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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 $8$.

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  • $\begingroup$ It also contains elements of certain even orders. $\endgroup$ – Tobias Kildetoft Apr 13 '16 at 18:48
  • $\begingroup$ 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 $2$. $\endgroup$ – Captain Lama Apr 13 '16 at 18:49
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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.

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