Symmetries of a CubeDate: 10/09/2003 at 02:27:34 From: Johny Subject: symmetries of cube Prove that the group of symmetries of a cube is isomorphic to S_4. Date: 10/09/2003 at 07:10:14 From: Doctor Jacques Subject: Re: symmetries of cube Hi Johny, Note that we must consider the group of rotations of the cube (i.e., reflections not included). Let G be that group. First, note that G has order 24. Indeed, if you place a cube on a table, you can put each of the 6 faces up, and rotate that face in 4 ways. The cube has 4 diagonals. If we label the vertices of the top face A, B, C, D (clockwise), and the vertices of the bottom face E, F, G, H, in such a way that A is above E, B above F, and so on, the diagonals are: (AG), (BH), (CE), and (DF) Note that we consider the diagonals as lines--(AB) is the same object as (BA). Every rotation of the cube maps a diagonal to a diagonal, i.e. permutes the diagonals between themselves. This shows that there is a homomorphism from G to S4. We must show that the homomorphism is bijective. As both groups are finite and have the same number of elements, the homomorphism will be injective (one-to-one) if and only if it is surjective (onto). Is is somewhat easier to show that it is surjective, i.e. that, for any permutation of the diagonals, there is a rotation of the cube the induces that permutation. Now, S4 is generated by the transpositions (the permutations that exchange two elements). Therefore, it is enough to show that any transposition of diagonals can be achieved by means of a rotation. Assume that we want to exchange the diagonals (AG) and (DF). Let M be the midpoint of AD and N the midpoint of GF. Can you figure out the effect of a rotation of 180 degrees with the line MN as axis? Please feel free to write back if you require further assistance. - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/ Date: 10/09/2003 at 14:39:59 From: Johny Subject: symmetries of cube Thank you for the answer. However, I still have some doubts. It is still not clear to me that every rotation of the cube permutes the diagonals. How do you show that? Moreover, why is it a group? It might be that I don't have a good spatial imagination, but it still not clear to me. I would be grateful if you could explain it in greater detail. Thank you very much! Date: 10/10/2003 at 02:36:13 From: Doctor Jacques Subject: Re: symmetries of cube Hi again Johny, A rotation of the cube maps a vertex to a vertex. Also, a rotation preserves distances. The distance between opposite vertices (the length of a diagonal) is not equal to any other distance between vertices (it is the largest such distance). The conclusion is that the image of a diagonal must be a diagonal. This means that any rotation permutes diagonals within themselves. The product of two rotations is obtained by executing the rotations one after the other. This will execute the diagonal permutations in sequence (this is just composition of functions). This means that, if r1 and r2 are rotations, and s1 and s2 the corresponding permutations, the permutation corresponding to (r2 o r1) will be (s2 o s1)--this is a homomorphism. Permutations of the diagonals are a group under composition--this is simply the group of permutations of 4 objects (in this case, the diagonals), i.e. S4. To see the effect of the particular rotation I told you, you should try this with a "physical" cube (a die, for example). With the labeling I used, the rotation through MN will induce the following permutation of the vertices: (AD)(BH)(CE)(FG) (You can verify this by noting that the points M and N are fixed by the rotation, and that distances are preserved.) If we consider the 4 diagonals AG, BH, CE, DF, we note that BH and CE will remain fixed *as diagonals* (the vertices will be exchanged, but the diagonals will remain the same), and AG and DF will be interchanged. This shows that the rotation induces a transposition of these two diagonals, and similar rotations can be used to generate any transposition, and therefore any permutation of the diagonals can be acheived by a composition of such rotations. This shows that the homomorphism is surjective (onto), and, as the groups have the same order, this is an isomorphism. Does this make sense? - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/ Date: 10/16/2003 at 01:07:42 From: Johny Subject: Thank you (symmetries of cube) Thank you very much for the answer! You helped me a lot! |
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