Question

In$\mathit{G}$is a normal subgroup of$\mathit{G}\mathbf{,}\mathit{N}$is solvable, and$\frac{\mathbf{G}}{\mathbf{N}}$is solvable. Prove that$\mathit{G}$is solvable.

Short answer

It is proved that G is solvable.

Step-by-step solution

Step 1:Explanation

Let$H_0\le H_1\le \mathrm{.......}\le H_n=N$ and

$J_0\le J_1\le \mathrm{.........}\le J_k=\frac{G}{N}$

Is a sequence of subgroups .which is showing that N and $\frac{G}{N}$are solvable.

Concept

Let$\phi :G\to \frac{G}{N}$ be the quotient homomorphism and we consider the sequence.

So,

$H_0\le H_1\le \mathrm{.......}\le H_n=N$

$N={\phi }^{-1}\left(J_0\right)\le {\phi }^{-1}\left(J_1\right)\le \mathrm{......}\le {\phi }^{-1}\left(J_k\right)=G$

$H_i$is subgroup follows the sequences that show N is solvable

Step3:Solution

Here first we need to satisfy the stability criteria of the${\phi }^{-1}\left(J_i\right)$.

Here we consider surjective homomorphism.

${\phi }^{-1}\left(J_{i+1}\right)\to \frac{J_{i+1}}{J_i}$.

Here${\phi }^{-1}\left(J\right)$ is a quotient homomorphism.Hence its kernel is.${\phi }^{-1}\left(J\right)$

Hence ${\phi }^{-1}\left(J\right)$must benormal. and its image is $\frac{J_{i+1}}{J_i}$by homomorphism first theorem.

Hence,

$\frac{{\phi }^{-1}\left(J_{i+1}\right)}{{\phi }^{-1}\left(J_i\right)}\cong \frac{J_{i+1}}{J_i}$is abelian.