Order of $(1 \,3)(2 \, 5 \, 4)$ in $S_5$

Question

What is the order of $(1 \,3)(2 \, 5 \, 4)$ in $S_5$?

From number theory, I remember that we defined the order to be the smallest positive integer $k$ for which $a^k \equiv 1 \pmod{n}$ and also $a$ and $n$ have to be relatively prime. In group theory, it seems to be quite similar I found that the order seems to be the smallest positive $n$ for which $g^n=e$ where $g \in G.$

I thought that in this case if I let $\sigma=(1 \,3)(2 \, 5 \, 4)$ and compute the powers of $\sigma$ I would have managed to find the order, but this didn't quite work. Is there another way to find this?

Answer 1

Here the numbers that $(1, 3)$ and $(2, 5, 4)$ operate on are mutually exclusive and so the two elements commute. Since the order of $(1, 2)$ is $2$ and $(2, 5, 4)$ is $3,$ the order of their product is $6$. More generally, if you have an element $g_1$ of order $m$ and $g_2$ of order $n$ and $g_1g_2 = g_2g_1$, the order of $g_1g_2$ is $LCM(m, n)$.

EDIT: As pointed out by Chris Custer, the generalization is wrong and does not hold when the two elements are inverses of each other. I believe it works in this case since the orders are co-prime. If you're interested you can find a proof here: https://yutsumura.com/order-of-the-product-of-two-elements-in-an-abelian-group/. Sorry for the false statement.

Answer 2

The component cycles $(13)$ and $(254)$ are disjoint, so, hence, they commute. Thus the order of your element $\sigma$ is six; observe:

$$\begin{align} \sigma&=(13)(254),\\ \sigma^2&=(13)(254)(13)(254)=(13)(13)(254)(254)=(245),\\ \sigma^3&=(245)(13)(254)=(245)(254)(13)=(13),\\ \sigma^4&=(13)(13)(254)=(254),\\ \sigma^5&=(254)(13)(254)=(254)(254)(13)=(245)(13),\\ \sigma^6&=(245)(13)(13)(254)=(245)(254)={\rm id}_{S_5}. \end{align}$$

There is, thus, an error in your computations.