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.
 +
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(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.
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(4 5 6 7 8 9 10) = (4 5)(4 6)(4 7)(4 8)(4 9)(4 10)
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 +
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 12:39, 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 $.

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