# HW2, Chapter 5 problem 8, Discussion, MA453, Prof. Walther

Problem Statement

"What is the maximum order of any element in A10?"

## Discussion

A10 is a an Alternating Group of degree n. An 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 A10. 21 is the maximum order of any element in A10.

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.

But can a permutation with a 5-cycle, 3-cycle, and a 2-cycle be even? Counting the transpositions...

1 for the 2-cycle
2 for the 3-cycle
4 for the 5-cycle

That adds up to 7, which means it is odd, and thus cannot be in A10.

This could be wrong, but theorem 5.7 in the book says the order of $A_n=n!/2$

I would use theorem 5.7 (page 104) as well to proof this. -Sarah