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20 Torsion Groups 20 Torsion groups Recall: ² tG = fx j nx = 0 for some n > 0g is the torsion subgroup of G. ² 0 ! tG ! G ! G=tG ! 0 |{z} | {z } torsion torsion¡free ² This sequence does not alwaysQ split (i.e., tG is not necessarily a direct summand of G), e.g., G = p prime Z=p. Now let G be any torsion group. Then we will show that there exists a short exact sequence M 0 ! cyclic ! |{z}G ! |{z}D ! 0; torsion divisible i.e., every torsion group is an extension of a direct sum of cyclic groups by a divisible group. M M Recall: Divisible D = ( Q) © Z=p1 | {z } | {z } torsion¡free torsion In order to prove the above theorem, we need the notions of pure and basic subgroups. Definition 20.1. S · G is pure if ng \ S = nS, (i.e., if x 2 S and x = ng then 9s 2 S x = ns). Example 20.2. S = 2Z=8 is not pure in G = Z=8 but S = 2Z=6 is pure in G = Z=6 since Z=6 »= Z=3 © Z=2. Facts: 1. Every direct summand of G is pure in G. 2. The converse is not true. (We will discuss when it is.) 3. If G=S is torsion-free then S · G is pure. 4. tG is pure in G (but not necessarily a direct summand). 5. S · T pure, T · G pure ) S · G pure. 6. S · T · G with S · G pure and T=S · G=S pure ) T · G pure.1 7. S · G pure, y 2 G=S. Then 9y 2 G y = y + S and order(y) = order(y).2 1If t = ng then t+ S = nx +S for some x 2 T since T=S is pure in G=S. Consequently, n(g ¡ x) 2 S so there is an s 2 S so that n(g ¡ x) = ns. But then t = n(x + s) 2 nT . 2If z is a representative of y and n = order(y) then ny = nz + S = S so nz 2 S. Since S is pure 9s 2 S nz = ns. But then z ¡ s is a representative of y of order n. 1 8. S1 · S2 · ¢ ¢ ¢ · G, Si · G pure )[Si · G pure. Recall: L ² G torsion ) Gp, Gp p-primary groups. ² If g 2 G has order m and (n; m) = 1 then g is divisible by n.(®m+¯n = 1 ) g = ®mg + ¯ng = n¯g) ² G[p] = fx 2 G j px = 0g. Lemma 20.3. Let G be p-primary and not divisible. Then G had a pure cyclic subgroup. Proof. Suppose first that G 6= pG then 9y 2 G which is not divisible by p. Claim hyi · G is pure cyclic (and p-primary). Proof: hyi is clearly cyclic so let ay 2 hyi and suppose ay = ng. (WTS: ay = n(by)) If (p; n) = 1 then n divides all elements of hyi as recalled above. Thus suppose n = p and ay = pg. Case 1. a = pa0. Then ay = p(a0y). Case 2. (a; p) = 1. Then ®a + ¯p = 1 ) y = ®ay + ¯py = ®pg + ¯py = p(®g + ¯y). Now suppose G = pG. Then G is divisible since every element is divisible by p and we already know that p-primary elements are divisible by all q relatively prime to p. In order to extend this to a direct sum of cyclic groups, we need pure independent sets. P Recall: X L⊆ G is independent if ( m®x® = 0 ) m®x® = 0) or equiva- lently, hXi = hx i. x®2X ® Definition 20.4. X ⊆ G is pure independent if X is independent and hXi · G is pure. Lemma 20.5. If G is p-primary and X ⊆ G is maximal pure independent then G= hXi is divisible. Proof. Suppose not, i.e., suppose G= hXi is not divisible. Then, by Lemma 20.3, there is a pure cyclic subgroup hyi · G= hXi. By Fact 6, 9y 2 G s.t. order(y) = order(y) and y + hXi = y. Claim X [ fyg is pure independent (contradicting the maximality of X). hX [ fygi pure: hXi is pure in G and hyi = hX [ fygi = hXi is pure in G= hXi. By Fact 6 this implies hX [P fygi is pure in G. X [ fygPindependent: Suppose m®x® + myy = 0. Then myy = 0 ) myy = 0 ) m®x® = 0 ) each m®x® = 0 since X is independent. Definition 20.6. B · G is a basic subgroup if 1. B is a direct sum of cyclic groups, 2 2. B · G is pure and 3. G=B is divisible. Theorem 20.7 (Kulikov). Every torsion group G has a basic subgroup, i.e., G is an extension of a direct sum of cyclic groups by a divisible group: M pure 0 ! cyclic ¡¡! G ! Ddivisible ! 0 L LProof. G torsion ) GL= Gp, Gp p-primary. If Bp · Gp is basic then Bp is basic in G = Gp. Thus we may assume that G is p-primary. If G is divisible let B = 0. If G is not divisible then G contains a pure cyclic subgroup hyi. Thus fyg is a pure independent subset of G. By Zorn’s lemma there is a maximal pure independent set X. (Given any tower X® of pure independent sets, [X® is independent and h[X®i = [ hX®i is pure in G by Fact 8.) ByL Lemma 20.5, G= hXi is divisible. By independence we also have: hXi = hx®i. So hXi is basic in G. Corollary 20.8. If G is a torsion group of bounded order (nG = 0) then G is a direct sum of cyclic groups. L Proof. 0 ! cyclic ! G ! D ! 0. nGL= 0 ) nD = 0. But D = nD since D is divisible. Thus D = 0 and G = cyclic. Now we want to show that any two basic subgroups of a torsion group are isomorphic. It suffices to consider the p-primary case. Definition 20.9. The Ulm invariants of a p-primary group G are given by: pnG \ G[p] U(n; G) = dimZ =p pn+1G \ G[p] Lemma 20.10. Let G be a direct sum of cyclic groups of p-power order. Then U(n; G) is equal to the number of cyclic summands of G of order pn+1. i Proof. Let Gi be the direct sum of all summands of G of order p . Then G = G1 © G2 © ¢ ¢ ¢ © Gn+1 © Gn+2 © ¢ ¢ ¢ n n+1 G[p] = G1 © pG2 © ¢ ¢ ¢ © p Gn+1 © p Gn+2 © ¢ ¢ ¢ n n n p G = 0 © 0 © ¢ ¢ ¢ © p Gn+1 © p Gn+2 © ¢ ¢ ¢ n n n+1 p G \ G[p] = 0 © 0 © ¢ ¢ ¢ © p Gn+1 © p Gn+2 © ¢ ¢ ¢ n+1 n+1 p G \ G[p] = 0 © 0 © ¢ ¢ ¢ © 0 © p Gn+2 © ¢ ¢ ¢ pnG \ G[p] »= pnG ) pn+1G \ G[p] n+1 n n+1 U(n; G) = dimZ=p p Gn+1 = #cyclic summands of G of order p 3 Corollary 20.11. G; H p-primary direct sums of cyclic groups, then G »= H , U(n; G) = U(n; H) for all n ¸ 0. Theorem 20.12. Any two basic subgroups of a p-primary group G are iso- morphic. Proof. Let B be a basic subgroup of G. ² The number of cyclic summands of order pn in B is equal to the number of cyclic summands of order p in B=pn+1B, i.e., U(n ¡ 1;B) = U(0; B=pn+1B) ² G = B + pnG Pf: G=B divisible so each g + B = g 2 G=B is divisible by pn, i.e., g + B = pnx + B ) g = b + pnx 2 B + pnG. ² B=pn+1B »= G=pn+1G G B+pn+1G » B B Pf: pn+1G = pn+1G = B\pn+1G = pn+1B since B is pure. So U(n ¡ 1;B) = U(0; G=pn+1G) depends only on G. We pointed out earlier that pure subgroups are not necessarily direct summands. Now we give a sufficient conditions when it is a direct summand and give some interesting corollaries. Corollary 20.13 (Pr¨ufer). A pure subgroup of bounded order is a direct summand. Proof. Let S · G pure with nS = 0. Let ¼ : G ³ G=(S + nG) be the quotient map. ² G=(S +nG) has bounded order n, so it is a sum of cyclic groups (20.8). L ² G=(S + nG) = hx®i, r® = order(x®) divides n. ² Choose x® 2 G ¼(x®) = x®. ² r®x® = s® + ng® ) s® = r®x® ¡ ng® = r®(x® ¡ (n=r®)g®). 0 0 ² S · G pure ) 9s® s® = r®s®. 0 0 ² Let y® = x® ¡ s®. Then ¼(y®) = ¼(x®) = x® so r®y® = r®x® ¡ r®s® = s® + ng® ¡ s® = ng® 2 nG. ² Let K = hnG [ fy®gi. 4 Claim 1: S \ PK = 0. P P Pf: s = ng + m®y® ) ¼(s) = ¼(ng) + m®¼(y®) ) m®x® = 0 ) m®x® = 0 ) r®jm® ) m®y® 2 nG ) s 2 nG \ S = nS = 0. Claim 2: S + K = G. P P P Pf: g 2 G ) ¼(g) =P `®x® = `®¼(y®) ) g ¡ `®y® 2 ker ¼ = S + nG ) g 2 s + (nG + `®y®) ⊆ S + K. Therefore, S © K = G. Corollary 20.14. If tG has bounded order then G »= tG © G=tG. Proof. tG · G is pure. Corollary 20.15. If G is torsion and not divisible then G has a p-primary cyclic direct summand. L Proof. Since G = Gp we may assume G is p-primary. By Lemma 20.3, G has a pure cyclic subgroup which must be a direct summand by Pr¨ufer.
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