Question

If G is a simple nonabelian group prove that G is not solvable.

Short answer

G is not solvable.

Step-by-step solution

Showing that every abelian is solvable.

Let G be a group and ${\mathrm{H}}_1,{\mathrm{H}}_2,.......{\mathrm{H}}_{\mathrm{n}}$ are sungroupof G. If

$$\begin{gathered} \mathrm{H}={\mathrm{H}}_1\le {\mathrm{H}}_2\le ,.......\le {\mathrm{H}}_{\mathrm{n}} \\ =\mathrm{G} \end{gathered}$$

Then G is solvable. Where ${\mathrm{H}}_{\mathrm{i}}$ are normal in ${\mathrm{H}}_{\mathrm{i}+1}$ and $\frac{{\mathrm{H}}_{\mathrm{i}+1}}{{\mathrm{H}}_{\mathrm{i}}}$ is abelian.

So every abelian is solvable.

Step-2: Showing that G is not solvable.

Let $G=S_n$. As $A_n$ is subgroup of $S_n$ for $n\ge 5$ $A_n$is non abelian and ${\mathrm{S}}_{\mathrm{n}}^{\text{'}}\subset {\mathrm{A}}_{\mathrm{n}}$so role="math" localid="1657972993738" $n\ge 5$any 3-cycle is a commutator of the form

$$\begin{gathered} \left[\left(\mathrm{jkv}\right),\left(\mathrm{ikr}\right)\right]\mathbf{=}\left(\mathrm{vkj}\right)\left(\mathrm{rki}\right)\left(\mathrm{jkv}\right)\left(\mathrm{ikr}\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\left(\mathrm{vkj}\right)\left(\mathrm{jiv}\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\left(\mathrm{ikj}\right) \end{gathered}$$

This implies ${\mathrm{S}}_{\mathrm{n}}^{\text{'}}={\mathrm{A}}_{\mathrm{n}}^{\mathrm{i}}={\mathrm{A}}_{\mathrm{n}}$

Hence proved $A_n$ and $S_n$ are not solvable for$\mathrm{n}\ge 5$ .

Hence proved G is not solvable.