Question

Let $\mathit{N}\mathbf{,}\mathit{H}$ be subgroups of a group$\mathit{G}$ .$\mathit{G}$ is called the semidirect product of$\mathit{N}$ and$\mathit{H}$ if$\mathit{N}$ is normal in $\mathit{G}$,$\mathit{G}\mathbf{=}\mathit{N}\mathit{H}$ , and$\mathit{N}\mathbf{\cap }\mathit{H}\mathbf{=}\mathbf{⟨}\mathit{e}\mathbf{⟩}$ . Show that each of the following groups is the semidirect product of two of its subgroups:${\mathbf{S}}_{\mathbf{4}}$

Short answer

It has been proved that,$S_4$ is the semidirect product of two of its subgroups.

Step-by-step solution

Step-by-Step SolutionStep 1: Definition of Semidirect Product

A group$G$ is called the semidirect product of its subgroups$N$ and$H$ if$N$ is normal in$G$ ,$G=NH$ , and$N\cap H=⟨e⟩$ .

Prove that S4 is the semidirect product of two of its subgroups

Consider $G$as the symmetric group$S_4$ .

Then,$\left|S_4\right|=24$ .

Consider its subgroups$N=A_4$ and $H=\left\{\right(),(1,2)\}$.

Here,$N$ is normal in $G$,$G=NH$ , and$N\cap H=⟨e⟩$ .

Thus, it can be concluded that$S_4$ is the semidirect product of two of its subgroups.