I want show $ S_{4} $ isn't supersoluble group. For this suppose $ 1 \leq B_{4} \leq A_{4} \leq S_{4} $ be a normal serie of $ S_{4} $, that $ B_{4} $ is Klein’s four-group. Since $ B_{4} $ isn't cyclic group, then $ S_{4} $ isn't supersoluble group. But i want search $ S_{4}/B_{4} $ is supersoluble or no? For this suppose $ 1 \leq A_{4}/B_{4} \leq S_{4}/B_{4} $ be a normal series of $ S_{4}/B_{4} $. We know $ A_{4}/B_{4} $ is cyclic group. $ (S_{4}/B_{4})/(A_{4}/B_{4}) \cong S_{4}/A_{4} $ is cyclic, hence $ S_{4}/B_{4} $ is supersoluble. Is it true or no ?
Let $ 1 \leq U \leq B_{4} $ be a normal series of $ B_{4} $, that $ U $ generalized by $ (1,2 )(3,4) $. Since $ B_{4}/U \cong C_{2} $, Then $ B_{4} $ is supersoluble.