# Permutations. Help needed!

1. Sep 11, 2008

### sutupidmath

1. The problem statement, all variables and given/known data

Here are two problems that i am striving with:
1.
Let $$\theta$$ be the s-cycle (123...s).

(i) what is the smallest positive integer m such that $$\theta^h=\theta^k$$ when

h\equiv k(mod m)?

NOte: This is not actually a homework problem, but since the other one that i am gonna write down, is i posted this one here two, because i feel that if i can get this one right, then i can get the next one too.

2. Let $$\theta$$ be the s-cycle (123....s). what is the smallest integer r such that the set

$$\{(1), \theta , \theta^2,........,\theta^{r-1}\}$$ is closed under multiplication?

Well, i think the key problem that i am having here is to prove, or show that if

$$\theta$$ is an s-cycle, then $$\theta^s=(1)$$ where (1) is the identity permutation in standard form.

I am not sure whether this one comes directly from the def. of an s-cycle or there is a special way of prooving it.

Anyways here are my first thoughts about this last one. i think that it somes directly as a result of the def. of an s-cycle.

Let $$\theta =(a_1,a_2,.....,a_s)$$ be an s- cycle. Then

since $$\theta (a_i)=a_{i+1},i=1,2,.....,s-1$$, and $$\theta(a_s)=a_1$$

SO, now the reason that $$\theta^s=(1)$$ i think is that, if we start operatin with $$\theta$$ in the firs element of the s-cycle $$a_1$$ then we get:

$$\theta(a_1)=a_2, \theta(a_2)=a_3,=>\theta(\theta(a_1))=\theta^2(a_1)=a_3,......,\theta^{s-1}(a_1)=a_s$$ but since $$\theta(a_s)=a_1=>\theta^s(a_1)=a_1$$

And similarly with other elements of the cycle.

This way we notice that $$\theta$$ has to opertate s times in each element of the cycle in order to go back to that same element. Hence i think that this is the reason that

$$\theta^s=(1)$$ , or am i wrong? Anyways try to clarify this a lill bit for me please.

So, now let's go back to the first problem. If what i just did holds, i mean if the proof is correct, then here it is how i tackled the first problem:

$$\theta^h=\theta^k$$ when

$$h\equiv k(mod m)$$=> we get that

$$h-k=nm, n\in Z=>h=k+nm$$ hence we get:

$$\theta^h=\theta^{k+nm}=\theta^k\theta^{nm}=\theta^k(\theta^m)^n$$ so in order for $$\theta^h=\theta^k$$ we need to have m=s, since then we would get

$$\theta^s\theta^k=(1)\theta^k=\theta^k$$ Is this even close to the right way of approaching this problem, or i am way off?

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 12, 2008

### sutupidmath

So no input on this one?

3. Sep 13, 2008

### morphism

You almost have the right idea. Proving that $\theta^s =1$ isn't enough (your proof is fine, by the way) - you also need to prove that if 0<r<s, then $\theta^r \neq 1$.

4. Sep 13, 2008

### sutupidmath

Holy, crap!! hehe.. this is exactly what i did! Only that i proved that $$\theta^r \neq 1$$ for any other r. But basically this is what i did.
I did this after i posted the problem here though, but before i turned in the hw.

Last edited: Sep 13, 2008