Determining order of an element in a symmetric group

Question

Specifically, how is an element in $S_5$ e.g $(1 2 3) (4 5)$ have order $6$? Can someone explain this?

Answer 1

Because you have disjoint cycles, the order of the permutation is just the LCM of the lengths of the cycles.

In this case, we have $\operatorname{lcm}(3,2)=6.$

Answer 2

Let $$\alpha=(1\ 2\ 3)(4\ 5).$$ I get $$\alpha^2=(1\ 3\ 2)(4)(5),$$ $$\alpha^3=(1)(2)(3)(4\ 5),$$ $$\alpha^4=(1\ 2\ 3)(4)(5),$$ $$\alpha^5=(1\ 3\ 2)(4\ 5),$$ $$\alpha^6=(1)(2)(3)(4)(5)=\mathrm{Id}.$$

Since $\alpha^k\ne\mathrm{Id}$ for $k=1,2,3,4,5$ while $\alpha^6=\mathrm{Id}$, it looks like $\alpha$ has order $6$.

Why, did you get something different?

Answer 3

$(123)$ is order $3$ element, and $(45)$ is order $2$ element, and since $gcd(2,3) = 1$ , their product $(123)(45)$ is order $2\cdot 3 = 6$

Answer 4

A general fact for groups: the order of the product of commuting elements $$\sigma= c_1 \cdot c_2 \ldots \cdot c_m$$

is the lowest common multiple of the orders of the $c_i$. Consider now the group to be $S_n$ and $c_i$ disjoint cycles therefore commuting.

Pairwise commuting factors is essential. As an example $(1,3)(1,2)= (1,2,3)$ is of order $3$ while the factors have order $2$. The factors do not commute since $(1,2)(1,3)= (1,3,2) \ne (1,2,3)$.