1
Rotational Groups of the
Five Platonic Solids
Hoveeg Boyadjian
Professor: Julien Paupert
Arizona State University
Boyadjian 2
The purpose of this Honors Contract is to investigate the groups of rigid motions of the
five platonic solids and to analyze which permutation group each is isomorphic to. The five
platonic solids are the tetrahedron, hexahedron, octahedron, dodecahedron and icosahedron. It is
important to note the concept of duality for polyhedra. Duality is defined as such: [1] For each
regular polyhedron, the dual polyhedron is defined to be the polyhedron constructed by
(1) placing a point in the center of each face of the original polyhedron,
(2) connecting each new point with the new points of its neighboring faces, and
(3) erasing the original polyhedron
As such, the tetrahedron is dual to itself, the hexahedron is dual to the octahedron, and the
dodecahedron is dual to the icosahedron. The explicit rotations for the tetrahedron, hexahedron
and the octahedron will be provided in the appendix; the dodecahedron and the icosahedron will
be excluded because of the size of both rotational groups and will have an implicit count.
Two theorems will be used to aid in the discussion:
Theorem 1 (Cayley’s Theorem): Every group is isomorphic to a group of permutations. [2]
Theorem 2: If the group of rigid motions of a regular solid, Gr, acts faithfully on some set of
features of the regular solid X, i.e. vertices, diagonals, inscribed solids etc, then Gr is a subgroup
of 𝑆|𝑋| . [3]
Proof: Let Gr be a group of rigid motions of a given regular solid and let X be the set of features
of this regular solid. Assume that Gr acts faithfully on X. Then, define the map 𝜑 = 𝐺𝑟 → 𝑆|𝑋|
𝑟 ↦ 𝜎, where r is a rotation element of Gr and 𝜎 is a permutation in S|X|. This map is a
homomorphism because permutation multiplication is function composition. Furthermore, this
homomorphism is well-defined because it involves the permutations on the set X. Since 𝜑 is a
homomorphism, we have that the image of 𝜑 is a subgroup of S|X|, namely 𝜑(𝐺𝑟 ) ≤ 𝑆|𝑋| . By the
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𝐺
𝑟
Fundamental Isomorphism Theorem, we have that 𝑘𝑒𝑟𝜑
≃ 𝜑(𝐺𝑟 ) . Since Gr acts faithfully on X,
the only element of Gr that keeps X fixed is the identity element; hence, ker𝜑 is trivial. Thus,
𝐺𝑟
𝑘𝑒𝑟𝜑
≃ 𝐺𝑟 ≃ 𝜑(𝐺𝑟 ) ≤ 𝑆|𝑋| implies that 𝐺𝑟 ≤ 𝑆|𝑋| .
∎
The Tetrahedron
Claim: The group of rigid motions of the tetrahedron is isomorphic to A4.
Let T denote a tetrahedron and let Gr(T) be the group of rigid motions of the tetrahedron
and let X = {1, 2, 3, 4} be the set of vertices of the tetrahedron. Let Gr(T) act on the set X.
Appendix A shows the rotations of the vertices of the tetrahedron. From Appendix A, it is clear
that the only element that fixes the four vertices of the tetrahedron is the identity element; thus,
by Theorem 2 𝐺𝑟 (𝑇) ≤ 𝑆4 . From Appendix A, the order of Gr(T) is 12 while the order of S4 is 24.
Hence, (S4: Gr(T)) = 2 implies that Gr(T) is a normal subgroup of S4. Since the only other normal
subgroup of order 12 in S4 is A4, the group of even permutations, Gr(T) must be A4. This can be
noted since all elements in Gr(T) are even permutations of the vertices of the tetrahedron. Hence,
𝐺𝑟 (𝑇) ≃ 𝐴4 .
∎
The Hexahedron and Octahedron
Claim: The group of rigid motions of both the hexahedron and octahedron are isomorphic to S4.
Let H denote the hexahedron and Gr(H) be the group of rigid motions of the hexahedron.
Similarly, let O denote the octahedron and Gr(O) be the group of rigid motions of the octahedron.
Appendix B and Appendix C give the explicit count to the number of elements in both Gr(H) and
Gr(O); note that both have the same order. Similarly, since the hexahedron is dual to the
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octahedron, both groups of rigid motions will be isomorphic to the same group. This proof will
be done for the group of rigid motions of the hexahedron.
Let X = {A, B} be the set of two inscribed tetrahedrons inside of the hexahedron. A
tetrahedron is inscribed by using the diagonals of each face of the hexahedron; the other
tetrahedron is inscribed by using the remaining diagonals. A picture is provided in Appendix B.
Let Gr(H) act on the set X. There are four different ways the elements of Gr(H) can act on X:
leaving the elements of X fixed with the identity, through rotation through the center of the face,
through rotation of opposite vertices, and through rotation through the midpoints of opposite
edges. All of the elements of Gr(H) are shown in Appendix B. Because the rigid motions of the
cube preserve the isometry, the diagonals of the faces are mapped to diagonals of other faces;
also, faces that are right next to each other are still next to each other after a rotation. As such,
the inscribed tetrahedrons have their isometries preserved as well through the action of Gr(H);
each element of Gr(H) uniquely rotates each of the inscribed tetrahedrons and preserves the
isometry. As such, the only element that leaves the inscribed tetrahedrons fixed is the identity
element. Note that the order of Gr(H) is 24. Since it was shown that 𝐺𝑟 (𝑇) ≃ 𝐴4 , and since Gr(H)
preserves the isometry of both inscribed tetrahedrons, Gr(H) is isomorphic to a group of order 2 ∗
|𝐴4 | = 24. Namely, 𝐺𝑟 (𝐻) ≃ 𝑆4 . Since the hexahedron is dual to the octahedron, we also have
that 𝐺𝑟 (𝑂) ≃ 𝑆4 .
∎
The Dodecahedron and the Icosahedron
Claim: The group of rigid motions of both the dodecahedron and icosahedron have order 60.
Boyadjian 5
Let D denote the dodecahedron and Gr(D) be the group of rigid motions of the
dodecahedron. Similarly, let I denote the icosahedron and Gr(I) be the group of rigid motions of
the icosahedron. The rigid motions of the dodecahedron are as follows:
(1) Identity
(2) 6 sets of rotations about the center of a face. Since each face is a pentagon, it can be
rotated 4 times before returning to its fixed position. Hence, there are 24 elements.
(3) 10 sets of rotations about opposite vertices. These rotations can be done twice per set
of opposite vertices. Hence, there are 20 elements.
(4) 15 sets of rotations about midpoints of opposite edges. This can be done only once
per set of opposite edges. Hence, there are 15 elements.
Therefore, Gr(D) has order 1 + 24 + 20 + 15 = 60. Similarly, the rigid motions of the icosahedron
are as follows.
(1) Identity
(2) 10 sets of rotations about the center of a face. Since each face is a triangle, it can be
rotated 2 times before returning to its fixed position. Hence, there are 20 elements.
(3) 6 sets of rotations about opposite vertices. These rotations can be done four times per
set of opposite vertices. Hence, there are 24 elements.
(4) 15 sets of rotations about midpoints of opposite edges. This can be done only once
per set of opposite edges. Hence, there are 15 elements.
Therefore, Gr(I) has order 1 + 20 + 24 + 15 = 60.
∎
Claim: The group of rigid motions of both the dodecahedron and icosahedron are isomorphic to
A5.
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Since the dodecahedron is dual to the icosahedron, this proof will be exercised for the
dodecahedron.
Proof: Let Gr(D) be the group of rigid motions of a dodecahedron and let X = {a, b, c, d, e} be
the set of inscribed hexahedrons inside the dodecahedron. A hexahedron is inscribed by taking
the “diagonal” of each of the pentagonal faces and connecting them to each other. An image is
provided for one such hexahedron in Appendix D. Let Gr(D) act on the set X. Similar to how
Gr(H) preserved the isometry of the two inscribed tetrahedrons, Gr(D) preserves the isometry of
the five inscribed hexahedrons. This is because each of the twelve edges of one hexahederon lies
on exactly one diagonal of each pentagonal face. Since the diagonals of the pentagonal faces are
rotated to other diagonals of pentagonal faces, the only element in Gr(D) that leaves the five
inscribed hexahedrons fixed is the identity. As such, by Theorem 2 𝐺𝑟 (𝐷) ≤ 𝑆5 . This can also be
observed in a similar way. Since the 𝐺𝑟 (𝐻) ≃ 𝑆4 and since there are five inscribed hexahedrons,
the action of Gr(D) on X produces 5 ∗ |𝑆4 | = 120 = |𝑆5 | elements; so again, 𝐺𝑟 (𝐷) ≤ 𝑆5 . By the
counting argument proposed above, |Gr(D)| = 60 implies that (S5: Gr(D)) = 2. As such, Gr(D) is a
normal subgroup of S5. Since the only normal subgroup of S5 is A5, the group of even
permutations, we have that Gr(D) = A5. Hence, 𝐺𝑟 (𝐷) ≃ 𝐴5 . Again, since the dodecahedron is
dual to the icosahedron we have that 𝐺𝑟 (𝐼) ≃ 𝐴5 .
∎
Boyadjian 7
Appendix
A: The rotations of the Tetrahedron
1
b
a
c
c
4
2
a
b
Rotation about Vertex 1
1. (234)
2. (243)
Rotation about Vertex 2
1. (134)
2. (143)
3
Rotation about Vertex 3
1. (124)
2. (142)
Rotation about Vertex 4
1. (123)
2. (132)
Rotation through Midpoint b
Rotation through Midpoint a
1. (14)(23)
1. (12)(34)
Rotation through Midpoint c
1. (13)(24)
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B1: The rotations of the Hexahedron
1
b
2
a
3
d
4
c
f
e
e
f
5
d
6
c
8
Rotation about Vertices 1 and 7
1. (245)(386)
2. (254)(368)
Rotation about Vertices 2 and 8
1. (163)(457)
2. (136)(475)
Rotation about x-axis
1. (1265)(4378)
2. (16)(25)(47)(38)
3. (1562)(4873)
Rotation about z-axis
1. (1234)(5678)
2. (13)(24)(57)(68)
3. (1432)(5876)
a
b
7
Rotation about Vertices 3 and 5
1. (186)(274)
2. (168)(247)
Rotation about Vertices 4 and 6
1. (183)(257)
2. (138)(275)
Rotation about y-axis
1. (1485)(2376)
2. (18)(45)(27)(36)
3. (1584)(2673)
Rotation through Midpoint a
1. (28)(35)(14)(67)
Rotation through Midpoint b
1. (35)(46)(12)(78)
Rotation through Midpoint c
1. (17)(46)(23)(58)
Rotation through Midpoint d
1. (17)(28)(34)(56)
9
Rotation through Midpoint e
Rotation through Midpoint f
1. (28)(46)(15)(37)
1. (17)(35)(26)(48)
B2: Tetrahedron inscribed in the hexahedron
C: The rotations of the Octahedron
1
a
c
b
d
2
5
4
3
d
c
b
a
6
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Rotation about Vertices 1 and 6
Rotation about Vertices 2 and 4
1. (2345)
2. (24)(35)
3. (2543)
Rotation about Vertices 3 and 5
1. (1264)
2. (16)(24)
3. (1462)
1. (1365)
2. (16)(35)
3. (1563)
Rotation through Midpoint a
1. (24)(15)(36)
Rotation through Midpoint b
1. (35)(12)(46)
Rotation through Midpoint c
Rotation through Midpoint d
1. (24)(13)(56)
1. (35)(14)(26)
Rotation through midpoints of triangle 154 and 236
1. (152)(346)
2. (125)(364)
Rotation through midpoints of triangle 123 and 546
1. (134)(265)
2. (143)(256)
Rotation through midpoints of triangle 143 and 526
1. (154)(263)
2. (145)(236)
Rotation through midpoints of triangle 152 and 436
1. (123)(456)
2. (132)(465)
Boyadjian 11
D: Hexahedron inscribed in a Dodecahedron
References
[1] Lim, Yongwhan. Symmetry Groups of Platonic Solids. MIT, Massachusetts. December 4,
2008.
[2] Fraleigh, John B. A First Course in Abstract Algebra 7th Edition. University of Rhode Island,
Rhode Island: Pearson Education, Inc., 2003. pp. 82.
[3] Siegel, Adam. Regular Solids and their Rotation Groups. Spring 2007.