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In $S_9$ I want to find the order for the permutation $(2,9,6,3,7)(4,8)$

My result is that the order should be 8. Wolfram Alpha tells me instead that it is 10.

Question: Why is the order of the given permutation 10?

According to this post I have done following:

$$ord \pi := min{\{ k \in \mathbb N | \pi^{k} = id_x\}}$$

so: $$ord((2,9,6,3,7)(4,8)) = lcm(n,m)$$ for: $$(2,9,6,3,7)^n=(1) \quad \& \quad (4,8)^m=(1)$$

for $(4,8)^n = (1):$ $${ \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 1 & 2 & 3 & 8 & 5 & 6 & 7 & 4 & 9 \\ \end{pmatrix} } \cdot {\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 1 & 2 & 3 & 8 & 5 & 6 & 7 & 4 & 9 \\ \end{pmatrix} } $$

$$={\begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ \end{pmatrix} } \quad n = 2$$

for $(2,9,6,3,7)^m = (1):$

$$ {\pi^0: \quad}{ \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 1 & 9 & 7 & 4 & 5 & 3 & 2 & 8 & 6 \\ \end{pmatrix}}$$

$$ {\pi^1: \quad}{ \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 1 & 6 & 2 & 4 & 5 & 7 & 9 & 8 & 3 \\ \end{pmatrix}}$$

$$ {\pi^2: \quad}{ \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 1 & 7 & 6 & 4 & 5 & 9 & 3 & 8 & 2 \\ \end{pmatrix}}$$

$$ {\pi^3: \quad}{ \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 1 & 3 & 9 & 4 & 5 & 2 & 6 & 8 & 7 \\ \end{pmatrix}}$$

$$ {\pi^4: \quad}{ \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \\ 1 & 9 & 7 & 4 & 5 & 3 & 2 & 8 & 6 \\ \end{pmatrix}} \quad m = 4$$

So according to:

So the order of σσ is exactly the smallest natural number n such that $(4,5)n=(1)(4,5)n=(1) and (2,3,7)n=(1)$$(2,3,7)n$$=(1)$ (think about this fact for a moment).

The order should be $n \cdot m = 2 \cdot 4$

Community
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jublikon
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    The order is $;10;$ : the product of the orders of the two cycles whose product is the permutation, because these cycles are disjoint – DonAntonio Jan 08 '17 at 15:24
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    You wrote the wrong exponents, where you wrote $\pi^0$, it's actually $\pi^1$ etc. So $\pi^5 = \operatorname{id}$. – Daniel Fischer Jan 08 '17 at 15:24
  • Some advice. When you have permutations written in cycle notation you should use that representation to do the arithmetic. The two line version is much harder to work with. – Ethan Bolker Jan 08 '17 at 15:26

2 Answers2

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Your calculation of the powers of $(2\;9\;6\;3\;7)$ is wrong. $\pi^0$ is always the identity permutation; only $\pi^1$ will be the cycle you have written down.

What you write as $\pi^0$ is actually $\pi$ itself; what you claim is $\pi^1$ is $\pi^2$, and what you claim is $\pi^2$ actually seems to be $\pi^4$; it looks like you're squaring each line to get the next line, instead of multiplying it by the original $(2\;9\;6\;3\;7)$.

Thus, the order of the 5-cycle should be $5$, not $4$.

hmakholm left over Monica
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$\sigma=(2~9~6~3~7)$ and $\phi = (4~8)$. Thus $ord(\phi)=2$ and $ord(\sigma)=5$

So $gcd(ord(\phi),ord(\sigma))=1$. Hence $ord(\phi\sigma)=ord(\phi) \cdot ord(\sigma)=2 \cdot 5 = 10$

BrianH
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idk
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