# Permutation group order

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• September 11th 2007, 07:45 PM
tttcomrader
Permutation group order
Show that $A_{8}$ contains an element of order 15.

Note: $A_{8}$ is the group of even permutation of 8.

I really don't know how to start this proof, please help.
• September 11th 2007, 07:50 PM
ThePerfectHacker
Quote:

Originally Posted by tttcomrader
Show that $A_{8}$ contains an element of order 15.

Note: $A_{8}$ is the group of even permutation of 8.

I really don't know how to start this proof, please help.

Let $\sigma$ be a cycle with length $n$. And let $\tau$ be a cycle with length $m$. Then the order of the permutation $\sigma \tau$ is equal to $\mbox{lcm}(n,m)$. So the idea is to find an even cycle of length 5 and another even cycle with length 3 and take their product.
• September 11th 2007, 08:24 PM
ThePerfectHacker
Quote:

Originally Posted by Bernhard Riemann
Let $\sigma$ be a cycle with length $n$. And let $\tau$ be a cycle with length $m$. Then the order of the permutation $\sigma \tau$ is equal to $\mbox{lcm}(n,m)$. So the idea is to find an even cycle of length 5 and another even cycle with length 3 and take their product.

Okay how about I give you my example and you confirm the details.

1)Consider $(1,2,3) = (1,2)(2,3)$ this is an even permutation, why? So $(1,2,3)\in A_8$.

2)Consider $(1,2,3,4,5) = (1,2)(2,3)(3,4)(4,5)$ this is an even permutation, why? So $(1,2,3,4,5)\in A_8$.

3)Define $\phi = (1,2,3)(1,2,3,4,5)$ this is an even permutation, why? So $\phi \in A_8$.

4)By my previous comment $\mbox{ord}(\phi) = \mbox{lcm}(3,5) = 15$.
• September 11th 2007, 08:26 PM
tttcomrader
So I pick O = (1 2 3) and r = (4 5 6 7 8 9), they are both even permutations, and O has length 3, r has length 5, so ord(Or) = 15.

Thank you!