$ K/N $ be a normal subgroup of $ (G/N)^{\prime} $. Now why $ K \cap G^{\prime} $ is a maximal subgroup of $ G^{\prime} $?

Question

Let $ G $ be a finite group and $ N $ is a normal subgroup of $ G $. Suppose $ K/N $ be a maximal subgroup of $ (G/N)^{\prime} $. Now why $ K \cap G^{\prime} $ is a maximal subgroup of $ G^{\prime} $?

Answer 1

$ ( G/N )^{\prime} = G^{\prime}N/N $. Let $ K/N $ be a maximal subgroup of $ G^{\prime}N/N $, then by Dedekind modular $ K/N = K/N \cap G^{\prime}N/N = (K \cap G^{\prime}N)/N = (K \cap G^{\prime})N/N $. Hence $ K \cap G^{\prime} $ is a maximal subgroup of $ G^{\prime} $.