Question

Prove that a subgroup H of a solvable group G is solvable.

Short answer

The subgroup H is sovable over the solvable group G .

Step-by-step solution

Defintion of homomorphism group. 

A homomorphism of a group is termed a monomorphism and injective homomorphism and it is satisfy these condition: injective as a map of sets,its kernel is trivial,it is a monomorphism with respect to the subgroup of the group.

Show that H is solvable.

Consider the group${\mathrm{H}}_{\mathrm{i}}=\mathrm{H}\cap {\mathrm{G}}_{\mathrm{i}}$. Here,$G$ is the group and$\mathrm{H}$is the subgroup over$G$. Let group is solvable that is

$\begin{array}{c}G=G_0\supseteq G_1\supseteq \mathrm{....}\supseteq G_n\\ =〈e〉\end{array}$

If the group is solvable so the series is also solvable and the series of group is the series of subgroup by the definition of solvbility. Now for each element${\mathrm{H}}_{\mathrm{i}}=\mathrm{f}\left({\mathrm{G}}_{\mathrm{i}}\right)$and the chain of subgroup is,

$\begin{array}{c}G=H_0\supseteq H_1\supseteq \mathrm{....}\supseteq H_n\\ =f〈e_0〉\\ =〈e_H〉\end{array}$

Assume that$H_i$is a normal subgroup of$H_{i-1}$for each element and$i=1,2,\mathrm{3.......}t$and consider the any two element$a,b$of a subgroup$H_{i-1}$and another two element$c,d$is the element of subgroup$G_{i-1}$. Therefore,$f\left(c\right)=a$and$f\left(d\right)=b$. Since the group$G_{i-1}/G_i$is abelian through the solvability now element$cd{c}^{-1}{d}^{-1}\in G_i$. Consequently

$\begin{array}{c}ab{a}^{-1}{b}^{-1}=f\left(c\right)f\left(d\right)f\left(c^{-1}\right)f\left(d^{-1}\right)\\ =f\left(cd{c}^{-1}{d}^{-1}\right)\in f\left(G_i\right)\\ =H_i\end{array}$

Hence, the subgroup${\mathrm{H}}_{\mathrm{i}-1}/{\mathrm{H}}_{\mathrm{i}}$is abelian if and only if$ab{a}^{-1}{b}^{-1}\in H$for all$a,b\in G$

And the subgroup $H$ of the group $G$ is solvable.