what is the maximum order one element could have in permutation group $S_5$?

Question

what is the maximum order one element could have in permutation group $S_5$?

Tried: $|S_5| = 5!=2^3\times3\times5$ but I don't think it has anything to do with the cardinality as often seen in $\mathbb{Z}$ congruence groups...

Can someone give a hint about it?

Answer 1

Hint: every permutation can be written as the product of disjoint cycles. The answer in this case will be $2 \times 3 = 6$.

Answer 2

You write the permutation as a product of disjoint cycles, and then use that the order of the product of disjoint cycles is just the lowest common multiple of order of the cycles.

That is, for disjoint cycles $\sigma_1,\sigma_2\ldots,\sigma_m\in S_n$, we have that $\mid\sigma_1,\sigma_2,\ldots,\sigma_m\mid=\operatorname{lcm}(|\sigma_1|,|\sigma_2|,\ldots,|\sigma_m|)$