Let $\mathrm{A}=\{1,2,3,…,9\}$ and $f:\mathrm{A \rightarrow A}$ is a bijection such that $(fofofo\cdots n$ times) $=f$ but $(fof),(fofof), \cdots, (fofofo\cdots (n-1)$ times) are not identical to $f$. Then largest value of $n$ is?
Answer given- 21
I thought of taking an apt function and getting the answer, but answer seems out of range, any hint?
Best Answer
The following answer is quite rough, and assume some acquaintance with permutation groups.
You can write $\{1,\dots,9\}$ as a disjoint union of cycles for $f$. For example, if $f(1)=3,f(3)=4,f(4)=1$, then $(1\,3\,4)$ is such a cycle, and we say its length is $3$. Then you can check that the smallest number $n>0$ such that $f^{\circ n}=\mathrm{id}_{\{1,\dots,9\}}$ is the least common multiple of the length of its cycles. From all the partition of $9$ (e.g. $(1,1,7),(2,2,2,3),\dots$), you can check that $(4,5)$ is the one maximizing this least common multiple, this latter being equal to $20$. Finally, since you want the smallest number $n>1$ such that $f^{\circ n}=f$, you have to add $1$ to $20$, thus leading to $21$.