Compositions in Symmetric Groups

A composition is the operation that combines the bijective correspondences of two symmetric groups within the same reference set X.

In simpler terms, a composition is the outcome of multiplying two permutations together.

Example

Consider a set X

$$ X = ( 1, 2, 3, 4, 5 ) $$

and two bijective correspondences

$$ \sigma:X \rightarrow X $$

$$ t:X \rightarrow X $$

where

$$ \sigma = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 4 & 5 & 2 & 1 & 3 \end{pmatrix} $$

$$ t = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 3 & 1 & 5 & 2 & 4 \end{pmatrix} $$

The composition σ * t is

$$ \sigma * t = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 2 & 4 & 1 & 3 & 5 \end{pmatrix} $$

For the sake of brevity, from now on I will write σ * t using the notation σt without indicating the operator symbol.

Explanation. Being a composition of the type f[g(x)], or σ[t(x)], I must first compute the inner permutation t (the second) and then the outer permutation σ (the first).
how to calculate the composition of two permutations
After determining the transition of the first element in the permutation t, I proceed sequentially to the second element of permutation t.
the second transition
Then to the third element of permutation t and so on, until a new position has been assigned to each element.
the third transition
the fourth transition
the fifth and final transition

Composition is a non-abelian operation because it does not respect the commutative property.

Switching the order of the permutations results in a different composition.

$$ t\sigma = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 \\ 2 & 4 & 1 & 3 & 5 \end{pmatrix} $$

To save time, I will only demonstrate the first result rather than going through all the steps to achieve the latter result.

verification

Note. Only if n ≤ 2, meaning if there are two or just one element in set X, is the composition an abelian (commutative) operation. For instance, consider these two permutations with two elements each: $$ \sigma = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} $$ $$ t = \begin{pmatrix} 1 & 2 \\ 1 & 2 \end{pmatrix} $$ The result of the composition σt $$ \sigma t = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} $$ is the same as obtained in the composition $$ t \sigma = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} $$ Hence, the operation satisfies the commutative property.

And so on.

 
 

Please feel free to point out any errors or typos, or share suggestions to improve these notes. English isn't my first language, so if you notice any mistakes, let me know, and I'll be sure to fix them.

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