Let $ M $ be a maximal subgroup of a solvable group $ G $, and assume that $ G=MC $, for some cyclic subgroup $ C $. Then $ \vert G : M \vert $ is a prime or $ 4 $. Also if $ \vert G : M \vert = 4 $ , then $ G/M_{G} = S_{4} $.
Proof: If let $ M_{G} \neq 1 $, how proof this case by induction on $ G/M_{G} $?
If $ M_{G}= 1 $, Let $ N $ be a minimal normal subgroup of $ G $. Then $ N $ is an elementary abelian $ p $-group and $ M $ is a complement for $ N $ in $ G $. $ M \cap C \leq M $, so $ M \cap C = 1 $ since $ M_{G} = 1 $.Thus $ \vert C \vert = \vert G : M \vert = \vert N \vert $ is a power of $ p $. Let $ P = NC $, so $ P $ is $ p $-group and let $ S = P \cap M $, so $ S $ complements $ N $ in $ P $. Also $ P = CS $ by Dedekind modular law. Then $ N $ is a prime or $ 4 $ and thus $ \vert G : M \vert $ is a prime or $ 4 $.