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Please help calculate the order of x and y. Let x and y denote permutations of $N(7)$ - Natural numbers mod 7 . Cycle notation:

$$x= (15)(27436) $$

$$y= (1372)(46)(5)$$

Thanks

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You may want to try to prove that the order of a product of disjoint cycles is the lowest common multiple of the cycles' lengths. –  DonAntonio Jan 30 '13 at 3:13

3 Answers 3

up vote 2 down vote accepted

Since you have the products of disjoint cycles, what do you know about the order of a cycle? For a single cycle, its order is equal to its length.

The order of a product of disjoint cycles is equal to the least common multiple $(\text{lcm})$ of the the orders of the cycles that form it, i.e., the least common multiple of the lengths of the cycles.

E.g. the order of $(1 2 3 4 5 6 7)$ is $7$. The order of $(123)(4567) = \text{lcm}\,(3, 4) = 12$.

The order of $(123)(456)(7) = \text{lcm}\,(3, 3, 1) = 3$.

Now, can you apply that to your permutations to find their orders?

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Okay, this makes more sense. So in X we have lcm(2,5) = 10. So I guess the order would be 10. My question is, is this saying it takes 10 cycles to get back to the identity (1,2,3,4,5,6,7)? I tried 10 cycles from 5 7 6 3 1 2 4, but didn't get the identify back. –  Allen Miller Jan 30 '13 at 4:21
1  
Yes, the order of X is indeed 10. applying x 10 times will get you to the identity. But (1,2,3,4,5,6,7) is not the identity. (1)(2)(3)(4)(5)(6)(7) is the identity. (1, 2, 3, 4, 5, 6, 7) is a 7-cycle which maps 1 to 2, 2 to 3, ... 6 to 7, and 7 to 1. The identity maps 1 to 1, 2 to 2...7 to 7. Note that for y, lcm(2, 4, 1) = 4. So if you take y^4, you will get the identity. –  amWhy Jan 30 '13 at 4:28
    
Thank you! That makes a lot of sense. Just one last question-- how do you combine cycles ex- (123) (123) . why does (123)(123)=(132). I understand it's the composite function. So you would end up with 1-> 3, 2->1, 3->2. Not sure how you get to (132). Thanks! –  Allen Miller Jan 30 '13 at 4:56
    
I'm a bit confused with your notation. Can you write out the steps to arrive at 132. Thanks so much. You have been very helpful! –  Allen Miller Jan 30 '13 at 5:05
    
Just accepted it. (sorry, i'm new to the site). If you could spell it out that would be great! –  Allen Miller Jan 30 '13 at 5:11

Hint: Find the least common multiple of the orders of each disjoint cycle.

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we start from left cycle. in the left cycle we have $1\to2$ and in the right cycle we have $2\to3$, so we deduce that $1\to3$.

now in the left cycle we have $3\to1$ and in the right cycle we have $1 \to 2$, so we deduce that 3$\to2$.

Finally in the left cycle we have$ 2\to3$ and in the right cycle we have $3 \to 1$, so we deduce that $2\to1$.

So we have in product of this two cycles $1\to3$ and $3\to2$ and $2\to1 $that is the cycle $(132)$.

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