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| + | You would probably want the most relatively prime numbers. In this case, 5+3+2=10 and the order would be 5*3*2=30 which is more than the order of 7 and 3 which is only 21. | ||
Revision as of 15:26, 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 $.
Another way to come to the conclussion of 7 and 3 is to look at the prime values that add to ten. From page 101 of the book, it defines the order as "the least common multiple of the lengths of the cycles." Since you are looking for the least common multiple, it is obvious that the lcm of prime numbers will be the product of those numbers. Thus, instead of writing out all of the options, it can be see that 7 and 3 are the only clear choices.
-Anna
Thanks Anna, that definitely helps!
You would probably want the most relatively prime numbers. In this case, 5+3+2=10 and the order would be 5*3*2=30 which is more than the order of 7 and 3 which is only 21.
