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Abelian Groups Abelian Groups A group is Abelian if xy = yx for all group elements x and y. The basis theorem An Abelian group is the direct product of cyclic p−groups. This direct product de- composition is unique, up to a reordering of the factors. n1 nk Proof: Let n = p1 · · · pk be the order of the Abelian group G, with pi's distinct primes. By Sylow's theorem it follows that G has exacly one Sylow p−subgroup for each of the k distinct primes pi: Consequently G is the direct product of its Sylow p−subgroups. [We call the Sylow p−subgroups the p − primary parts of G:] It remains to show that an Abelian p−group (corresponding to a p−primary part of G) is the direct product of cyclic groups. We prove this by induction on the power m of the order pm of the p−group. Assume that the result is true for m: Let P be an Abelian group of order pm+1 and Q a subgroup m of P of order p (such Q exists by Sylow's theorem). By induction Q =< a1 > × < a2 > × · · · × < ar > with r ki j < ai > j = p ; k1 ≥ k2 ≥ · · · ≥ kr; X ki = m: i=1 p p s1 s2 sr t1 t2 tr Let a 2 P − Q: Then a 2 Q and therefore a = a1 a2 · · · ar : Taking b = aa1 a2 · · · ar p d1 d2 dr for suitable ti, we obtain b 2 P − Q; and b = a1 a2 · · · ar ; where for each i either di = 0 or (di; p) = 1: [Take a deep breath and convince ourself that such b exists.] 1 p If all the di are 0, then b = 1 and, by a cardinality argument, P =< a1 > × < a2 > × · · · × < ar > × < b > : kj +1 If not, let j be the first index for which dj =6 0; hence j < b > j = p : We show that P =< a1 > × · · · × < aj−1 > × < b > × < aj+1 > × · · · × < ar > : To prove this it suffices to show that < b > \(< a1 > × · · · × < aj−1 > × < aj+1 > × · · · × < ar >) = 1; or that pkj b 2=< a1 > × · · · × < aj−1 > × < aj+1 > × · · · × < ar >; k since every nontrivial subgroup of < b > contains bp j : However, from the choice of b − k kj 1 p j dj p we see that b contains in its decomposition aj which is different from 1 because (dj; p) = 1: It follows that pkj b 2=< a1 > × · · · × < aj−1 > × < aj+1 > × · · · × < ar >; and this shows that P is the direct product of cyclic groups. To see that this direct product decomposition is unique, it suffices to show that it is unique for each of the p−primary parts. Indeed, if pm1 is the highest order of an element in P; then any direct product decomposition necessarily contains a cyclic group of order pm1 as a direct factor. The factor group of P to this cyclic subgroup is of lesser order than P; and the uniqueness of its factors follows by induction. This ends the proof. A consequence of the above result is that if (x1; : : : ; xs) are the generators of the cyclic direct factors of the Abelian group G, then any element x 2 G can be written uniquely j1 js as x = x1 · · · xs ; for positive integers ji: We say that (x1; : : : ; xs) is a basis for G: The orders of the basis elements xi are the invariants of G: 2 For an Abelian p−group P of order pm with invariants pm1 ≥ · · · ≥ pmk we call the vector (m1; : : : ; mk) of positive integers m1 ≥ · · · ≥ mk, which satisfy P mi = m, the type of P: The type of P is a partition of m: Examining the extremes, a group of type m is cyclic of order pm; while a group of type (1, 1, : : : , 1) is a direct product of m cyclic groups each of order p: We call the latter group elementary Abelian. Graphically represent the type of P by k rows of dots, the first of which contains m1 dots, the second m2 dots, and so on; we call this graph the Ferrer diagram of P: Flip the Ferrer diagram in its main diagonal. What results is another Ferrer diagram with si dots in row i: We call the vector (s1; s2; : : :) the signature of P: [For example, if the type is (4, 2, 1), then the signature is (3, 2, 1, 1).] It is occasionally helpful to see the type and signature as vectors of infinite lengths with all but the first finitely many entries being zero. In view of the Basis Theorem proved above, it is clear that there is a bijection between the nonisomorphic Abelian p-groups of order pm and the partitions of m. There are, therefore, as many Abelian p−groups of order pm as there are partitions of m: A determination in "closed form" of the number of such partitions was made by Hardy and Ramanujan by a process called the "circle method". On the other hand, recurrences for such partitions can easily be found. !!!Include Delsarte's result here 3 Composition series A subnormal series of a group G is a sequence of subgroups G = G0 ≥ G1 ≥ · · · ≥ Gn such that Gi+1 is normal in Gi; for 0 ≤ i < n: The factors of the series are the quotient groups Gi=Gi+1. By the length of a series we understand the number of strict inclusions that occur in the series (or, equivalently, the number of nonidentity factors). If each of the groups that occur in a subnormal series are normal in G we call the series normal: A one term refinement of a subnormal series G = G0 ≥ G1 ≥ · · · ≥ Gn is a subnormal series G = G0 ≥ G1 ≥ · · · ≥ Gi ≥ H ≥ Gi+1 ≥ · · · ≥ Gn or a subnormal series G = G0 ≥ G1 ≥ · · · ≥ Gn ≥ H; the subgroup H is the inserted term. A refinement of a subnormal series is a subnormal series obtained by succesively performing a finite number of one term refinements. Note that a refinement may contain more terms than the original series without having greater length. This happens if the groups that are inserted are repetitions of groups already in the series. A subnormal series G = G0 ≥ G1 ≥ · · · ≥ Gn = 1 is a composition series if each factor Gi=Gi+1 is a simple group. Two subnormal series are equivalent if there is a bijective correspondence between the nonidentity factors of the two series such that the corresponding factors are isomorphic groups. [In particular, equivalent series must necessarily have the same length.] Lemma (Zassenhaus) 4 If A1 ¢ A and B1 ¢ B are subgroups of a group G; then (a) A1(A \ B1) is normal in A1(A \ B) (b) B1(A1 \ B) is normal in B1(A \ B) (c) A1(A\B) ∼ B1(A\B) : A1(A\B1) = B1(A1\B) Proof: Observe that B1 ¢ B implies that A \ B1 = (A \ B) \ B1 is normal in A \ B: Similarly (A1 \ B) ¢ (A \ B): It follows that D = (A1 \ B)(A \ B1) is a normal subgroup of A \ B: Define f : A1(A \ B) ! (A \ B)=D as follows. For a 2 A1 and c 2 A \ B let f(ac) = cD: The function f is a well-defined surjective homomorphism. [Indeed, f((a1c1)(a2c2)) = f(a1a3c1c2) = c1c2D = (c1D)(c2D) = f(a1c1)f(a2c2); where ai 2 A1; ci 2 A \ B; and −1 c1a2c1 = a3 since A1 ¢ A:] The kernel of f is kerf = A1(A \ B1): This shows that A (A \ B ) is normal in A (A \ B) and, by the First Isomorphism Theorem, A1(A\B) ∼ 1 1 1 A1(A\B1) = A\B D : An entirely parallel argument yields B1(A\B) ∼ A\B : The isomorphism written in (c) B1(A1\B) = D now follows. This concludes the proof. And now, a fundamental Theorem of Schreier. Theorem (Schreier) Any two subnormal series of a group have subnormal refinements that are equivalent. Proof: Take two subnormal series G = G0 ≥ G1 ≥ · · · ≥ Gn and G = H0 ≥ H1 ≥ · · · ≥ Hm: Append each with 1, that is, define Gn+1 = Hm+1 = 1: Consider the groups 5 Gi = Gi+1(Gi \ H0) ≥ · · · ≥ Gi+1(Gi \ Hj) ≥ · · · ≥ Gi+1(Gi \ Hm+1) = Gi+1; for each 0 ≤ i ≤ n: The Zassenhaus Lemma applied to Gi+1 ¢ Gi and Hj+1 ¢ Hj shows that Gi+1(Gi \ Hj+1) is normal in Gi+1(Gi \ Hj), for each 0 ≤ j ≤ m: Inserting the above groups between Gi and Gi+1, and denoting Gi+1(Gi \ Hj) by Gij; yields a subnormal refinement of G = G0 ≥ G1 ≥ · · · ≥ Gn. Specifically, G = G00 ≥ · · · ≥ G0m ≥ G10 ≥ · · · ≥ G1m ≥ · · · ≥ Gn0 ≥ · · · ≥ Gnm; where Gi0 = Gi: A parallel argument yields a refinement of G = H0 ≥ H1 ≥ · · · ≥ Hm; specifically, G = H00 ≥ · · · ≥ Hn0 ≥ H01 ≥ · · · ≥ Hn1 ≥ · · · ≥ H0m ≥ · · · ≥ Hnm; where Hij = Hj+1(Gi \ Hj) and H0j = Hj: Both refinements have (n + 1)(m + 1) not necessarily distinct terms. For each (i; j); the Zassenhaus Lemma applied to Gi+1 ¢Gi and Hj+1 ¢Hj yields the isomorphism: Gij = Gi+1(Gi\Hj ) ∼= Hj+1(Gi\Hj ) = Hij : Gi;j+1 Gi+1(Gi\Hj+1) Hj+1(Gi+1\Hj ) Hi+1;j This provides the bijective correspondence on factors, proving the equivalence of the two subnormal series. Note: An analogous result can be obtained for normal series in a similar manner. Let us examine Schreier's Theorem in the case of composition series. Since the factors of a composition series are simple groups, it follows that any refinement of a composi- tion series is equivalent to that series. Schreier's Theorem, therefore, implies that two composition series are necessarily equivalent. This result is stated below.
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