Order of $\alpha=(124)(5439)(328)(1378542)\in S_{10}$ I know that the order of an element in a symmetric group is given by $\text{lcm}(a,b)$ where a and b are the length of the cycle. But not sure what to do with more then 2  cycles.
 A: By direct computation, and reading left to right,  the product of the permutations you have written decomposes into disjoint cycles as:  $$
(1 5 2) (3 9 4) (7 8) (6)$$
Two of these have length $3$, the third has length $2$ (the fixed point has length $1$ of course), and since these commute, the order of this product is $6$.
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Note:  as remarked in the comments, the problem says this is taking place in $S_{10}$, so we have to include $X=10$ in the cycles (using Roman Numerals since, say, writing $(10)$ is ambiguous).  Thus the disjoint cycle representation should have been $$(1 5 2) (3 9 4) (7 8) (6) (X)$$  It is also standard to simply omit fixed points, and write $$(1 5 2) (3 9 4) (7 8)$$
This convention comes in handy with permutations on infinite sets which fix all but finitely many elements. To be sure, none of this changes the calculation of the order.
A: It is still the $\DeclareMathOperator{lcm}{lcm} \lcm$, but this time the $\lcm(a_1, a_2, \dots, a_k)$ where $a_1, \dots a_k$ are the lengths of all disjoint cycles of $\alpha$.
Note that, in general, the $\lcm$ of the cycle lengths does not yield the order of $\alpha$ if those cycles have common elements: $\iota = (1,2,3)(1,3,2)$ but $\DeclareMathOperator{ord}{ord} \ord(\iota) = 1$
