I know $A_{10}$ has elements of order 1,3,5,7,9,15,21 where the odd numbers arise form single cycles being even and 15, 21 come from $lcm(3,5)$ and $lcm(3,7)$ respectively. I'm not sure how to justify the remaining orders of 2,4,6,8,10,12. Any help is appreciated.
2026-04-01 21:00:13.1775077213
Orders of elements in alternating group $A_{10}$
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12=lcm(2,3,4), and the perm has cycle structure 10=1+2+3+4, is even.
10 = lcm(2,2,5) and the perm has cycle structure 10= 1+2+2+5 is even.
To get an element of order 2, e.g., take two disjoint transpositions Others can be done similarly.