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Topic: Generators of the Alternating Group (Read 6178 times) |
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william wu
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Generators of the Alternating Group
« on: Feb 2nd, 2004, 9:35pm » |
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Show that when n is odd, the permutations (123) and (12...n) together generate An. Similarly, show that if n is even, (12) and (23....n) together generate An.
Hint: For n>=3, the 3-cycles generate An.
Background Information:
An is called the alternating group of degree n. It is the subgroup formed by the even permutations of Sn. A permutation is "even" if it can be written as the product of an even number of transpositions. Transpositions are permutations of length 2 (e.g. (12) is a transposition).
Source: Groups and Symmetry by M.A. Armstrong, p. 31
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| « Last Edit: Feb 2nd, 2004, 9:36pm by william wu » |
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Barukh
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Re: Generators of the Alternating Group
« Reply #1 on: Feb 5th, 2004, 8:28am » |
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on Feb 2nd, 2004, 9:35pm, william wu wrote:| Show that when n is odd, the permutations (123) and (12...n) together generate An. |
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As both given permutations are even, they may potentially generate An…
[smiley=blacksquare.gif]
As William pointed out, An is generated by the 3-cycles of Sn.1 So, it is to show that the 2 given permutations generate every 3-cycle.
First, note that (12…n)(123)(12…n)n-1 = (12…n)(123)(12…n)-1 = (234), similarly, (12…n)(234)(12…n)-1 = (345) etc. 2
Next, assume we’ve got all 3-cycles from the set {1,2,…,k} together with the 3-cycle L=(k-1 k k+1). Then, for every i,
(i k k+1) = L (i k-1 k) L-1,
(i k-1 k+1) = L-1 (i k k-1) L,
(i j k+1) = L (i j k) L-1, j < k-1.
This – together with the inverses – gives all 3-cycles from the set {1, 2, …, k+1}. Because we’ve got originally (123), we may gradually build all 3-cycles.
[smiley=blacksquare.gif]
Quote:| Smilarly, show that if n is even, (12) and (23....n) together generate An. |
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Since (12) is an odd permutation, these two generate something different… Did you mean (123) instead?
1 To prove this, observe that there are just two distinct pairs of transpositions – (ij)(jk) and (ij)(kl). The first one is a single 3-cycle (ijk), and the second is a composition (ijk)(ilk).
2 Given two elements g, x of a group, the third element gxg-1 is called a conjugate of x w.r.t. g. Here’s an easy way to obtain conjugates in Sn: if x = (i1i2… im), then gxg-1 = (g(i1) g(i2)… g(im)).
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| « Last Edit: Feb 5th, 2004, 8:35am by Barukh » |
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