Lagrange’s Theorem: Statement and Proof

1 lagrange s theorem : Statement and ProofPaul D. HumkeApril 5, 2002 AbstractLagrange s theorem is one of the central theorems of Abstract Algebra and it s Proof usesseveral important ideas. This is some good stuff to know!Before proving lagrange s theorem , we state and prove three a group with subgroupH, then there is a one to one correspondence betweenHand any coset a left coset ofHinG. Then there is ag Gsuch thatC=g :H Cbyf(x) =g one to , then asGhas cancellation,g x16=g x2. Hence,f(x1)6=f(x2). C, then sinceC=g H, there is anh Hsuch thaty=g h. It follows thatf(h) =yand asywas arbitrary,fis completes the Proof of Lemma a group with subgroupH, then the left coset relation,g1 g2if and only ifg1 H=g2 His an equivalence essence of this Proof is that is an equivalence relation because it is defined in termsofset equalityand equality for sets is an equivalence relation.

2 The details are below. is Gbe given. Then,g H={g h:h H}and as this set is well defined,g H=g use to represent the binary operation is ,g2 Gwithg1 g2. Then by the definition of ,g1 H=g2 H. That is,{g1 h:h H}={g2 h:h H}and as set equality is symmetric,{g2 h:h H}={g1 h:h H}.Hence,g2 g1and asg1andg2were arbitrary, is symmetric. is ,g2,g3 Gwithg1 g2andg2 g3. Then,g1 H={g1 h:h H}={g2 h:h H}=g2 Handg2 H={g2 h:h H}={g3 h:h H}=g3 set equality is transitive, it follows thatg1 H={g1 h:h H}={g3 h:h H}=g3 H,org1 H=g3 H. That is,g1 g3, and asg1,g2,g3 Gare arbitrary, is complete the Proof of the a set and be an equivalence relation onS. IfAandBare two equivalenceclasses withA B6= , thenA= prove the lemma, we show thatA BandB A. AsAandBare arbitrarily labeled,it suffices to show the A. AsA B6= , there is ac A B. AsAis an equivalence class of and bothaandcare inA, it follows thata c.

3 But asa c,c BandBis an equivalence class of , it followsthata with these three lemmas we proceed to the main 1.[ lagrange s theorem ] IfGis a finite group of ordernandHis a subgroup ofGoforderk, thenk|nandnkis the number of distinct cosets be the left coset equivalence relation defined in Lemma 2. It follows from Lemma 2that is an equivalence relation and by Lemma 3 any two distinct cosets of are disjoint. Hence,we can writeG= (g1 H) (g2 H) (g` H)where thegi H,i= 1,2,..,`are the disjoint left cosets ofHguaranteed by Lemma Lemma 1, the cardinality of each of these cosets is the same as the order ofH, and so|G|=|g1 H|+|g2 H|+ +|g` H|=|H|+|H|+ +|H| `summands=` |H|=` completes the