Question

Cayley’s Theorem 7.21 represents a group G as a subgroup of the permutation group A(G). A more efficient way of representing G as a permutation group arises from the following generalized Cayley’s Theorem. Let K be a subgroup of G and let T be the set of all distinct right cosets of K.

If$\mathit{a}\mathbf{\in }\mathit{G}$ , show that the map${\mathit{f}}_{\mathbf{a}}\mathbf{:}\mathit{T}\mathbf{\to }\mathit{T}$ given by${\mathit{f}}_{\mathbf{a}}\mathbf{\left(}\mathbf{K}\mathbf{b}\mathbf{\right)}\mathbf{=}\mathit{K}\mathit{b}\mathit{a}$ is a permutation set of the set T.

Short answer

It is proved that the map$f_a:T\to T$ given by$f_a\left(Kb\right)=Kba$ is a permutation set of the set T.

Step-by-step solution

Step by Step Solution Step 1: Cayley’s Theorem

Every group G is Isomorphic to a group of permutations.

Proving that map fa:T→ T given by fa(Kb) = Kba is a permutation set of the set T

Let a be any arbitrary constant such that $a\in G$.

Consider the map $f_a:T\to T$given by role="math" localid="1652767775009" ${\mathrm{f}}_{\mathrm{a}}\left(\mathrm{Kb}\right)=\mathrm{Kba}$.

To show that it is a permutation set of set T we have to prove that it is isomorphic to set T.

1. Proving homomorphism:

Consider arbitrary elements $b,c\in G$

$\begin{array}{c}{f}_a\left(K(b\circ c)\right)=K(b\circ c)a\\ =Kba\circ Kca\\ =f_a\left(Kb\right)\circ f_a\left(Kc\right)\end{array}$

Therefore,$f_a$is a homomorphism.

2. Proving Surjective:

Consider $Kb\in T$be an arbitrary element.

Take the right coset $Kb{a}^{-1}$as:

$\begin{array}{c}f\left(kb{a}^{-1}\right)=Kb{a}^{-1}a\\ =Kb\end{array}$

This implies,$f_a$is surjective.

Since $f_a$ is a surjective homomorphism, by First Isomorphism Theorem, it is an isomorphism.

Therefore, byCayley’s Theorem, we can conclude that$f_a\left(Kb\right)=Kba$ is a permutation set of the set T.