For which $n\geq2$ does there exist $\sigma \in A_n$ with $|\sigma|>n$?
I have found this same question here, But I have not been able to understand how it works.
On my own I came to the conclusion that for $n$ less than $8$ the result is not satisfied, but for $n = 8$, I found the permutation $\sigma$ formed by a $5$-cycle and a $3$-cycle. Which is in $A_n$ and $|\sigma|=15>8$. In particular, I cannot understand how the proof follows, to conclude that the result is true for all $n\geq 8$. Could you help me? If there is an easier way to see this, please, I would appreciate if you shared it.
For each $n$ consider taking a $3$-cycle and a cycle of length $l$ where $l$ is the largest number less than or equal to $n-3$ that is not a multiple of $3$ nor a multiple of $2$. Notice we have $l\geq n-6$ ( because it is not possible for $n-3,n-4,n-5$ and $n-6$ to all be even or multiples of $3$).
It follows that there is an element of order at least $3(n-6)$, and we have $3(n-6)> n$ for $n>9$.
For $n=9$ we can take a $3$-cycle and a $5$-cycle.
For $n=8$ we can take a $3$-cycle and a $5$-cycle.
Hence the only cases we must check are $n=1,2,3,4,5,6,7$ (which apparently is what you did).