup of S8?
– Yaroslav Bulatov
Dec 21 '10 at 2:22
I don't understand your question, but here is some random information: it is well known that the group of rotations of the cube is $S_4$, the full symmetry group on 4 letters. This can be seen concretely by letting the rotations permute the 4 long diagonals of the cube. Of course, by labelling the vertices and writing down what each rotation does to them, you can realise it as a subgroup of $S_8$, or by doing the same with the faces you get a subgroup of $S_6$, and so on.
– Alex B.
Dec 21 '10 at 1:54
I don't understand your question, but here is some random information: it is well known that the group of rotations of the cube is $S_4$, the full symmetry group on 4 letters. This can be seen concretely by letting the rotations permute the 4 long diagonals of the cube. Of course, by labelling the vertices and writing down what each rotation does to them, you can realise it as a subgroup of $S_8$, or by doing the same with the faces you get a subgroup of $S_6$, and so on.
– Alex B.
Dec 21 '10 at 1:54
I also don't understand your question. Can you say more precisely what you want?
– Qiaochu Yuan
Dec 21 '10 at 1:59
I also don't understand your question. Can you say more precisely what you want?
– Qiaochu Yuan
Dec 21 '10 at 1:59
reworded for clarity
– Yaroslav Bulatov
Dec 21 '10 at 2:06
reworded for clarity
– Yaroslav Bulatov
Dec 21 '10 at 2:06
For the reworded question: just label the vertices, rotate and see how the vertices are permuted. It's enough to do it for a set of generators of $S_4$ (which is the group of rotations), so e.g. do it for a rotation of order 4 and one of order 3. What exactly are you having difficulties with?
– Alex B.
Dec 21 '10 at 2:17
For the reworded question: just label the vertices, rotate and see how the vertices are permuted. It's enough to do it for a set of generators of $S_4$ (which is the group of rotations), so e.g. do it for a rotation of order 4 and one of order 3. What exactly are you having difficulties with?
– Alex B.
Dec 21 '10 at 2:17
Is it possible to do it without referring to geometry of the cube? IE, S4 has generators (1,2) and (1,2,3,4), what would be the corresponding generators when viewing it as a subgroup of S8?
– Yaroslav Bulatov
Dec 21 '10 at 2:22
Is it possible to do it without referring to geometry of the cube? IE, S4 has generators (1,2) and (1,2,3,4), what would be the corresponding generators when viewing it as a subgroup of S8?
– Yaroslav Bulatov
Dec 21 '10 at 2:22
add a comment |
1 Answer
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Observation number one: the group we are talking about is $S_4$.
Observation number two: $S_4$ is generated by any 4-cycle and any 3-cycle. Indeed, the group generated by them must be of size at least 12, and it cannot be $A_4$, since the 4-cycle is an odd permutation.
So now, just write down what a rotation of order 4 and a rotation of order 3 do to your vertices. E.g. if the vertices around one face are labelled by 1,...,4, and the vertices around the opposite face are 5,...,8, then a rotation of order 4 sits as (1 2 3 4)(5 6 7 8) inside $S_8$. I am sure you can write down yourself what a rotation about a long diagonal looks like.
Of course, you can do it for any other set of generators, but others might be harder to spot. As I wrote in a comment, the group of rotations is realised as $S_4$ by its action on the 4 long diagonals of the cube. This should enable you to write down any given element of $S_4$ in terms of its action on the vertices.
Edit: Since there still seems to be some confusion, I will add this for clarification: when I say that the rotations permute the diagonals, I regard a diagonal as a line segment, which is unchanged if you swap the two end points. In particular, a rotation about an axis that goes through the mid points of two edges (and of course through the centroid of the cube) doesn't fix any vertices, but it fixes two diagonals by swapping the end points, so it induces a 2-cycle in $S_4$. A rotation about a long diagonal of the cube fixes only that diagonal and induces a 3-cycle.
answered Dec 21 '10 at 2:22
Alex B.Alex B.
16.3k13567
I see...is there a way to express it as rotation of order 2 and rotation of order 4?
– Yaroslav Bulatov
Dec 21 '10 at 3:38
@Yaroslav Bulatov: Yes; in general, $S_n$ can be generated by the transpositions $(1,2)$, $(2,3)$, $(3,4),ldots,(n-1,n)$ (by conjugating appropriately, you get all transpositions of the form $(1,i)$, and from there you can get all transpositions). So you can instead start with $(1,2)$ and $(1,2,3,4)$; these two together can get you the permutations $(1,2)$, $(2,3)$, and $(3,4)$, so from there you can get all of $S_4$. So now you can write them down inside $S_8$ by doing much the same thing that Alex Bartel did above for the 4-cycle and a 3-cycle.
– Arturo Magidin
Dec 21 '10 at 4:44
@Arturo Magidin: I'm guessing 4-cycle is a rotation around center of the face, but what does 2-cycle correspond to?
– Yaroslav Bulatov
Dec 21 '10 at 4:50
@Yaroslav Bulatov: The $4$-cycle can be the same thing it was in the answer above (since it corresponds to the same element or $S_4$); so you can realize it as $(1,2,3,4)(5,6,7,8)$; that is what you suggest. For a $2$-cycle, you can either cheat and use your old representation to figure it out, or try to find some nice transposition. For example, how about rotation about a large diagonal of the cube (joining opposite vertices)? I don't know for sure that will work, but that's the kind of thing you can try.
– Arturo Magidin
Dec 21 '10 at 4:55
@Yaroslav A 2-cycle is a rotation in an axis that goes through the midpoints of two edges and through the centre of the cube. (check!)
– Alex B.
Dec 21 '10 at 5:02
|
show 2 more comments
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Observation number one: the group we are talking about is $S_4$.
Observation number two: $S_4$ is generated by any 4-cycle and any 3-cycle. Indeed, the group generated by them must be of size at least 12, and it cannot be $A_4$, since the 4-cycle is an odd permutation.
So now, just write down what a rotation of order 4 and a rotation of order 3 do to your vertices. E.g. if the vertices around one face are labelled by 1,...,4, and the vertices around the opposite face are 5,...,8, then a rotation of order 4 sits as (1 2 3 4)(5 6 7 8) inside $S_8$. I am sure you can write down yourself what a rotation about a long diagonal looks like.
Of course, you can do it for any other set of generators, but others might be harder to spot. As I wrote in a comment, the group of rotations is realised as $S_4$ by its action on the 4 long diagonals of the cube. This should enable you to write down any given element of $S_4$ in terms of its action on the vertices.
Edit: Since there still seems to be some confusion, I will add this for clarification: when I say that the rotations permute the diagonals, I regard a diagonal as a line segment, which is unchanged if you swap the two end points. In particular, a rotation about an axis that goes through the mid points of two edges (and of course through the centroid of the cube) doesn't fix any vertices, but it fixes two diagonals by swapping the end points, so it induces a 2-cycle in $S_4$. A rotation about a long diagonal of the cube fixes only that diagonal and induces a 3-cycle.
answered Dec 21 '10 at 2:22
Alex B.Alex B.
16.3k13567
I see...is there a way to express it as rotation of order 2 and rotation of order 4?
– Yaroslav Bulatov
Dec 21 '10 at 3:38
@Yaroslav Bulatov: Yes; in general, $S_n$ can be generated by the transpositions $(1,2)$, $(2,3)$, $(3,4),ldots,(n-1,n)$ (by conjugating appropriately, you get all transpositions of the form $(1,i)$, and from there you can get all transpositions). So you can instead start with $(1,2)$ and $(1,2,3,4)$; these two together can get you the permutations $(1,2)$, $(2,3)$, and $(3,4)$, so from there you can get all of $S_4$. So now you can write them down inside $S_8$ by doing much the same thing that Alex Bartel did above for the 4-cycle and a 3-cycle.
– Arturo Magidin
Dec 21 '10 at 4:44
@Arturo Magidin: I'm guessing 4-cycle is a rotation around center of the face, but what does 2-cycle correspond to?
– Yaroslav Bulatov
Dec 21 '10 at 4:50
@Yaroslav Bulatov: The $4$-cycle can be the same thing it was in the answer above (since it corresponds to the same element or $S_4$); so you can realize it as $(1,2,3,4)(5,6,7,8)$; that is what you suggest. For a $2$-cycle, you can either cheat and use your old representation to figure it out, or try to find some nice transposition. For example, how about rotation about a large diagonal of the cube (joining opposite vertices)? I don't know for sure that will work, but that's the kind of thing you can try.
– Arturo Magidin
Dec 21 '10 at 4:55
@Yaroslav A 2-cycle is a rotation in an axis that goes through the midpoints of two edges and through the centre of the cube. (check!)
– Alex B.
Dec 21 '10 at 5:02
|
show 2 more comments
Observation number one: the group we are talking about is $S_4$.
Observation number two: $S_4$ is generated by any 4-cycle and any 3-cycle. Indeed, the group generated by them must be of size at least 12, and it cannot be $A_4$, since the 4-cycle is an odd permutation.
So now, just write down what a rotation of order 4 and a rotation of order 3 do to your vertices. E.g. if the vertices around one face are labelled by 1,...,4, and the vertices around the opposite face are 5,...,8, then a rotation of order 4 sits as (1 2 3 4)(5 6 7 8) inside $S_8$. I am sure you can write down yourself what a rotation about a long diagonal looks like.
Of course, you can do it for any other set of generators, but others might be harder to spot. As I wrote in a comment, the group of rotations is realised as $S_4$ by its action on the 4 long diagonals of the cube. This should enable you to write down any given element of $S_4$ in terms of its action on the vertices.
Edit: Since there still seems to be some confusion, I will add this for clarification: when I say that the rotations permute the diagonals, I regard a diagonal as a line segment, which is unchanged if you swap the two end points. In particular, a rotation about an axis that goes through the mid points of two edges (and of course through the centroid of the cube) doesn't fix any vertices, but it fixes two diagonals by swapping the end points, so it induces a 2-cycle in $S_4$. A rotation about a long diagonal of the cube fixes only that diagonal and induces a 3-cycle.
answered Dec 21 '10 at 2:22
Alex B.Alex B.
16.3k13567
I see...is there a way to express it as rotation of order 2 and rotation of order 4?
– Yaroslav Bulatov
Dec 21 '10 at 3:38
@Yaroslav Bulatov: Yes; in general, $S_n$ can be generated by the transpositions $(1,2)$, $(2,3)$, $(3,4),ldots,(n-1,n)$ (by conjugating appropriately, you get all transpositions of the form $(1,i)$, and from there you can get all transpositions). So you can instead start with $(1,2)$ and $(1,2,3,4)$; these two together can get you the permutations $(1,2)$, $(2,3)$, and $(3,4)$, so from there you can get all of $S_4$. So now you can write them down inside $S_8$ by doing much the same thing that Alex Bartel did above for the 4-cycle and a 3-cycle.
– Arturo Magidin
Dec 21 '10 at 4:44
@Arturo Magidin: I'm guessing 4-cycle is a rotation around center of the face, but what does 2-cycle correspond to?
– Yaroslav Bulatov
Dec 21 '10 at 4:50
@Yaroslav Bulatov: The $4$-cycle can be the same thing it was in the answer above (since it corresponds to the same element or $S_4$); so you can realize it as $(1,2,3,4)(5,6,7,8)$; that is what you suggest. For a $2$-cycle, you can either cheat and use your old representation to figure it out, or try to find some nice transposition. For example, how about rotation about a large diagonal of the cube (joining opposite vertices)? I don't know for sure that will work, but that's the kind of thing you can try.
– Arturo Magidin
Dec 21 '10 at 4:55
@Yaroslav A 2-cycle is a rotation in an axis that goes through the midpoints of two edges and through the centre of the cube. (check!)
– Alex B.
Dec 21 '10 at 5:02
|
show 2 more comments
Observation number one: the group we are talking about is $S_4$.
Observation number two: $S_4$ is generated by any 4-cycle and any 3-cycle. Indeed, the group generated by them must be of size at least 12, and it cannot be $A_4$, since the 4-cycle is an odd permutation.
So now, just write down what a rotation of order 4 and a rotation of order 3 do to your vertices. E.g. if the vertices around one face are labelled by 1,...,4, and the vertices around the opposite face are 5,...,8, then a rotation of order 4 sits as (1 2 3 4)(5 6 7 8) inside $S_8$. I am sure you can write down yourself what a rotation about a long diagonal looks like.
Of course, you can do it for any other set of generators, but others might be harder to spot. As I wrote in a comment, the group of rotations is realised as $S_4$ by its action on the 4 long diagonals of the cube. This should enable you to write down any given element of $S_4$ in terms of its action on the vertices.
Edit: Since there still seems to be some confusion, I will add this for clarification: when I say that the rotations permute the diagonals, I regard a diagonal as a line segment, which is unchanged if you swap the two end points. In particular, a rotation about an axis that goes through the mid points of two edges (and of course through the centroid of the cube) doesn't fix any vertices, but it fixes two diagonals by swapping the end points, so it induces a 2-cycle in $S_4$. A rotation about a long diagonal of the cube fixes only that diagonal and induces a 3-cycle.
answered Dec 21 '10 at 2:22
Alex B.Alex B.
16.3k13567
Observation number one: the group we are talking about is $S_4$.
Observation number two: $S_4$ is generated by any 4-cycle and any 3-cycle. Indeed, the group generated by them must be of size at least 12, and it cannot be $A_4$, since the 4-cycle is an odd permutation.
So now, just write down what a rotation of order 4 and a rotation of order 3 do to your vertices. E.g. if the vertices around one face are labelled by 1,...,4, and the vertices around the opposite face are 5,...,8, then a rotation of order 4 sits as (1 2 3 4)(5 6 7 8) inside $S_8$. I am sure you can write down yourself what a rotation about a long diagonal looks like.
Of course, you can do it for any other set of generators, but others might be harder to spot. As I wrote in a comment, the group of rotations is realised as $S_4$ by its action on the 4 long diagonals of the cube. This should enable you to write down any given element of $S_4$ in terms of its action on the vertices.
Edit: Since there still seems to be some confusion, I will add this for clarification: when I say that the rotations permute the diagonals, I regard a diagonal as a line segment, which is unchanged if you swap the two end points. In particular, a rotation about an axis that goes through the mid points of two edges (and of course through the centroid of the cube) doesn't fix any vertices, but it fixes two diagonals by swapping the end points, so it induces a 2-cycle in $S_4$. A rotation about a long diagonal of the cube fixes only that diagonal and induces a 3-cycle.
answered Dec 21 '10 at 2:22
Alex B.Alex B.
16.3k13567
edited Dec 21 '10 at 5:13
answered Dec 21 '10 at 2:22
Alex B.Alex B.
16.3k13567
answered Dec 21 '10 at 2:22
Alex B.Alex B.
16.3k13567
answered Dec 21 '10 at 2:22
Alex B.Alex B.
16.3k13567
16.3k13567
I see...is there a way to express it as rotation of order 2 and rotation of order 4?
– Yaroslav Bulatov
Dec 21 '10 at 3:38
@Yaroslav Bulatov: Yes; in general, $S_n$ can be generated by the transpositions $(1,2)$, $(2,3)$, $(3,4),ldots,(n-1,n)$ (by conjugating appropriately, you get all transpositions of the form $(1,i)$, and from there you can get all transpositions). So you can instead start with $(1,2)$ and $(1,2,3,4)$; these two together can get you the permutations $(1,2)$, $(2,3)$, and $(3,4)$, so from there you can get all of $S_4$. So now you can write them down inside $S_8$ by doing much the same thing that Alex Bartel did above for the 4-cycle and a 3-cycle.
– Arturo Magidin
Dec 21 '10 at 4:44
@Arturo Magidin: I'm guessing 4-cycle is a rotation around center of the face, but what does 2-cycle correspond to?
– Yaroslav Bulatov
Dec 21 '10 at 4:50
@Yaroslav Bulatov: The $4$-cycle can be the same thing it was in the answer above (since it corresponds to the same element or $S_4$); so you can realize it as $(1,2,3,4)(5,6,7,8)$; that is what you suggest. For a $2$-cycle, you can either cheat and use your old representation to figure it out, or try to find some nice transposition. For example, how about rotation about a large diagonal of the cube (joining opposite vertices)? I don't know for sure that will work, but that's the kind of thing you can try.
– Arturo Magidin
Dec 21 '10 at 4:55
@Yaroslav A 2-cycle is a rotation in an axis that goes through the midpoints of two edges and through the centre of the cube. (check!)
– Alex B.
Dec 21 '10 at 5:02
|
show 2 more comments
I see...is there a way to express it as rotation of order 2 and rotation of order 4?
– Yaroslav Bulatov
Dec 21 '10 at 3:38
@Yaroslav Bulatov: Yes; in general, $S_n$ can be generated by the transpositions $(1,2)$, $(2,3)$, $(3,4),ldots,(n-1,n)$ (by conjugating appropriately, you get all transpositions of the form $(1,i)$, and from there you can get all transpositions). So you can instead start with $(1,2)$ and $(1,2,3,4)$; these two together can get you the permutations $(1,2)$, $(2,3)$, and $(3,4)$, so from there you can get all of $S_4$. So now you can write them down inside $S_8$ by doing much the same thing that Alex Bartel did above for the 4-cycle and a 3-cycle.
– Arturo Magidin
Dec 21 '10 at 4:44
@Arturo Magidin: I'm guessing 4-cycle is a rotation around center of the face, but what does 2-cycle correspond to?
– Yaroslav Bulatov
Dec 21 '10 at 4:50
@Yaroslav Bulatov: The $4$-cycle can be the same thing it was in the answer above (since it corresponds to the same element or $S_4$); so you can realize it as $(1,2,3,4)(5,6,7,8)$; that is what you suggest. For a $2$-cycle, you can either cheat and use your old representation to figure it out, or try to find some nice transposition. For example, how about rotation about a large diagonal of the cube (joining opposite vertices)? I don't know for sure that will work, but that's the kind of thing you can try.
– Arturo Magidin
Dec 21 '10 at 4:55
@Yaroslav A 2-cycle is a rotation in an axis that goes through the midpoints of two edges and through the centre of the cube. (check!)
– Alex B.
Dec 21 '10 at 5:02
I see...is there a way to express it as rotation of order 2 and rotation of order 4?
– Yaroslav Bulatov
Dec 21 '10 at 3:38
I see...is there a way to express it as rotation of order 2 and rotation of order 4?
– Yaroslav Bulatov
Dec 21 '10 at 3:38
@Yaroslav Bulatov: Yes; in general, $S_n$ can be generated by the transpositions $(1,2)$, $(2,3)$, $(3,4),ldots,(n-1,n)$ (by conjugating appropriately, you get all transpositions of the form $(1,i)$, and from there you can get all transpositions). So you can instead start with $(1,2)$ and $(1,2,3,4)$; these two together can get you the permutations $(1,2)$, $(2,3)$, and $(3,4)$, so from there you can get all of $S_4$. So now you can write them down inside $S_8$ by doing much the same thing that Alex Bartel did above for the 4-cycle and a 3-cycle.
– Arturo Magidin
Dec 21 '10 at 4:44
@Yaroslav Bulatov: Yes; in general, $S_n$ can be generated by the transpositions $(1,2)$, $(2,3)$, $(3,4),ldots,(n-1,n)$ (by conjugating appropriately, you get all transpositions of the form $(1,i)$, and from there you can get all transpositions). So you can instead start with $(1,2)$ and $(1,2,3,4)$; these two together can get you the permutations $(1,2)$, $(2,3)$, and $(3,4)$, so from there you can get all of $S_4$. So now you can write them down inside $S_8$ by doing much the same thing that Alex Bartel did above for the 4-cycle and a 3-cycle.
– Arturo Magidin
Dec 21 '10 at 4:44
@Arturo Magidin: I'm guessing 4-cycle is a rotation around center of the face, but what does 2-cycle correspond to?
– Yaroslav Bulatov
Dec 21 '10 at 4:50
@Arturo Magidin: I'm guessing 4-cycle is a rotation around center of the face, but what does 2-cycle correspond to?
– Yaroslav Bulatov
Dec 21 '10 at 4:50
@Yaroslav Bulatov: The $4$-cycle can be the same thing it was in the answer above (since it corresponds to the same element or $S_4$); so you can realize it as $(1,2,3,4)(5,6,7,8)$; that is what you suggest. For a $2$-cycle, you can either cheat and use your old representation to figure it out, or try to find some nice transposition. For example, how about rotation about a large diagonal of the cube (joining opposite vertices)? I don't know for sure that will work, but that's the kind of thing you can try.
– Arturo Magidin
Dec 21 '10 at 4:55
@Yaroslav Bulatov: The $4$-cycle can be the same thing it was in the answer above (since it corresponds to the same element or $S_4$); so you can realize it as $(1,2,3,4)(5,6,7,8)$; that is what you suggest. For a $2$-cycle, you can either cheat and use your old representation to figure it out, or try to find some nice transposition. For example, how about rotation about a large diagonal of the cube (joining opposite vertices)? I don't know for sure that will work, but that's the kind of thing you can try.
– Arturo Magidin
Dec 21 '10 at 4:55
@Yaroslav A 2-cycle is a rotation in an axis that goes through the midpoints of two edges and through the centre of the cube. (check!)
– Alex B.
Dec 21 '10 at 5:02
@Yaroslav A 2-cycle is a rotation in an axis that goes through the midpoints of two edges and through the centre of the cube. (check!)
– Alex B.
Dec 21 '10 at 5:02
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I don't understand your question, but here is some random information: it is well known that the group of rotations of the cube is $S_4$, the full symmetry group on 4 letters. This can be seen concretely by letting the rotations permute the 4 long diagonals of the cube. Of course, by labelling the vertices and writing down what each rotation does to them, you can realise it as a subgroup of $S_8$, or by doing the same with the faces you get a subgroup of $S_6$, and so on.
– Alex B.
Dec 21 '10 at 1:54
I also don't understand your question. Can you say more precisely what you want?
– Qiaochu Yuan
Dec 21 '10 at 1:59
reworded for clarity
– Yaroslav Bulatov
Dec 21 '10 at 2:06
For the reworded question: just label the vertices, rotate and see how the vertices are permuted. It's enough to do it for a set of generators of $S_4$ (which is the group of rotations), so e.g. do it for a rotation of order 4 and one of order 3. What exactly are you having difficulties with?
– Alex B.
Dec 21 '10 at 2:17
Is it possible to do it without referring to geometry of the cube? IE, S4 has generators (1,2) and (1,2,3,4), what would be the corresponding generators when viewing it as a subgroup of S8?
– Yaroslav Bulatov
Dec 21 '10 at 2:22