If N is a normal subgroup of G and both N and G/N are solvable groups, then G is also a solvable group. This is proven by constructing a solvable series for G using the solvable series of N and G/N.
To show that the group G is solvable, given that N is a normal subgroup of G and both N and G/N are solvable, we need to prove that G has a solvable series.
A solvable group is a group that has a series of subgroups such that each subgroup is normal in the previous subgroup, and the factor groups are all abelian.
Here are the steps to prove that G is solvable:
1. Consider the series of subgroups:
{e} ⊲ N ⊲ G
2. Since N is a normal subgroup of G, and N is solvable, we know that there exists a solvable series for N:
{e} = N₀ ⊲ N₁ ⊲ N₂ ⊲ ... ⊲ Nₙ = N
3. Additionally, since G/N is solvable, there exists a solvable series for G/N:
{N} = G/N₀' ⊲ G/N₁' ⊲ G/N₂' ⊲ ... ⊲ G/Nₘ' = G/N
4. Now, define the following series for G:
{e} = N₀ ⊲ N₁ ⊲ N₂ ⊲ ... ⊲ Nₙ = N ⊲ N₀' ⊲ N₁' ⊲ N₂' ⊲ ... ⊲ Nₘ' = G
5. It can be shown that this series satisfies the conditions for a solvable group:
- Each subgroup is normal in the previous subgroup, as N is normal in G and N is normal in G/N.
- The factor groups are all abelian, as Nᵢ₊₁/Nᵢ is isomorphic to (Nᵢ/N)/(Nᵢ₊₁/Nᵢ), which is an abelian group.
Therefore, G is solvable.