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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?

Shaun
  • 44,997
  • Yes. Note that disjoint cycles commute and consider the order of each cycle. You should be able to find or discover a theorem relating the lengths of the cycles in a disjoint cyclic representation of a permutation to the order of the permutation itself. – JMoravitz Aug 11 '20 at 19:02
  • "but this didn't quite work" It would have if you were patient enough and avoided mistakes. This, however, is not the recommended approach as it does not generalize nicely to much larger questions. Using the approach alluded to before you could find the order of $(1~2~3~\cdots 49)(50~51~52~\cdots 100)$ in $S_{100}$ almost instantaneously rather than having to go through a few thousand iterations. – JMoravitz Aug 11 '20 at 19:08
  • https://math.stackexchange.com/q/1008610/648587 , https://math.stackexchange.com/q/67180/648587 , https://math.stackexchange.com/q/3481018/648587 , https://math.stackexchange.com/q/290281/648587 – Kumar Aug 12 '20 at 18:52

2 Answers2

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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.

Akash Kulkarni
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    The generalization is incorrect. Suppose $g_1g_2=e$. –  Aug 11 '20 at 19:33
  • @ChrisCuster I do think your concern may be addressed by a simple Modification $\frac{\text{LCM}(m,n)}{\text{GCD}(m,n)}$. Akash, your belief is true. – Kumar Aug 11 '20 at 19:57
  • https://math.stackexchange.com/q/67180/648587 – Kumar Aug 11 '20 at 20:02
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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.

Shaun
  • 44,997
  • Thanks! By disjoint you mean what exactly here? –  Aug 11 '20 at 20:37
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    You're welcome, @Daniel. Two permutations, $(\sigma_{1, 1}\dots\sigma_{1,s_1})\dots(\sigma_{n, 1}\dots\sigma_{n,s_s})$ and $(\tau_{1,1}\dots\tau_{1,t_1})\dots(\tau_{m, 1}\dots\tau_{m,t_t})$, are disjoint if the sets ${\sigma_{1, 1},\dots,\sigma_{1,s_1},\dots ,\sigma_{n, 1}\dots\sigma_{n,s_s}}$ and ${\tau_{1,1},\dots,\tau_{1,t_1},\dots,\tau_{m, 1},\dots,\tau_{m,t_t}}$ share no elements. – Shaun Aug 12 '20 at 00:20