Past exam of the mathematics course of the University of Cambridge / 2018 / ia / Paper 3 / 1D (Groups)

Composing the permutation cycles from right to left giveswith $6$ fixed. It therefore has order $7$ and, as a seven-cycle, sign $(-1{)}^6=+1$.

An element of $S_6$ has order $6$ precisely for cycle type $(6)$ or $(3,2,1)$. These contributeso there are $240$ elements of order $6$. Order $3$ has cycle type $(3,1,1,1)$ or $(3,3)$, contributingso there are $80$ elements of order $3$.

Inspecting the partitions of $9$ shows that the largest cycle-length least common multiple available in $S_9$ is $20$, from the partition $5+4$, but such a permutation is odd. Among integers from $16$ through $20$, orders $17$ and $19$ would require cycles of those prime lengths, order $16$ requires a cycle length divisible by $16$, and order $18$ requires disjoint cycles using at least $9+2>9$ letters. The disjoint product of a five-cycle and a three-cycle is even and has order $15$. Hence the greatest order in $A_9$ is $15$.