# Permutation Algebra

1. Sep 9, 2010

### roam

1. The problem statement, all variables and given/known data

This is a worked problem:

[PLAIN]http://img409.imageshack.us/img409/4821/14091194.gif [Broken]

3. The attempt at a solution

In the answer, how did they get from $$(1 3 4 9)^7(2 6 8)^7$$ to $$(1 3 4 9)^{-1}(2 6 8)$$?

I know that $$\tau^7$$ means the permutation $$\tau$$ repeated 7 times. But I just don't understand why they changed "7" to "-1" on the first cycle and changed "7" to "1" on the second. They are probably using a shortcut but I can't follow it...

Last edited by a moderator: May 4, 2017
2. Sep 9, 2010

### HallsofIvy

(1249) involves 4 numbers and so has order 4- that is (1249)^4= identity. It is generally true that a cycle with n numbers has order n (that is used in the first problem) because doing it n time just cycles through all the numbers. Of course (1249)^8= (1249)^4(1249)^4= identity and 7 is just 1 less than 8: (1249)^7= (1249)^-1 because (1249)^7(1249)= (1249)^8= identity.

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