= Solution
The set $S(\mathbb N)$ consists of all <bijections> $\mathbb N\to\mathbb N$. The composition of two bijections is a bijection, composition is associative, the identity map is an identity, and every bijection has an inverse, so $S(\mathbb N)$ is a <group>.
An element of $S_{\mathrm{fin}}(\mathbb N)$ has finite <support of a permutation>[support]. The support of a composition is contained in the union of the two supports, and a permutation and its inverse have the same support. The identity has empty support. Hence $S_{\mathrm{fin}}(\mathbb N)$ is a <subgroup>.
Let $\sigma$ be a cycle and choose $n$ that it moves. Because $\sigma$ has finite support, the sequence
$$
n,\sigma(n),\sigma^2(n),\ldots
$$
must repeat. Since $\sigma$ is invertible, its first repetition returns to $n$; let the least positive return time be $l$. The cycle condition says that every moved point occurs in this orbit, so $\sigma^l$ fixes every point. No smaller positive power fixes $n$. Therefore
$$
\boxed{\operatorname{ord}(\sigma)=l},
$$
which is finite.
For $\tau\in S_{\mathrm{fin}}(\mathbb N)$, partition its finite support into the orbits of the <cyclic group> $\langle\tau\rangle$. On each orbit, let $\sigma_i$ agree with $\tau$ and fix every point outside that orbit. Then each $\sigma_i$ is a <permutation cycle>, their supports are pairwise disjoint, and
$$
\tau=\sigma_1\cdots\sigma_k.
$$
Disjoint cycles commute because at every point at most one of them acts nontrivially.
Writing $l_i=\operatorname{ord}(\sigma_i)$, a power $\tau^m$ is the identity exactly when every $\sigma_i^m$ is the identity, equivalently when every $l_i$ divides $m$. Thus
$$
\boxed{\operatorname{ord}(\tau)
=\operatorname{lcm}(l_1,\ldots,l_k)}.
$$
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