# Integers for which the alternating group contains an element of that order

I'm asked to find the integers $m$ for which the alternating group $A_8$ contains elements of order $m$.

Is this simply $1,3,5,7$?

My thinking is that the alternating group contains only even permutations, and a $k$-cycle is even is $k$ is odd so I guess it's just all of the odd numbers up to $8$.

• It also contains elements of certain even orders. – Tobias Kildetoft Apr 13 '16 at 18:48
• Any group of order divisible by a prime $p$ contains an element of order $p$, so since $2$ divides $8!/2$, $A_8$ must contain an element of order $2$. – Captain Lama Apr 13 '16 at 18:49

The list of possible orders of elements in $A_8$ is $$\{1, 2, 3, 4, 5, 6, 7, 15\}.$$ The idea is to look at partitions of $8$, e.g. $8=1+2+2+3$, gives for example $(23)(45)(678)$of order $lcm(1,2,2,3)=6$. Or $8=3+5$ gives, say, $(123)(45678)$ of order $15$.
If you prefer finite matrix groups, then the isomorphism $A_8\cong GL(4,\mathbb{F}_2)$ is useful.