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I have an issue with a question from some homework for my introduction to group theory course.

For which integers $d$ does the alternating group $A_8$ have elements of order $d$?

So through some rough work, I think the answer is $d=1,2,3,4,5,6,7,15$ but I got this by finding examples of elements with these orders and so obviously I am not sure if these are the only possible orders, only that they are possible orders.

Is there a better way of finding the solution?

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Note that $A_8$ consists of only even permutations and hence the maximum order of any element of $A_8$ can be $15$ which is a small number so it's not that hard to check for each number between $1$ and $15$ that whether there exists an element of that order or not. So for example for odd numbers between $1$ and $8$ $\exists$ an element of that order, just consider the element $(12345)$ for exmaple for $5$.

$\nexists$ any element of order $8$ because the only such element in $S_8$ is $(12345678)$ but it is not in $A_8$. The possibilities of an element of order 11 and 13 are easily ruled out. Now since the order of any two disjoint cycles is the $lcm$ of their orders hence the possibilities of $9,10,14$ is also ruled out. Finally for orders $12$ and $15$ consider the elements $(123456)(78)$ and $(12345)(678)$ respectively.

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Hint: Every element $\sigma$ in $S_8$ can be written as cycles with disjoint orbits. and the order of $\sigma$ is simply the least common multiple of the lengths of this cycles. You need only list all different partitions of $\{1,2,\cdots,8\}$, discard the ones corresponding to odd permutation and calculate the least common multiple of the lengths of the orbits.

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  • $\begingroup$ "you need only list all different partitions of {1,2,⋯,8}" - are there not a huge number of partitions of this set? $\endgroup$ – user195486 Mar 26 '15 at 23:29
  • $\begingroup$ @user195486 I think it's OK, but if you'd like to reduce the calculation, there is some improvements: If the length of all cycles are $\leqslant4$, the possible least common multiples are :$1,2,4,3,6,$($12$ is impossible becasue permutaions of type (1,3,4) are odd permutations. The remained case, that is, when there is a cycle with length $>4$, are rather rare. $\endgroup$ – Censi LI Mar 26 '15 at 23:45

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