Permutations: Explaining & Finding A, A^-1 & 2 Examples of σ

[katesmith410]

Let n ∈ N. Explain what is meant by saying that π is a permutation of n = {1, 2, . . . , n}.
The permutation A is given below in two line notation. Write A in disjoint cycle notation.

1 2 3 4 5
3 5 1 2 4

Find A^-1, writing your answer in disjoint cycle notation. Give two examples of permutations σ of 5 which satisfy σ = σ^-1 but which have different cycle structure.

I am able to write A in disjoint cycle notation and find the inverse. However I am having problems with the final part. Would it make sense for σ to be the identity permutation since
1->1, 2->2, 3->3 etc. Unsure of another example though, any guidance towards an answer would be great, thanks!

 

[micromass]

Hi katesmith!

The identity permutation is good here! Let's find another one. Let's assume for simplicity that the disjoint cycle notation of sigma has only one nontrivial cycle. For example,

\sigma=(1~2)~\text{or}~\sigma=(1~2~3~4)

For which such permutations does it holds that \sigma=\sigma^{-1}??

 

[katesmith410]

Hi :)

Is it right to say that for your given example that,

σ=(12)=σ^-1

When σ=(1234), I know that σ^-1=(1432). But I am unsure what permutation holds for σ=σ^-1.

 

[micromass]

So, you've figured out that \sigma^{-1}=\sigmafor transpositions (1 2). Does it hold for every transposition? Can you use that information?

 

[katesmith410]

Surely it can't hold for every transposition, as in the case for the transposition σ=(1234), σ≠σ^-1?

 

[micromass]

Surely it can't hold for every transposition, as in the case for the transposition σ=(1234), σ≠σ^-1?

A transposition is something that just exchanges two elements. So (1 2) and (2 3) are transpositions, but (1 2 3 4) is not. I should have said that...

 

[katesmith410]

Oh right, I think i understand now! So am I right in saying then that for every transposition σ=σ^-1?
So for my question, σ=(13)(254) but can i also say that σ=(13)(25)(54)(42)? Therefore σ would be equal to its inverse?