Show that $\displaystyle A_{8}$ contains an element of order 15.

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

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

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- Sep 11th 2007, 06:45 PMtttcomraderPermutation group order
Show that $\displaystyle A_{8}$ contains an element of order 15.

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

I really don't know how to start this proof, please help. - Sep 11th 2007, 06:50 PMThePerfectHacker
Let $\displaystyle \sigma$ be a cycle with length $\displaystyle n$. And let $\displaystyle \tau$ be a cycle with length $\displaystyle m$. Then the order of the permutation $\displaystyle \sigma \tau$ is equal to $\displaystyle \mbox{lcm}(n,m)$. So the idea is to find an even cycle of length 5 and another even cycle with length 3 and take their product.

- Sep 11th 2007, 07:24 PMThePerfectHackerQuote:

Originally Posted by**Bernhard Riemann**

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

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

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

4)By my previous comment $\displaystyle \mbox{ord}(\phi) = \mbox{lcm}(3,5) = 15$. - Sep 11th 2007, 07:26 PMtttcomrader
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!