Question

Prove that for $\mathit{n}\mathbf{\ge }\mathbf{5}\mathbf{,}{\mathit{A}}_{\mathbf{n}}$ is not solvable.

Short answer

$\mathit{n}\mathbf{\ge }\mathbf{5}\mathbf{,}{\mathit{A}}_{\mathbf{n}}$is not solvable.

Step-by-step solution

Definition of the abelian group.

An abelian group is a commutative group and satisfy the condition for all $\mathit{a}\mathbf{,}\mathit{b}\mathbf{}$in the field extension is$\mathbf{}\mathit{a}\mathbf{\cdot }\mathit{b}\mathbf{=}\mathit{b}\mathbf{\cdot }\mathit{a}$ .

Step-2: Showing that n≥5,An is not solvable.

Consider the n element of the same ordern which consist in the different subgroup let assume the order of the subgroup and the chain of the of the subgroup is

Let consist the three element $\mathbf{\left(}\mathbf{rst}\mathbf{\right)}$be any 3 cycle in ${\mathbf{S}}_{\mathbf{n}}\mathbf{}\mathbf{}$and the two element $\mathbf{}\mathbf{u}\mathbf{,}\mathbf{v}\mathbf{}\mathbf{}$is the element of the set of the subgroup of order role="math" localid="1657971913247" $\mathit{n}$ and element of the subgroup is role="math" localid="1657971882943" $\left\{1,2,.....n\right\}$ and thsese two element are the exist because $\mathbf{n}\mathbf{\ge }\mathbf{5}$. Since ${\mathit{s}}_{\mathbf{n}}$over ${\mathbf{G}}_{\mathbf{i}}$ is abelian.

Now the subgroup one is contain two element $\mathit{a}\mathbf{,}\mathit{b}$ and $\mathbf{a}\mathbf{=}\left\{\mathrm{tus}\right\}\mathbf{,}\mathbf{b}\mathbf{=}\left\{\mathrm{srv}\right\}$

Then

$$\begin{gathered} \mathbf{\left\{}\mathbf{tus}\mathbf{\right\}}\mathbf{\left\{}\mathbf{srv}\mathbf{\right\}}{\mathbf{\left\{}\mathbf{tus}\mathbf{\right\}}}^{\mathbf{-}\mathbf{1}}\mathbf{=}{\mathbf{\left\{}\mathbf{srv}\mathbf{\right\}}}^{\mathbf{-}\mathbf{1}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{\left\{}\mathbf{tus}\mathbf{\right\}}\mathbf{\left\{}\mathbf{srv}\mathbf{\right\}}\mathbf{\left\{}\mathbf{tus}\mathbf{\right\}}\mathbf{\left\{}\mathbf{srv}\mathbf{\right\}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{\left(}\mathbf{rst}\mathbf{\right)} \end{gathered}$$

Therefore the first subgroup ${\mathbf{G}}_{\mathbf{1}}$ contains all the 3 cycle since ${\mathbf{G}}_{\mathbf{2}}$ is abelianover ${\mathbf{G}}_{\mathbf{1}}$. Repeat the process for $\mathit{n}$ order. Hence for each subgroup is abelian and all are contain the 3 cycle which is opposition. Therefore ${\mathit{s}}_{\mathbf{n}}$ is not solvable.