Abstract Algebra Theory and Applications - Computer Science ...

198 CHAPTER 11 THE STRUCTURE OF GROUPSSince H i ∩K m−1 is normal in H i+1 ∩K m−1 , the Second Isomorphism Theoremimplies that(H i+1 ∩ K m−1 )/(H i ∩ K m−1 ) = (H i+1 ∩ K m−1 )/(H i ∩ (H i+1 ∩ K m−1 ))∼= H i (H i+1 ∩ K m−1 )/H i ,where H i is normal in H i (H i+1 ∩ K m−1 ). Since {H i } is a composition series,H i+1 /H i must be simple; consequently, H i (H i+1 ∩ K m−1 )/H i is eitherH i+1 /H i or H i /H i . That is, H i (H i+1 ∩ K m−1 ) must be either H i or H i+1 .Removing any nonproper inclusions from the seriesH n−1 ⊃ H n−1 ∩ K m−1 ⊃ · · · ⊃ H 0 ∩ K m−1 = {e},we have a composition series for H n−1 . Our induction hypothesis says thatthis series must be equivalent to the composition seriesHence, the composition seriesandH n−1 ⊃ · · · ⊃ H 1 ⊃ H 0 = {e}.G = H n ⊃ H n−1 ⊃ · · · ⊃ H 1 ⊃ H 0 = {e}G = H n ⊃ H n−1 ⊃ H n−1 ∩ K m−1 ⊃ · · · ⊃ H 0 ∩ K m−1 = {e}are equivalent. If H n−1 = K m−1 , then the composition series {H i } and {K j }are equivalent and we are done; otherwise, H n−1 K m−1 is a normal subgroupof G properly containing H n−1 . In this case H n−1 K m−1 = G and we canapply the Second Isomorphism Theorem once again; that is,Therefore,andK m−1 /(K m−1 ∩ H n−1 ) ∼ = (H n−1 K m−1 )/H n−1 = G/H n−1 .G = H n ⊃ H n−1 ⊃ H n−1 ∩ K m−1 ⊃ · · · ⊃ H 0 ∩ K m−1 = {e}G = K m ⊃ K m−1 ⊃ K m−1 ∩ H n−1 ⊃ · · · ⊃ K 0 ∩ H n−1 = {e}are equivalent and the proof of the theorem is complete.□