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.
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.
Okay how about I give you my example and you confirm the details.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 $\displaystyle \mbox{ord}(\phi) = \mbox{lcm}(3,5) = 15$.