Find the number of cyclic subgroups groups of the alternating group $A_8$.
I don't know how to even begin approaching this question. Is there a faster way to do this besides explicitly list each and every cyclic subgroup?
2026-04-02 13:13:02.1775135582
Number of cyclic subgroups of the alternating group $A_8$
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I can tell you how to begin approaching the problem. All of them will involve combinatorial counting arguments. The cycle decomposition types that can occur should be listed first: the cyclic permutations of odd order namely, 7, 5 and 3. Then non-cyclic permutations of type $6+2,4+4, 4+2, 2+2, 5+3,3+3, 3+2+2,
Now you have to deal with each kind. In each kind count how many elements are there; then check if the cyclic subgroups generated by them have intersections or not; this is the trickiest part. I'll do this for the easiest kind and leave the others to you.
There are $6!\times 8$ cyclic permutations of order 7. The cyclic subgroups generated by each of them will use 6 elements, and they can have only the identity element in common. So the number of cyclic subgroups of $A_8$ consisting of $7$-cycles is $6!\times8/6=960$.