Galois Theory and Finite Permutation GroupsDate: 06/01/2007 at 15:18:35 From: Rolando Subject: finite permutation groups Do the elements (123456)(78) and (17) generate the symmetric group of order eight? It is apparently a problem difficult to answer without the aid of a computer: the symmetric group of order eight consists of 1x2x3x4x5x6x7x8 = 40320 permutations. I know a cycle of order 7 like (1234567) and a cycle of order two like (18) do generate the symmetric group. The method is to take the products (1234567)^t(18)(1234567)^(-t), that is, the elements conjugate to (18) with respect to the powers of the cycle (1234567). You get: (18),(28),(38),(48),(58),(68),(78). Take now (18)(28)(18) = (12) and similarly get (ij) and so, the complete symmetric group. I think the answer in the present case is not, and I guess it should not be very difficult to prove it. For example, if I get the transposition (12) it is yes, else it is not. I've tried to get (12) but I've failed. Yet I don't have a proof. I can get the transpositions (13),(15),(17),(35),(37),(57) and also (24),(26),(28),(46),(48),(68) by the method just described at the beginning; that is, (ij) with i and j of the same parity. I've failed for one like (12). This would decide the question. Date: 06/02/2007 at 04:30:44 From: Doctor Jacques Subject: Re: finite permutation groups Hi Rolando, You are closer to the solution than you think: you have noticed that you can make transpositions (i,j) if i and j have the same parity, but not otherwise. Look at the action of the generators on the sets O and E of odd and even points, respectively. Notice that the first generator (1,2,3,4,5,6)(7,8) interchanges O and E as sets; under its action, *all* odd points become even, and conversely. On the other hand, the second generator (1,7) leaves O and E invariant as sets; *all* odd points remain odd and the same holds for even points. This means that, in any case, either the sets O and E are preserved, or they are exchanged. (In technical terms, we say that the sets O and E constitute a block system.) Can you finish from here? Please feel free to write back if you require further help. - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/ Date: 06/05/2007 at 09:20:06 From: Rolando Subject: Thank you (finite permutation groups) Many thanks for your answer, Dr. Jacques. A second part of the problem comes to me at once. The elements (123456)(78) and (17) do not generate the symmetric group of order eight S8. Do they generate H? Call H the proper subgroup of S8 whose elements are all those permutations which preserve the set O or else exchange the sets O and E (O = {1,3,5,7}, E = {2,4,6,8}). I just put down what comes right now to my head. Date: 06/06/2007 at 03:14:00 From: Doctor Jacques Subject: Re: Thank you (finite permutation groups) Hi again Rolando, The short answer is yes. Any permutation of H can be decomposed as: (a) A permutation that exchanges O and E, if required (b) A permutation of the odd elements, that leaves the even elements fixed (c) A permutation of the even elements, that leaves the odd elements fixed. If you only use one permutation for (a) (like the first generator of the group), this decomposition is unique. As there are two choices for (a) and 24 choices for each of (b) and (c), the size of the group is: 2 * 24 * 24 = 1152 Using the GAP software, it is possible to confirm that this is the size of G (the size can also be computed manually, using something called the Schreier-Sims algorithm--this is rather tedious). We should try to prove this directly. As shown before, G is a subgroup of H. All that remains to prove is that G contains a generating set of H. Now, you can generate H with the following permutations: (a) The first generator of the group : (1,2,3,4,5,6)(7,8). This permutation exchanges O and E. (b) All transpositions (x,y) with x and y odd. One such transposition is (1,7), the second generator. You should be able to express the others from the generators of G. Note that, when you have this, you also have the transpositions of even elements (by conjugating with (a)), and this allows you to express any element of H. Please feel free to write back if you require further assistance. - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/ |
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