I was doing mini-test involving a True/False section and came across the following statement.
Every element of $Sym(n)$ has order at most $n$
I admit I had gotten this incorrect as I had thought that the statement is true, but the answer was that the statement is false.
However, I am having trouble seeing a counter example, and I would appreciate one for the sake of understanding symmetric groups further.
In the future it will be worthwhile for you to know how to compute the order of an element of $\mathrm{Sym}(n)$, so I will detail it here.
Recall that every permutation is a product of disjoint cycles, for example $\sigma=35412$ is $(134)(25)$. Disjoint cycles commute with each other, so exponentiating a permutation amounts to raising each disjoint cycle in a decomposition to the same power. You can verify that $\sigma^3=(134)^3(25)^3=(25)$.
A cycle of length $k$, say $(i_1i_2i_3\cdots i_k)$, has order exactly $k$, and the order of a permutation will be the smallest number that is divisible by the orders of all of the cycles, in other words their least common multiple. So the order of $\sigma$ is $\mathrm{lcm}(2,3)=6$, which is small enough to verify manually.