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 kathrynag Oct12-10 05:13 PM

Permutations, cycles

1. The problem statement, all variables and given/known data
Let t be an element of S be the cycle (1,2....k) of length k with k<=n.
a) prove that if a is an element of S then ata^-1=(a(1),a(2),...,a(k)). Thus ata^-1 is a cycle of length k.
b)let b be any cycle of length k. Prove there exists a permutation a an element of S such that ata^-1=b.

2. Relevant equations

3. The attempt at a solution
We assume t is an element of S and a is an element S.
By definition of elements of S if t is in S, we have a determined by t(1), t(2),...,t(n)
Furthermore if a is in S, we have a(1), a(2)....a(n).
That's as far as I get.

 tiny-tim Oct12-10 05:21 PM

hi kathrynag! :smile:

just read out that equation in English …
Quote:
 Quote by kathrynag (Post 2930146) ata^-1=(a(1),a(2),...,a(k))
… so what, for example, does ata^-1 do to a(2) ? :wink:

 kathrynag Oct12-10 05:33 PM

Re: Permutations, cycles

Does it go to 2?

 kathrynag Oct12-10 05:49 PM

Re: Permutations, cycles

Ok if a is defined by 1--->a(1), 2-----a(2),k--->a(k), then a^-1 is defined as a(1)-->1,a(2)--->2, a(k)--->k
Then ata^-1 for a(2) is a(t(2))=a(2)

 tiny-tim Oct12-10 05:49 PM

Quote:
 Quote by kathrynag (Post 2930170) Does it go to 2?

a-1
sends it to 2 …

so what does ata-1 send it to?

 kathrynag Oct12-10 05:51 PM

Re: Permutations, cycles

a(t(2))
t(2)=2
a(2)

 tiny-tim Oct12-10 05:58 PM

Quote:
 Quote by kathrynag (Post 2930146) Let t be an element of S be the cycle (1,2....k)
Quote:
 Quote by kathrynag (Post 2930201) a(t(2)) t(2)=2 a(2)
ah! no, you're misunderstanding the notation for a cycle …

(1,2....k) means that it sends 1 to 2, 2 to 3, … and k to 1. :wink:
so t(2) = … ?

 kathrynag Oct12-10 05:59 PM

Re: Permutations, cycles

3
so we are left with a(3)=4

 tiny-tim Oct12-10 06:09 PM

Quote:
 Quote by kathrynag (Post 2930218) 3 so we are left with a(3)=4
no!!

you're not told anything about a, are you?

a(3) is just a(3) ! :bigginr:

(and I'm off to bed :zzz: … see you tomorrow!)

 kathrynag Oct12-10 06:21 PM

Re: Permutations, cycles

I thought I could do this:
By definition of elements of S if t is in S, we have a determined by t(1), t(2),...,t(n)
Furthermore if a is in S, we have a(1), a(2)....a(n).

 kathrynag Oct12-10 07:44 PM

Re: Permutations, cycles

So I have ata^-1=at(n)
because a^-1 sends a(1)--->1,a(2)--->2,.....a(n)--->n
By defininition of t, we have a(1), a(2), a(3)...a(k). We go to k because our definition of t says k<=n.

For b,
Let b be any cycle of length k.
We have (1,2.....k). I'm not sure how to show the rest.

 tiny-tim Oct13-10 03:28 AM

hi kathrynag! :smile:

(just got up :zzz: …)
Quote:
 Quote by kathrynag (Post 2930353) So I have ata^-1=at(n) because a^-1 sends a(1)--->1,a(2)--->2,.....a(n)--->n By defininition of t, we have a(1), a(2), a(3)...a(k).
do you mean we have (a(1), a(2), a(3),...a(k))?

anyway, before we go any further, i need to know: what exactly is S? :confused:

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