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Consider the group $S_8$. What is the order of $\sigma = (4,5)(2,3,7)$ and $\tau = (1,4)(3,5,7,8)$?

My book says I should just use a trick by the order of a permutation expressed as a product of disjoint cycles is the least common multiple of the lengths of the cycles.

I don't understand the trick very well and would like to see how the counting process here actually works. I can count for non disjoint ones like say $(1,2,3,8)$ has order 4

I don't know how to count $\sigma = \begin{pmatrix} 1 &2 &3 &4 &5 &6 &7 &8 \\ 1 &3 &7 &5 &4 &6 &2 &8 \end{pmatrix}$

EDIT Just so I am clear, the answer for the order $\sigma$ is 6 and $\tau$ is 4. I am still confused as to how they got this

I have an extremely poor understanding of this subject so please help me! Thank you

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2  
It's not a trick, it's a perfectly respectable, efficient, elegant way to determine the order of a permutation, and learning about it will teach you a lot more than winding your way through numbers by brute force. –  joriki Nov 6 '12 at 3:09
    
No I am asking the question because I want to know how on earth did the answer come up with this conclusion in the first place –  jip Nov 6 '12 at 3:24

2 Answers 2

up vote 8 down vote accepted

So take $\sigma = (4,5)(2,3,7)$. The order, by definition, is the the smallest natural number $n$ such that $\sigma^n = (1)$ (i.e. the identity element in the group, i.e. the element that sends ever number to itself). Since the cycles $(4,5)$ and $(2,3,7)$ are disjoint you have

$$\begin{align} \sigma^n &= (4,5)(2,3,7)(4,5)(2,3,7)\dots (4,5)(2,3,7)(4,5)(2,3,7)\\ &= (4,5)(4,5)\dots (4,5)(4,5)(2,3,7)(2,3,7)\dots (2,3,7)(2,3,7)\\ &= (4,5)^n(2,3,7)^n \end{align}$$

(note that the two elements $(4,5)$ and $(2,3,7)$ commute).

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

But what is the order of a each of $(4,5)$ and $(2,3,7)$?

When the order of $(4,5)$ is two exactly because $(4,5)^2 = (4,5)(4,5) = (1)$. The order of $(2,3,7)$ is $3$ because $$\begin{align} (2,3,7)^1 &= (2,3,7) \\ (2,3,7)^2 &= (1,7,3) \\ (2,3,7)^3 &= (1) \end{align} $$

Now it is not to hard to see that the order of $\sigma$ is exactly the least common multiple of $2$ and $3$ (since we need both both $(4,5)^m = (1)$ and $(2,3,7)^m = (1)$ and the smallest $m$ where this happens is exactly the least common multiple). Hence the final answer is $6$.


Addendum: I just wanted to add a bit about orders of these elements. First note that for example the lement $(4,5)$ is just the element $$ (4,5) = \begin{pmatrix} 1 & 2 &3 & 4 & 5 & 6 & 7 & 8 \\ 1 & 2 & 3 &5 & 4& 6 & 7 & 8 \end{pmatrix}. $$ (Hence $4$ maps to $5$ and $5$ to $4$). So when you compose (multiply) the element with itself, then you get $$ \begin{align} (4,5)(4,5) &= \begin{pmatrix} 1 & 2 &3 & 4 & 5 & 6 & 7 & 8\\ 1 & 2 & 3 &5 & 4& 6 & 7 & 8 \end{pmatrix}\begin{pmatrix} 1 & 2 &3 & 4 & 5 & 6 & 7 & 8\\ 1 & 2 & 3 &5 & 4& 6 & 7 & 8 \end{pmatrix} \\ &= \begin{pmatrix} 1 & 2 &3 & 4 & 5 & 6 & 7 & 8\\ 1 & 2 & 3 &4 & 5& 6 & 7 & 8 \end{pmatrix} = (1) \end{align} $$ (I usually write the identity as $(1)$).

This means that the order of $(4,5)$ is $2$. Likewise you find that $$ (2,3,7) = \begin{pmatrix} 1 & 2 &3 & 4 & 5 & 6 & 7 & 8\\ 1 & 3 & 7 &4 & 5& 6 & 2 & 8 \end{pmatrix}. $$

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Sorry if I am confusing something here, but what do you mean by smallest natural number s.t. $\sigma^n = (4,5)^n (2,3,7)^n = (1)$? I know you are leading me to the hint the book is giving, but (maybe this is because I am still not understanding this) doesn't it only like one cycle to map to 1? That is one maps to one? –  jip Nov 6 '12 at 3:28
    
@jak: Sorry about that. All I mean is that the order of an element $\sigma (\neq 1)$ is exactly the smallest natural number $n$ such that $\sigma^n = \sigma\circ\dots\circ\sigma = (1)$ (i.e. is equal to the identity element in the group.) –  Thomas Nov 6 '12 at 3:30
    
I am still probably not understanding this correctly, but no matter how many times you nest $\sigma$, it's still only going to take exactly once to map back to one since one is fixed –  jip Nov 6 '12 at 3:35
    
OKay this is the same where I was confused "$(4,5)(4,5) = (1)$" How did you get this? To me, it reads like this, 5 maps to 4, 4 maps back to 5. How did (1) appear? –  jip Nov 6 '12 at 3:39
    
@jak: Ahh I see. What I mean by $(4,5)(4,5)$ is exactly that $4$ maps to $4$ and $5$ maps to $5$ (and also $1$ to $1$ and ...). Hence the product of the two sends ever number to itself. Hence the product is the identity element in the group $S_8$. I often denote the identity element by $(1)$, but this might not be standard. –  Thomas Nov 6 '12 at 3:43

Prove that easy lemma: the order of a permutation expressed as a product of disjoint cycles is the minimal common multiple of the lengths of the cycles.

To prove the above you may want to use the fact (also easily provable) that two disjoint cycles commute...

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Easy is in the eye of the beholder. –  joriki Nov 6 '12 at 3:12
    
Indeed: I'm trying to push forward the OP into trying his strength since, imo, the beholder here is anyone having a minimal good understanding of cycles and their orders. –  DonAntonio Nov 6 '12 at 3:13
    
The OP claims to "have an extremely poor understanding of this subject." –  Jesse Madnick Nov 6 '12 at 3:18
    
Yes, he does...and thus he needs to make an effort to enhance that understanding of his of the basics, otherwise it'd be like he's expecting to get a complete class on this subject here, which imo would be non-realistic. –  DonAntonio Nov 6 '12 at 3:20
    
One further hint for the OP: in any group $\,G\,$, if there are two elements $\,x\,,\,y\in G\,$ that commute, then $\,(xy)^n=x^ny^n\,\,,\,\,\forall\,n\in\Bbb N\,$ –  DonAntonio Nov 6 '12 at 3:21

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