Suppose that a permutation $f$ is the product of disjoint cycles $f_1,f_2,\dots, f_m$. Show that $o(f)$ is the least common multiple of $o(f_1), o(f_2),\dots, o(f_m)$.
Really lost with the question..
Say if there are 3 disjoint cycles, then $f=f_1*f_2*f_3$ say $f=(a,b)(c,d)(e,f)$ would that mean that $o(f)=\text{lcm}(o(f_1),o(f_2),o(f_3))$? How do you go about proving that?
I also have a second part that wants to know the order of a product of $k$-cycles;
$(1,6,4,9)(2,7,11)(3,5,8)(10,12)$? I got 12 because $\text{lcm}(4,3,3,2)$ is that true?
Disjoint cycles commute, so, if $f_1,\dots f_m$ are disjoint, then $$(f_1f_2\dots f_m)^k=f_1^kf_2^k\dots f_m^k.$$ Since the cycles are disjoint, this product is $1$ if and only if all $f_i^k$ are $1$, so $k$ has to be a multiple of all of $o(f_1),\dots o(f_m)$. Then, the rank would be the smallest of those natural numbers, so it has to be their least common multiple.
In your first example, the rank is $2$, and the least common multiple of $2,2$ and $2$ is $2$. In your second example, $12$ is indeed the correct answer.