Why rotation of dodecahedron corresponds to an even permutation of inscribed five tetrahedra?Equidecomposability of a Cube into 6 Trirectangular TetrahedraSymmetry group of the dodecahedron and its subsetsCleverest construction of a dodecahedron / icosahedron?Rotation of a regular tetrahedronStudying the symmetry group of the dodecahedron by introducing axes.Platonic solids calculating the relations between the diameter of the sphere and the sides lengthsTetrahedron into similar tetrahedronAll the tetrahedra that can be inscribed in the unit sphere up to isometryDraw five tetrahedral in a drawing of 3,5Deduce that the symmetry group of the dodecahedron is a subgroup of $S_5$ of order 60.
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Why rotation of dodecahedron corresponds to an even permutation of inscribed five tetrahedra?
Equidecomposability of a Cube into 6 Trirectangular TetrahedraSymmetry group of the dodecahedron and its subsetsCleverest construction of a dodecahedron / icosahedron?Rotation of a regular tetrahedronStudying the symmetry group of the dodecahedron by introducing axes.Platonic solids calculating the relations between the diameter of the sphere and the sides lengthsTetrahedron into similar tetrahedronAll the tetrahedra that can be inscribed in the unit sphere up to isometryDraw five tetrahedral in a drawing of 3,5Deduce that the symmetry group of the dodecahedron is a subgroup of $S_5$ of order 60.
$begingroup$
I’m reading Arnold’s Abel’s Theorem in Problems and Solutions, where it says:
We now prove that the alternating group $A_5$ is not soluble. One of the
possible proofs uses the following construction. We inscribe in the
dodecahedron five regular tetrahedra, numbered by the numbers 1, 2, 3,
4 and 5 in such a way that to every rotation of the dodecahedron there
corresponds an even permutation of the tetrahedra, and that to
different rotations there correspond different permutations. So we
have defined an isomorphism between the group of rotations of the
dodecahedron and the group $A_5$ of the even permutations of degree 5. The
non-solubility of the group $A_5$ will thus follow from the
non-solubility of the group of rotations of the dodecahedron.
I’m a bit confused. Why rotation of dodecahedron corresponds to an even permutation of inscribed five tetrahedra?
The answer just mentioned
The tetrahedra are‚ for example‚ those with the vertices chosen in the
following way1: (1‚ 8‚ 14‚ 16)‚ (2‚ 9‚ 15‚ 17)‚ (3‚ 10‚ 11‚ 18)‚ (4‚
6‚ 12‚ 19)‚ (5‚ 7‚ 13‚ 20).
, regarding to the notation
The translator provided an image of the inscribed cube:
, and added some explanation
The inscription of the five Kepler cubes inside the dodecahedron helps
us to find the five tetrahedra.
The edges of the cubes are the diagonals of the dodecahedron faces.
Every pair of opposite vertices of the dodecahedron is a pair of two
opposite vertices of two Kepler cubes. Each cube has thus only one
pair of vertices in common with any one of the others. (Two Kepler
cubes — black and white — having two opposite vertices in common are
shown in the figure). In each cube one can inscribe two tetrahedra
(see Problems 126 and 127). Since each tetrahedron is defined by four
vertices of the cube‚ and any two Kepler cubes have only 2 vertices in
common‚ all tetrahedra inscribed in the Kepler cubes are distinct. So
there are in all 10 tetrahedra‚ two for every vertex of the
dodecahedron. Any two of such tetrahedra either have no vertices in
common‚ or they have only one vertex in common. Indeed‚ if two
vertices belonged to two tetrahedra‚ the edge of such tetrahedra
joining them should be the diagonal of a face of two different cubes‚
but we know that any two cubes have in common only opposite vertices.
There are two possible choices of 5 tetrahedra‚ without common
vertices‚ inside the dodecahedron: indeed‚ when we choose one of them‚
we have to reject the four tetrahedra having a vertex in common with
the chosen tetrahedron. The remaining tetrahedra are five. Amongst
them four tetrahedra have disjoint sets of vertices‚ whereas the
remaining tetrahedron has one vertex in common with each of the four
disjoint tetrahedra. The choice of the first tetrahedron thus forces
the choice of the others‚ so there are in all only two choices.
(Translator’s note)
, but I still couldn’t see why.
abstract-algebra geometry
asked Mar 12 at 11:50
athosathos
97411340
$endgroup$
add a comment |
$begingroup$
I’m reading Arnold’s Abel’s Theorem in Problems and Solutions, where it says:
We now prove that the alternating group $A_5$ is not soluble. One of the
possible proofs uses the following construction. We inscribe in the
dodecahedron five regular tetrahedra, numbered by the numbers 1, 2, 3,
4 and 5 in such a way that to every rotation of the dodecahedron there
corresponds an even permutation of the tetrahedra, and that to
different rotations there correspond different permutations. So we
have defined an isomorphism between the group of rotations of the
dodecahedron and the group $A_5$ of the even permutations of degree 5. The
non-solubility of the group $A_5$ will thus follow from the
non-solubility of the group of rotations of the dodecahedron.
I’m a bit confused. Why rotation of dodecahedron corresponds to an even permutation of inscribed five tetrahedra?
The answer just mentioned
The tetrahedra are‚ for example‚ those with the vertices chosen in the
following way1: (1‚ 8‚ 14‚ 16)‚ (2‚ 9‚ 15‚ 17)‚ (3‚ 10‚ 11‚ 18)‚ (4‚
6‚ 12‚ 19)‚ (5‚ 7‚ 13‚ 20).
, regarding to the notation
The translator provided an image of the inscribed cube:
, and added some explanation
The inscription of the five Kepler cubes inside the dodecahedron helps
us to find the five tetrahedra.
The edges of the cubes are the diagonals of the dodecahedron faces.
Every pair of opposite vertices of the dodecahedron is a pair of two
opposite vertices of two Kepler cubes. Each cube has thus only one
pair of vertices in common with any one of the others. (Two Kepler
cubes — black and white — having two opposite vertices in common are
shown in the figure). In each cube one can inscribe two tetrahedra
(see Problems 126 and 127). Since each tetrahedron is defined by four
vertices of the cube‚ and any two Kepler cubes have only 2 vertices in
common‚ all tetrahedra inscribed in the Kepler cubes are distinct. So
there are in all 10 tetrahedra‚ two for every vertex of the
dodecahedron. Any two of such tetrahedra either have no vertices in
common‚ or they have only one vertex in common. Indeed‚ if two
vertices belonged to two tetrahedra‚ the edge of such tetrahedra
joining them should be the diagonal of a face of two different cubes‚
but we know that any two cubes have in common only opposite vertices.
There are two possible choices of 5 tetrahedra‚ without common
vertices‚ inside the dodecahedron: indeed‚ when we choose one of them‚
we have to reject the four tetrahedra having a vertex in common with
the chosen tetrahedron. The remaining tetrahedra are five. Amongst
them four tetrahedra have disjoint sets of vertices‚ whereas the
remaining tetrahedron has one vertex in common with each of the four
disjoint tetrahedra. The choice of the first tetrahedron thus forces
the choice of the others‚ so there are in all only two choices.
(Translator’s note)
, but I still couldn’t see why.
abstract-algebra geometry
asked Mar 12 at 11:50
athosathos
97411340
$endgroup$
add a comment |
$begingroup$
I’m reading Arnold’s Abel’s Theorem in Problems and Solutions, where it says:
We now prove that the alternating group $A_5$ is not soluble. One of the
possible proofs uses the following construction. We inscribe in the
dodecahedron five regular tetrahedra, numbered by the numbers 1, 2, 3,
4 and 5 in such a way that to every rotation of the dodecahedron there
corresponds an even permutation of the tetrahedra, and that to
different rotations there correspond different permutations. So we
have defined an isomorphism between the group of rotations of the
dodecahedron and the group $A_5$ of the even permutations of degree 5. The
non-solubility of the group $A_5$ will thus follow from the
non-solubility of the group of rotations of the dodecahedron.
I’m a bit confused. Why rotation of dodecahedron corresponds to an even permutation of inscribed five tetrahedra?
The answer just mentioned
The tetrahedra are‚ for example‚ those with the vertices chosen in the
following way1: (1‚ 8‚ 14‚ 16)‚ (2‚ 9‚ 15‚ 17)‚ (3‚ 10‚ 11‚ 18)‚ (4‚
6‚ 12‚ 19)‚ (5‚ 7‚ 13‚ 20).
, regarding to the notation
The translator provided an image of the inscribed cube:
, and added some explanation
The inscription of the five Kepler cubes inside the dodecahedron helps
us to find the five tetrahedra.
The edges of the cubes are the diagonals of the dodecahedron faces.
Every pair of opposite vertices of the dodecahedron is a pair of two
opposite vertices of two Kepler cubes. Each cube has thus only one
pair of vertices in common with any one of the others. (Two Kepler
cubes — black and white — having two opposite vertices in common are
shown in the figure). In each cube one can inscribe two tetrahedra
(see Problems 126 and 127). Since each tetrahedron is defined by four
vertices of the cube‚ and any two Kepler cubes have only 2 vertices in
common‚ all tetrahedra inscribed in the Kepler cubes are distinct. So
there are in all 10 tetrahedra‚ two for every vertex of the
dodecahedron. Any two of such tetrahedra either have no vertices in
common‚ or they have only one vertex in common. Indeed‚ if two
vertices belonged to two tetrahedra‚ the edge of such tetrahedra
joining them should be the diagonal of a face of two different cubes‚
but we know that any two cubes have in common only opposite vertices.
There are two possible choices of 5 tetrahedra‚ without common
vertices‚ inside the dodecahedron: indeed‚ when we choose one of them‚
we have to reject the four tetrahedra having a vertex in common with
the chosen tetrahedron. The remaining tetrahedra are five. Amongst
them four tetrahedra have disjoint sets of vertices‚ whereas the
remaining tetrahedron has one vertex in common with each of the four
disjoint tetrahedra. The choice of the first tetrahedron thus forces
the choice of the others‚ so there are in all only two choices.
(Translator’s note)
, but I still couldn’t see why.
abstract-algebra geometry
asked Mar 12 at 11:50
athosathos
97411340
$endgroup$
I’m reading Arnold’s Abel’s Theorem in Problems and Solutions, where it says:
We now prove that the alternating group $A_5$ is not soluble. One of the
possible proofs uses the following construction. We inscribe in the
dodecahedron five regular tetrahedra, numbered by the numbers 1, 2, 3,
4 and 5 in such a way that to every rotation of the dodecahedron there
corresponds an even permutation of the tetrahedra, and that to
different rotations there correspond different permutations. So we
have defined an isomorphism between the group of rotations of the
dodecahedron and the group $A_5$ of the even permutations of degree 5. The
non-solubility of the group $A_5$ will thus follow from the
non-solubility of the group of rotations of the dodecahedron.
I’m a bit confused. Why rotation of dodecahedron corresponds to an even permutation of inscribed five tetrahedra?
The answer just mentioned
The tetrahedra are‚ for example‚ those with the vertices chosen in the
following way1: (1‚ 8‚ 14‚ 16)‚ (2‚ 9‚ 15‚ 17)‚ (3‚ 10‚ 11‚ 18)‚ (4‚
6‚ 12‚ 19)‚ (5‚ 7‚ 13‚ 20).
, regarding to the notation
The translator provided an image of the inscribed cube:
, and added some explanation
The inscription of the five Kepler cubes inside the dodecahedron helps
us to find the five tetrahedra.
The edges of the cubes are the diagonals of the dodecahedron faces.
Every pair of opposite vertices of the dodecahedron is a pair of two
opposite vertices of two Kepler cubes. Each cube has thus only one
pair of vertices in common with any one of the others. (Two Kepler
cubes — black and white — having two opposite vertices in common are
shown in the figure). In each cube one can inscribe two tetrahedra
(see Problems 126 and 127). Since each tetrahedron is defined by four
vertices of the cube‚ and any two Kepler cubes have only 2 vertices in
common‚ all tetrahedra inscribed in the Kepler cubes are distinct. So
there are in all 10 tetrahedra‚ two for every vertex of the
dodecahedron. Any two of such tetrahedra either have no vertices in
common‚ or they have only one vertex in common. Indeed‚ if two
vertices belonged to two tetrahedra‚ the edge of such tetrahedra
joining them should be the diagonal of a face of two different cubes‚
but we know that any two cubes have in common only opposite vertices.
There are two possible choices of 5 tetrahedra‚ without common
vertices‚ inside the dodecahedron: indeed‚ when we choose one of them‚
we have to reject the four tetrahedra having a vertex in common with
the chosen tetrahedron. The remaining tetrahedra are five. Amongst
them four tetrahedra have disjoint sets of vertices‚ whereas the
remaining tetrahedron has one vertex in common with each of the four
disjoint tetrahedra. The choice of the first tetrahedron thus forces
the choice of the others‚ so there are in all only two choices.
(Translator’s note)
, but I still couldn’t see why.
abstract-algebra geometry
abstract-algebra geometry
asked Mar 12 at 11:50
athosathos
97411340
asked Mar 12 at 11:50
athosathos
97411340
asked Mar 12 at 11:50
athosathos
97411340
asked Mar 12 at 11:50
athosathos
97411340
asked Mar 12 at 11:50
athosathos
97411340
97411340
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let's label the tetrahedra $12345$ according to the lowest numbered vertex they contain.
A clockwise rotation by 72° about the axis passing through the centers of top and bottom face of the dodecahedron corresponds to a permutation of tetrahedra $(12345)to(51234)$, which is an even permutation. The same goes, of course, for a similar rotation about the axis of another couple of opposite faces. But any rotation mapping the dodecahedron to itself is the composition of such rotations, and thus corresponds to an even permutation of the five tetrahedra.
answered Mar 12 at 15:02
AretinoAretino
25.2k21445
$endgroup$
$begingroup$
I finally understood it by writing down the tetrahedra’s rotation results . Amazing Arnold figured this out! Thank you!
$endgroup$
– athos
19 hours ago
add a comment |
Your Answer
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let's label the tetrahedra $12345$ according to the lowest numbered vertex they contain.
A clockwise rotation by 72° about the axis passing through the centers of top and bottom face of the dodecahedron corresponds to a permutation of tetrahedra $(12345)to(51234)$, which is an even permutation. The same goes, of course, for a similar rotation about the axis of another couple of opposite faces. But any rotation mapping the dodecahedron to itself is the composition of such rotations, and thus corresponds to an even permutation of the five tetrahedra.
answered Mar 12 at 15:02
AretinoAretino
25.2k21445
$endgroup$
$begingroup$
I finally understood it by writing down the tetrahedra’s rotation results . Amazing Arnold figured this out! Thank you!
$endgroup$
– athos
19 hours ago
add a comment |
$begingroup$
Let's label the tetrahedra $12345$ according to the lowest numbered vertex they contain.
A clockwise rotation by 72° about the axis passing through the centers of top and bottom face of the dodecahedron corresponds to a permutation of tetrahedra $(12345)to(51234)$, which is an even permutation. The same goes, of course, for a similar rotation about the axis of another couple of opposite faces. But any rotation mapping the dodecahedron to itself is the composition of such rotations, and thus corresponds to an even permutation of the five tetrahedra.
answered Mar 12 at 15:02
AretinoAretino
25.2k21445
$endgroup$
$begingroup$
I finally understood it by writing down the tetrahedra’s rotation results . Amazing Arnold figured this out! Thank you!
$endgroup$
– athos
19 hours ago
add a comment |
$begingroup$
Let's label the tetrahedra $12345$ according to the lowest numbered vertex they contain.
A clockwise rotation by 72° about the axis passing through the centers of top and bottom face of the dodecahedron corresponds to a permutation of tetrahedra $(12345)to(51234)$, which is an even permutation. The same goes, of course, for a similar rotation about the axis of another couple of opposite faces. But any rotation mapping the dodecahedron to itself is the composition of such rotations, and thus corresponds to an even permutation of the five tetrahedra.
answered Mar 12 at 15:02
AretinoAretino
25.2k21445
$endgroup$
Let's label the tetrahedra $12345$ according to the lowest numbered vertex they contain.
A clockwise rotation by 72° about the axis passing through the centers of top and bottom face of the dodecahedron corresponds to a permutation of tetrahedra $(12345)to(51234)$, which is an even permutation. The same goes, of course, for a similar rotation about the axis of another couple of opposite faces. But any rotation mapping the dodecahedron to itself is the composition of such rotations, and thus corresponds to an even permutation of the five tetrahedra.
answered Mar 12 at 15:02
AretinoAretino
25.2k21445
answered Mar 12 at 15:02
AretinoAretino
25.2k21445
answered Mar 12 at 15:02
AretinoAretino
25.2k21445
answered Mar 12 at 15:02
AretinoAretino
25.2k21445
25.2k21445
$begingroup$
I finally understood it by writing down the tetrahedra’s rotation results . Amazing Arnold figured this out! Thank you!
$endgroup$
– athos
19 hours ago
add a comment |
$begingroup$
I finally understood it by writing down the tetrahedra’s rotation results . Amazing Arnold figured this out! Thank you!
$endgroup$
– athos
19 hours ago
$begingroup$
I finally understood it by writing down the tetrahedra’s rotation results . Amazing Arnold figured this out! Thank you!
$endgroup$
– athos
19 hours ago
$begingroup$
I finally understood it by writing down the tetrahedra’s rotation results . Amazing Arnold figured this out! Thank you!
$endgroup$
– athos
19 hours ago
add a comment |
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