Question

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

Short answer

Answer:

The subgroup H is sovable over the solvable group.

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.

Step-2: Show that H is solvable

Consider the group $H_i=H\cap G_i$. Here, G is the group and H is the subgroup overG . Let group is solvable that is

f 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 $H_i=f\left(G_i\right)$and the chain of subgroup is,

Assume that$H_i$is a normal subgroup of$H_{i-1}$for each element and$i=1,2,3.......t$and consider the any two element a,b of a subgroup$H_{i-1}$and another two elementis the element of subgroup$G_{i-1}$. Therefore,and. Since the group$\raisebox{1ex}{$G_{i-1}$}\!\left/ \!\raisebox{-1ex}{$G_i$}\right.$is abelian through the solvability now element. Consequently

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

Hence, the subgroup $\raisebox{1ex}{$H_{i-1}$}\!\left/ \!\raisebox{-1ex}{$H_i$}\right.$ is abelian if and only if $ab{a}^{-1}{b}^{-1}\in H$ for all

$a,b\in G$