- #1
sutupidmath
- 1,630
- 4
Homework Statement
Here are two problems that i am striving with:
1.
Let [tex]\theta[/tex] be the s-cycle (123...s).
(i) what is the smallest positive integer m such that [tex] \theta^h=\theta^k[/tex] when
h\equiv k(mod m)?
NOte: This is not actually a homework problem, but since the other one that i am going to 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 [tex] \theta[/tex] be the s-cycle (123...s). what is the smallest integer r such that the set
[tex] \{(1), \theta , \theta^2,...,\theta^{r-1}\}[/tex] is closed under multiplication?
Well, i think the key problem that i am having here is to prove, or show that if
[tex] \theta[/tex] is an s-cycle, then [tex] \theta^s=(1)[/tex] 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 [tex] \theta =(a_1,a_2,...,a_s)[/tex] be an s- cycle. Then
since [tex] \theta (a_i)=a_{i+1},i=1,2,...,s-1[/tex], and [tex] \theta(a_s)=a_1[/tex]
SO, now the reason that [tex] \theta^s=(1)[/tex] i think is that, if we start operatin with [tex]\theta[/tex] in the firs element of the s-cycle [tex] a_1[/tex] then we get:
[tex]\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[/tex] but since [tex] \theta(a_s)=a_1=>\theta^s(a_1)=a_1[/tex]
And similarly with other elements of the cycle.
This way we notice that [tex]\theta[/tex] 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
[tex] \theta^s=(1)[/tex] , 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:
[tex] \theta^h=\theta^k[/tex] when
[tex] h\equiv k(mod m)[/tex]=> we get that
[tex] h-k=nm, n\in Z=>h=k+nm[/tex] hence we get:
[tex] \theta^h=\theta^{k+nm}=\theta^k\theta^{nm}=\theta^k(\theta^m)^n[/tex] so in order for [tex] \theta^h=\theta^k[/tex] we need to have m=s, since then we would get
[tex] \theta^s\theta^k=(1)\theta^k=\theta^k[/tex] Is this even close to the right way of approaching this problem, or i am way off?
SO, any advice?