Order of $\alpha=(124)(5439)(328)(1378542)\in S_{10}$

Question

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.

Answer 1

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 $6$. '

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.

Answer 2

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 $\DeclareMathOperator{ord}{ord} \ord(\iota) = 1$