Order of a permutation in $S_n$ I'm currently in first year math and got given this question, but I have absolutely no idea how to go about it. 
After asking me to write the permutations given by $f$ as a product of disjoint cycles, it asked me to find the least number $m\in\mathbb{N}$ such that $f^m=\mathrm{id}$, where id denotes the identity function.
I tried composing it again and again so I saw what was obtained for $m=2,m=3,...$, but it seems like it's going nowhere: I cannot spot any patterns in the lengths of the cycles corresponding to the value of $m$. I was wondering if I could get a hint that could take me in the correct direction?
Thank you!
 A: The answer has been given in the comments, with many helpful remarks from other users. Having said that, the way I think about this problem is in the following way:
You now have your permutation written down as a product of two disjoint cycles:
$$f=(5\ 6\ 7\ 10)(1\ 8\ 2\ 3\ 4\ 9).$$
Lets look at these two cycles individually. Lets denote $f_1=(5\ 6\ 7\ 10)$ and $f_2=(1\ 8\ 2\ 3\ 4\ 9)$ so that $f=f_1 \circ f_2$. (WARNING: Do not get subscripts confused with superscripts. Here I am just labelling things with subscripts. Superscripts (used later) are for multiplication.)
You either know, or will be able to figure out, that if you have a cycle $\alpha$ of length $n$, then $\alpha^n = Id$. Lets be clear about what I mean: If you have a cycle of length $n$, and you do it $n$ times, then that is the same as the identity permutation. In plain english: doing a cycle of length $n$ $n$ times is the same as doing nothing at all. Everything goes back to where it started.
Now, that means that if you do $f_1$ four times you get the identity (i.e. $(f_1)^4 = Id$), and similarly $(f_2)^6 = Id$.
Initially, that may seem like it's not important and it doesn't help. After all, we are doing $f_1$ and $f_2$ "at the same time" when we do $f$. The key thing to realise here is that if, for example, $(f_1)^4=Id$, then for any multiple of four, doing $f_1$ that many times is the same as doing the identity. The same goes for $f_2$: doing $f_2$ any number of times which is a multiple of $6$ results in doing the identity. 
What does that all add up to? Well, when we do $f$, we are doing $f_1$ and $f_2$ at the same time. The conclusion is that if you do $f$ a number of times which is both a multiple of four and six, that will be the same as doing nothing, because that is the same as doing nothing for both of the cycles which we put together to make $f$.
For example, $f^{24} = Id$.
Now, if you check the definition of what the lowest common multiple of two numbers is vs the above logic, perhaps that will bring the insight which you're after to understand Crostul's answer in the comments.
This exercise actually demonstrates some of the motivation for why we like to write permutations as products of disjoint cycles.
