| Line 15: | Line 15: | ||
4 12 20 (from 4, 5, 1 cycles) 4 (from 4 2 4 cycles) 4 (from 4 3 3) 12 | 4 12 20 (from 4, 5, 1 cycles) 4 (from 4 2 4 cycles) 4 (from 4 3 3) 12 | ||
| + | |||
| + | |||
| + | But there are also the 3, 7 cycles which give an order of 21, and can be even. For example: | ||
| + | |||
| + | (1 2 3)(4 5 6 7 8 9 10) has order 21. | ||
| + | |||
| + | (1 2 3) can be broken down into 2 transpositions. | ||
| + | (1 2 3) = (1 2)(1 3) | ||
| + | |||
| + | (4 5 6 7 8 9 10) can be broken down into 6 transpositions. | ||
| + | (4 5 6 7 8 9 10) = (4 5)(4 6)(4 7)(4 8)(4 9)(4 10) | ||
| + | |||
| + | This gives total of 8 permutations that the permutation can be broken into, making it even, and thus, in <math>A_10</math>. 21 is the maximum order of any element in <math>A_10</math>. | ||
Revision as of 11:39, 10 September 2008
The problem asks, "What is the maximum order of any element in$ A_10 $?"
$ A_10 $ is a an Alternating Group of degree n. $ A_n $ is the group of only even permutations of n symbols.
We have 10 numbers: 1 2 3 4 5 6 7 8 9 10
Size Order
10 10
8 8
6 12 6 6 6 6
4 12 20 (from 4, 5, 1 cycles) 4 (from 4 2 4 cycles) 4 (from 4 3 3) 12
But there are also the 3, 7 cycles which give an order of 21, and can be even. For example:
(1 2 3)(4 5 6 7 8 9 10) has order 21.
(1 2 3) can be broken down into 2 transpositions. (1 2 3) = (1 2)(1 3)
(4 5 6 7 8 9 10) can be broken down into 6 transpositions. (4 5 6 7 8 9 10) = (4 5)(4 6)(4 7)(4 8)(4 9)(4 10)
This gives total of 8 permutations that the permutation can be broken into, making it even, and thus, in $ A_10 $. 21 is the maximum order of any element in $ A_10 $.
