Could someone give me some help with this proof?
Show that $A_8$ contains an element of order $15$.
I know that by Theorem (5.7) in my textbook: for $n>1$, $A_n$ has order $n!/2$.
So by that theorem, $A_8$ would have order in the vicinity of $20,160$, right? How would I proceed from here?
Consider $x=(1\ 2 \ 3)(4 \ 5 \ 6 \ 7 \ 8)\in S_8$. Show that it is an even permutation (i.e., it is in $A_8$) and show that $x^{15}=1$.
How did I find that element? First, two disjoint cycle $x$ and $y$ commute ($xy=yx$). Thus, if $x$ has order $n$ and $y$ has order $m$ with $gcd(n,m)=1$, $xy$ has order $nm$. So you just have to take two disjoint cycles with order 3 and 5. $(1\ 2 \ 3)$ and $(4 \ 5 \ 6 \ 7 \ 8 )$ were the first that came to my mind. Now you just have to show that their product is an even permutation (easy).