Order of a cycle. (basic group theory) I thought the length of a cycle was the order of the cycle + I thought I knew how to compute transpositions, but I can't seem to get this one to equal the answer sheet:
let $\tau = (4,5)(2,3,7)$ be permutation of $S_8$. How do I find the order of $\tau$?
 A: Suppose we have two disjoint cycles $C_1$ and $C_2$ with lengths $l_1$ and $l_2$. We know that $l_1$ and $l_2$ are the orders of $C_1$ and $C_2$ respectively, thus, $C_1^{l_1} = e$ and $C_2^{l_2} = e$ where $e$ is the identity permutation.
What if we want to find the order of the product of the two cycles: $C_1C_2$? Notice that if we take any power of this cycle:
$$(C_1C_2)^n = \underbrace{(C_1C_2)(C_1C_2)\dots(C_1C_2)}_{n \text{ times }} = C_1^nC_2^n$$
since disjoint cycles commute.
Also, since $C_1^n$ and $C_2^n$ are disjoint from each other, if $C_1^nC_2^n = e$, then it must be the case that $C_1^n = e$ and $C_2^n = e$ (e.g. if $C_2^n$ wasn't $e$, then it permutes some element which $C_1^n$ won't permute back meaning $C_1^nC_2^n$ couldn't possibly be $e$).
Thus, in order to find the order of $C_1C_2$, we must find the smallest $n$ such that $C_1^n = e$ and $C_2^n = e$. It is not too difficult to show that for a element $g$ in a group $G$ with order $n$, that if $g^k = e$, then the order of $g$, $n$, must divide $k$. For our problem, this means that $n$ needs to be a multiple of both $l_1$ and $l_2$. Since we want to make $n$ as small as possible, we'll take the least common multiple of $l_1$ and $l_2$, i.e., $$n = \text{lcm}(l_1,l_2)$$.
