Powers of Permutations.

DgrayMan · **Joined:** Sat, 21 Jan 2012 03:59:22 UTC **Posts:** 182

I'm doing #3. sovalpha is of length s and its a cycle, alpha=(a1 a2....aS). so alpha inverse=(aS...a2 a1).
And now its saying, alpha inverse=alpha^(S-1) Now maybe im misunderstanding, but for example lets use Beta is a cycle of length 5 so B=(b1 b2 b3 b4 b5), does this mean that the inverse of Beta is also equal to (b1 b2 b3 b4) ,since its 5-1=4?

Shadow · **Posted:** Tue, 21 Feb 2012 01:58:13 UTC

DgrayMan wrote:

I'm doing #3. sovalpha is of length s and its a cycle, alpha=(a1 a2....aS). so alpha inverse=(aS...a2 a1).
And now its saying, alpha inverse=alpha^(S-1) Now maybe im misunderstanding, but for example lets use Beta is a cycle of length 5 so B=(b1 b2 b3 b4 b5), does this mean that the inverse of Beta is also equal to (b1 b2 b3 b4) ,since its 5-1=4?

No, one has order 5 the other has order 4, just because you raise something to a certain power doesn't mean you lose what things go to, if you look at the original permutation:

$(a_1\;\ldots\; a_s)^{s-1}$ by definition this means skip s-2

things and go to the next one, you should be able to see that this means the same thing as "go to the previous thing in the cycle listed" i.e. the inverse is $(a_s\; a_{s-1}\;\ldots\; a_1)$ , since cycles are the same up to cyclic permutation.

DgrayMan · **Joined:** Sat, 21 Jan 2012 03:59:22 UTC **Posts:** 182

so a^(s-1)= (a1 a3... as-2 as) ?

Shadow · **Posted:** Tue, 21 Feb 2012 03:57:03 UTC

DgrayMan wrote:

so a^(s-1)= (a1 a3... as-2 as) ?

No, it's what I wrote in my last post, and what you said you suspected it was in your first post.

DgrayMan · **Joined:** Sat, 21 Jan 2012 03:59:22 UTC **Posts:** 182

hmm. Is there somewhere where i can find examples. im still confused?

Shadow · **Posted:** Tue, 21 Feb 2012 05:02:54 UTC

DgrayMan wrote:

hmm. Is there somewhere where i can find examples. im still confused?

Examples of what?

DgrayMan · **Joined:** Sat, 21 Jan 2012 03:59:22 UTC **Posts:** 182

Well what about a permutation of length s= 5 . B=(1 2 3 4 5), and its inverse B^(-1)=(5 4 3 2 1)
what would B^(s-1) look like?

Shadow · **Posted:** Tue, 21 Feb 2012 07:46:37 UTC

DgrayMan wrote:

Well what about a permutation of length s= 5 . B=(1 2 3 4 5), and its inverse B^(-1)=(5 4 3 2 1)
what would B^(s-1) look like?

It would look like the same thing, we already established that for an s-cycle $\alpha$ , $\alpha^{-1}=\alpha^{s-1}$ , and inverses in a group are unique, so they must be the same thing.

outermeasure · **Posted:** Tue, 21 Feb 2012 14:43:35 UTC

Shadow wrote:

DgrayMan wrote:

so a^(s-1)= (a1 a3... as-2 as) ?

No, it's what I wrote in my last post, and what you said you suspected it was in your first post.

In fact, $(a_1\;a_3\;\dots\;a_{s-2}\;a_s)$ is not a power of a unless s=2.

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