Question

Let $\mathit{k}\mathbf{=}\mathbf{\{}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{,}\mathbf{\left(}\mathbf{12}\mathbf{\right)}\mathbf{\left(}\mathbf{34}\mathbf{\right)}\mathbf{,}\mathbf{\left(}\mathbf{13}\mathbf{\right)}\mathbf{\left(}\mathbf{24}\mathbf{\right)}\mathbf{,}\mathbf{\left(}\mathbf{14}\mathbf{\right)}\mathbf{\left(}\mathbf{23}\mathbf{\right)}\mathbf{\}}$. Show that$\mathit{k}$ is a subgroup of${\mathit{A}}_{\mathbf{4}}$ and hence, a subgroup of ${\mathit{S}}_{\mathbf{4}}$.

Short answer

It is shown that$k$ is subgroup of $A_4$, thus, subgroup of $S_4$.

Step-by-step solution

Determine k is subgroup of S4

Consider the given function,

$k=\{\left(1\right),\left(12\right)\left(34\right),\left(13\right)\left(24\right),\left(14\right)\left(23\right)\}$

Result

By using theorem 7.12, Let $H$be a nonempty finite subset of a group $G$. If$H$ is closed under the operation in $G$, then $H$is a subgroup of $G$.

Then, therefore $k$is subgroup of $S_4$.