Question

Prove Cayley’s Theorem by applying parts (b) and (c) with $\mathit{K}\mathbf{=}\mathbf{⟨}\mathbf{e}\mathbf{⟩}$.

Short answer

Cayley’s Theorem is proved.

Step-by-step solution

Step by Step Solution Step 1: Cayley’s Theorem

Every group G is Isomorphic to a group of permutations.

Referring to results of parts (b) and (c) of the question

• $\mathit{\phi }\mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{=}{\mathit{f}}_{{\mathbf{a}}^{\mathbf{-}\mathbf{1}}}$is a homomorphism of groups whose kernel is contained in K.
• $\mathit{K}\mathbf{=}\mathit{K}\mathit{e}\mathit{r}\mathit{n}\mathit{e}\mathit{l}\mathit{\phi }$
  

Proving Cayley’s Theorem

As we already proved in part (b) that $\phi$is a homomorphism if $K=⟨e⟩$.

Then, $G=A\left(T\right)$.

Since $K=⟨e⟩$, from the result of part (c) $\ker \phi =K=⟨e⟩$.

Therefore, it is surjective.

So, from theFirst Isomorphism Theorem, group G is isomorphic to A(T).

This implies that group G is isomorphic to the group of permutations.

Hence, Cayley’s Theorem is proved.