Jordan Holder Theorem

THE JORDAN-HOLDER THEOREM Let G be a finite group. Suppose we have two chains of subgroups 0 = G0 ( G1 ( G2 ( · · · ( G r = G and 0 = H0 ( H1 ( H2 ( · · · ( Hs = G so that Gi is normal in Gi+1 , with Gi+1 /Gi simple and Hj is normal in Hj+1 , with Hj+1 /Hj simple. Theorem (Jordan-Holder): In the above setting, we have r = s, and the list of quotients (G1 /G0 , G2 /G1 , . . . , Gr /Gr−1 ) is a rearrangement of the list of quotients (H1 /H0 , H2 /H1 , . . . , Hs /Hs−1 ). The set up of the theorem is easy to achieve: Theorem 2: Let G be any finite group. Then we can find a chain of normal subgroups 0 = G0 ( G1 ( G2 ( · · · ( Gr = G so that Gi is normal in Gi+1 and Gi+1 /Gi is simple. Proof: Let r be the largest possible number so that there is a chain 0 = G0 ( G1 ( G2 ( · · · ( Gr = G with Gi is normal in Gi+1 . We can always take r = 1, and (G0 , G1 ) = ({1}, G), so such an r exists and, since G is finite, there is a largest such r. We claim that Gi+1 /Gi is simple. If not, let H be a normal subgroup of Gi /Gi+1 and let π be the projection map Gi → Gi /Gi+1 . Then π −1 (H) is normal in Gi+1 , and Gi is normal in H. So 0 = G0 ( G1 ( G2 ( · · · ( Gi ( π −1 (H) ( Gi+1 ( · · · ( Gr = G is a longer chain, contradiction.  Such a chain is called a composition series for G. Preliminary lemmas Lemma 1: Let G be a group, H a normal subgroup and A an arbitrary subgroup of G. Then AH is a subgroup of G, where AH is the set of products {ah : a ∈ A, h ∈ H}. Proof 1: We need to show that AH is closed under multiplication and inverses. For the first, let a1 , a2 ∈ A and h1 , h2 ∈ H. Notice that (a1 h1 )(a2 h2 ) = (a1 a2 )(a−1 2 h1 a2 h2 ). −1 Since H is normal, we have a−1 2 h1 a2 ∈ H. So (a2 h1 a2 h2 ) ∈ H and, clearly, a1 a2 ∈ A. Similarly, for a ∈ A and h ∈ H, we have (ah)−1 = h−1 a−1 = a−1 (ah−1 a−1 ) and we see that (ah)−1 ∈ AH.  1 2 THE JORDAN-HOLDER THEOREM – NOTES FOR WEDNESDAY, JANUARY 23 Proof 2: Let π be the map G → G/H. Then AH = π −1 (A), which is clearly a subgroup of G.  Lemma 2 (Theorem 9 in the textbook): Let G be a group, with A and B normal subgroups and A ∩ B = {1G }. Then A × B ∼ = AB, and the isomorphism is (a, b) 7→ ab. This was done in class on Jan. 14; we repeat the proof: Proof: We want to show that (a1 b1 )(a2 b2 ) = (a1 a2 )(b1 b2 ). So we must show that b1 a2 = a2 b1 . In other words, we must check that, for a ∈ A and b ∈ B, we have ab = ba. Set c = aba−1 b−1 . Then c = (aba−1 )b−1 is in B. But, since c = a(ba−1 b−1 ), we also have a ∈ A. So c ∈ A ∩ B, and we have c = 1G . We have shown aba−1 b−1 = 1; rearranging gives ab = ba. Lemma 3: Let G be a group, let M and N be normal subgroups, M 6= N , with G/M and G/N simple. Then G/(M ∩N ) = G/M ×G/N and we have M/(M ∩N ) ∼ = G/N and N/(M ∩N ) ∼ = G/M . Proof: Notice that we cannot have M ( N . If we did, then N/M is normal in G/M and is neither G/M nor {1}, contradicting that G/M is simple. So M 6⊆ N . Similarly, N 6⊆ M . Set K = M ∩ N . Since it is an intersection of two normal subgroups, it is normal. Replace (G, M, N ) by (G/K, M/K, N/K). This preserves the truth of all the hypotheses and all the conclusions, and gives us the additional tool of letting us assume that M ∩ N = {1}. From the preceding lemma, we have M N = M × N . Also, since M and N are normal, the subgroup M N is normal in G. We claim that G = M N . Proof: Since M N is normal in G, the subgroup M N/M is normal in G/M . Since G/M is simple, we have M N/N = G/M or M N/M = {1}. The second possibility is wrong because N 6⊆ M , so M N 6= M . The first possibility, must then be right: M N/M = G/M . So M N = G. We have deduced that G ∼ = M ×N. That specific statement is only right with the additional tool of M ∩ N = {1}. But the consequence G/(M ∩ N ) ∼ = G/M × G/N lifts back to the original case, as do the others.  The Jordan-Holder theorem Our proof is by induction on |G|. The base case, |G| = 1, is trivial. We use the symbol ∼ to denote “are rearrangements of each other”. Fix two chains 0 = G0 ( G1 ( G2 ( · · · ( Gr = G and 0 = H0 ( H1 ( H2 ( · · · ( Hs = G as in the statement of the theorem. If Gr−1 = Hs−1 then we are done by induction, applying the inductive hypothesis to the group Gr−1 . So we may assume that Gr−1 6= Hs−1 . We set M = Gr−1 N = Hs−1 K = M ∩ N. Using Lemma 2, we can find a composition series 0 = K0 ( K1 ( K2 ( · · · ( Kt = K for K. Then 0 = G0 ( G1 ( G2 ( · · · ( Gr−1 = M and 0 = K0 ( K1 ( K2 ( · · · ( Kt = K ( M are both composition series for M . By induction, we have r − 1 = t + 1 and we have (G1 /G0 , G2 /G1 , . . . , Gr−2 /Gr−1 ) ∼ (K1 /K0 , K2 /K1 , . . . , Kt /Kt−1 , M/K). (∗) THE JORDAN-HOLDER THEOREM – NOTES FOR WEDNESDAY, JANUARY 23 3 Similarly, 0 = H0 ( H1 ( H2 ( · · · ( Hs−1 = N and 0 = K0 ( K1 ( K2 ( · · · ( Kt = K ( N are both composition series for N . So s − 1 = t + 1 and the successive quotients (H1 /H0 , H2 /H1 , . . . , Hs−2 /Hs−1 ) ∼ (K1 /K0 , K2 /K1 , . . . , Kt /Kt−1 , N/K). (∗∗) From the equalities r − 1 = t + 1 = s − 1 we immediately deduce r = s. Taking (∗) and appending the quotient G/M to both lists, we have (G1 /G0 , G2 /G1 , . . . , Gr−2 /Gr−1 , G/Gr−1 ) ∼ (K1 /K0 , K2 /K1 , . . . , Kt /Kt−1 , M/K, G/M ). Similarly (∗∗) gives, (H1 /H0 , H2 /H1 , . . . , Hs−2 /Hs−1 , G/Hs−1 ) ∼ (K1 /K0 , K2 /K1 , . . . , Kt /Kt−1 , N/K, G/N ). The right hand sides of the above equations are identical except for the last two elements. And, by Lemma 3, we have (M/K, G/M ) ∼ (N/K, G/N ).