Theorem 4.2 (Order of a Permutation): The order of a permutation written in disjoint cycle form is the least common multiple of the lengths of the cycles.
Proof: One cycle: As we noted above, a cycle of length m has order m. (See Exercise 9.)
Two disjoint cycles: Two disjoint cycles: Now suppose $\alpha$ and $\beta$ are disjoint cycles of lengths a and b. Let k be the least common multiple of a and b, that is, k is the smallest positive integer which is divisible by both a and b. Since $\alpha$ and $\beta$are disjoint cycles, hence they commute. Then $(\alpha\beta)^k = \alpha^k \beta^k = ee = e$. Here we used Lemma 3.4 and that fact that $\rho|k$ implies $\rho^k = e$ for all permutations $\rho$.
It follows from Theorem 3.6 that the order of $\alpha\beta = |\alpha\beta|$divides $k$. We now wish to show $|\alpha\beta|$ = k. From $e = (\alpha\beta)^{|\alpha\beta|} = \alpha^{|\alpha\beta|}\beta^{|\alpha\beta|} \iff \alpha^{-|\alpha\beta|} = \beta^{|\alpha\beta|}$. However, $\alpha$ and $\beta$ have no symbol in common, and since raising a cycle to a power does not introduce new symbols, $\alpha^{-|\alpha\beta|}$ and $\beta^{|\alpha\beta|}$ also have no symbol in common. Since $\alpha^{-|\alpha\beta|} = \beta^{|\alpha\beta|}$ and have no common symbols then they both must be the identity: $\alpha^{-|\alpha\beta|} = \beta^{|\alpha\beta|} = e.$ If follows from Theorem 3.6 that $|\alpha\beta| $ is divisible by a and b. This means that k = lcm(a; b) must also divide $|\alpha\beta| $ . Therefore $|\alpha\beta| $ = k, as desired.
Another explanation. Reference: Joseph Gallian p.102 in Contemporary Abs Algebra
More than two disjoint cycle: The general case involving more than two cycles is handled in an analogous way.
How does $\alpha^{-|\alpha\beta|} = \beta^{|\alpha\beta|} = e $ ? What's the intuition? I understand this comment:
If $\alpha,\beta$ are permutations of $X$ and $X=A\cup B$ is a disjoint union, and if $$ \alpha|_A = \operatorname{id}_A \quad \beta|_B=\operatorname{id}_B $$ then what happens if you restrict the relation $$ 1=(\alpha \beta)^n=\alpha^n \beta^n $$ to $A$ or $B$?
edit: $$ \operatorname{id}_A = \operatorname{id}_X|_A =(\alpha \beta)^n|_A =\alpha|_A^n \beta|_A^n = \beta|_A^n $$ but on $B$ you have $\beta|_B^n=\operatorname{id}_B$. So $\beta^n=1$, because it does nothing on $A$ and $B$.