Question

(a) Show that${\mathbf{S}}_{\mathbf{4}}$is a solvable group. [Hint: Consider the subgroup $\mathbf{H}\mathbf{=}\left\{\left(12\right)\left(34\right),\left(13\right)\left(24\right),\left(14\right)\left(23\right),\left(1\right)\right\}\mathbf{}\mathbf{of}\mathbf{}{\mathbf{A}}_{\mathbf{4}}$of ]

(b) Show that ${\mathit{D}}_{\mathbf{4}}$ is a solvable group.

Short answer

(a) It is proved that$S_4$ is a solvable group.

(b) It is proved that$D_4$ is a solvable group.

Step-by-step solution

Solvable Group

Suppose that a group be Gand its subgroups are ${\mathit{H}}_{\mathbf{1}}\mathbf{,}{\mathit{H}}_{\mathbf{2}}\mathbf{,}{\mathit{H}}_{\mathbf{3}}\mathbf{,}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{,}{\mathit{H}}_{\mathbf{n}}$So, if the condition $\mathit{H}\mathbf{=}{\mathit{H}}_{\mathbf{1}}\mathbf{\le }{\mathit{H}}_{\mathbf{2}}\mathbf{\le }{\mathit{H}}_{\mathbf{3}}\mathbf{\le }\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{\le }{\mathit{H}}_{\mathbf{n}}\mathbf{=}\mathit{G}$, then we can say that Gis a solvable group, and also Hare normal in${\mathit{H}}_{\mathbf{i}\mathbf{+}\mathbf{1}}$. We can say also suggest that$\frac{{\mathbf{H}}_{\mathbf{i}\mathbf{+}\mathbf{1}}}{{\mathbf{H}}_{\mathbf{i}}}$is abelian.

Proof of part (a)

To prove that$s_4$is solvable, we have to suppose that .

$H_0=S_{4,}{H}_1=A_4,H_2=\left\{r,\left(12\right)\left(34\right),\left(13\right)\left(24\right),\left(14\right)\left(23\right)\right\},H_3=\left\{e\right\}$

Here, if we look at the order of the group $\frac{H_0}{H_1}$, which is equal to 2, the order of the group $\frac{H_2}{H_3}$is equal to 3, and the order of the group$\frac{H_2}{H_3}\simeq H_2$is equal to 4.

Each and every group written above are quotient groups and we can say that these groups are abelian.

Therefore, it is proved that$S_4$is solvable.

Proof of part (b)

Suppose there is a rotation subgroup R in $D_4$ This means R is a normal subgroup of $D_4$.

The above statements imply that $\frac{D_4}{R}\cong Z_2$. This means the order of $\frac{D_4}{R}$is 4 and this means $\frac{D_4}{R}$is an abelian group and R is also an abelian group.

Therefore, we can say that $D_4\supseteq R\supseteq \left\{e_{D_4}\right\}$.

Hence, it is proved that $D_4$is solvable group.