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, for G is itself a permutation group on {I, 2, 3}. (iii) Let G be the cyclic group of order 2 and let X be the set of all integers..., -1, 0, 1,.... Let a be the permutation of X defined by a : 2i..... 2i + 1, 2i + 1..... 2i for i = 0, ±1,... Thus a sends an even integer to the succeeding odd integer and an odd integer to the preceding even integer. Let 1 denote the identity permutation. Then 1 and a together constitute a subgroup r of the symmetric group on X since Now suppose G consists of the elements 1 and a. Then the mapping J1: a..... a, is actually an isomorphism since both G and r are cyclic of order two. 1..... 1 So J1 is a permutational representation. (iv) Suppose n is a positive integer and that G is the cyclic group of order n, G = {I, a, a2,..., an-I} Let a be the permutation of X = Z the integers defined by jn..... jn + 1, jn + 1..... jn + 2,..., jn + (n - 1)..... jn a: (j = 0, ±1,... ) In particular we have Oa = 1, la = 2,..., (n -1)a = 0 Note that a cyclically permutes the integers taken in blocks of n. It is not difficult to see that a is of order n. First of all, Oa2 = 2, I = n - 1. This means that the permutations I, a, a2, •••, an- I Oa3 = 3,..., Oan are all distinct since they act differently on O. If j is any integer, jan = j. So an = 1 and {1,a,a2, •••,an- I } is a cyclic group r of order n. Hence the mapping J1 of G into r defined by is an isomorphism and therefore a permutational representation of G. Sec. 7.3] DEGREE OF A REPRESENTATION AND FAITHFUL REPRESENTATIONS 217 (v) Suppose that G is the dihedral group of degree 4. G is the group of sym metries of the square '------~e |
Let g E G. By Lemma 3.12, page 75, g takes each vertex of ABeD to a vertex. Since g is one-to-one, Ag, Bg, eg and Dg are distinct vertices. If we define X = {A, B, e, D} and Yg by xYg = xg for all x E X, then Yg E Sx. Let fJ: G -> Sx be defined by gfJ = Yg. Then if x E X, g, hE G, x(gh)fJ = XYgh = x(gh) = (xg)h = (XYg)Yh = X(YgYh) = x(gfJ)(hfJ) Thus (gh)fJ = gfJhfJ, and so fJ is a permutational representation of G. If gfJ is the identity permutation of X, then g leaves every vertex of ABeD un changed. But by Lemma 3.7, page 71, g = I. Hence Ker fJ = {I} and fJ is actually an isomorphism. 7.3 DEGREE OF A REPRESENTATION AND FAITHFUL REPRESENTATIONS a. Degree of a representation Definition: The degree of a permutational representation on X (more briefly referred to as a representation) is the number of elements in X. Example 2: We inspect Example l(i)-(v). (i) This representation is of degree 6. (ii) This representation is of degree 3. Notice that in (i) and (ii) we have two representations of the same group, namely the symmetric group on {I, 2, 3}, of different degrees. (iii) This representation is of infinite degree. Notice that here G is cyclic of order 2. Hence there are representations of finite groups which are of infinite degree. (iv) This representation is of infinite degree. (v) This representation is of degree 4. Problems 7.2. Find a representation of degree 3 for a cyclic group of order 2. Solution: Let G be the cyclic group of order 2, say G = {I, a}. Let X = {I, 2, 3}. Then there are several possible representations of G on X. First of all there is the rather trivial representation 'T: 1->1, a->I where I |
is as usual the identity permutation. elements g, h in G we choose, 'T is clearly a representation. For no matter which (gh)'T = I and (g'T)(h'T) = II = I from which (gh)'T = (g'T)(h'T). Another representation of G involves choosing a different homomor phism. Now observe that if /L is a homomorphism of G into the symmetric group Sx, then either /L = 'T or G/L is of order 2. Note also that G/L must be a subgroup. So in deciding on a choice of /L, we need subgroups of Sx of order 2. There are actually 3 of them. To see this let al: 1 -> 2, 2 -> 1, 3 -> 3 a2: 1 -+ 3, 2 -+ 2, 3 -+ 1 aa: 1 -+ 1, 2 -+ 3, 3 -+ 2 Then {I, al}' {I, az}, {I, aa} are subgroups of Sx of order two, since a7 = I for mappings /La: 1 -+ I, a -> aa /Ll: 1 -+ I, a -+ al /L2: 1 -+ I, a -+ a2 i = 1,2,3. Thus the are representations of G on X. 218 PERMUTATIONAL REPRESENTATIONS [CHAP. 7 The proof that there are precisely three subgroups of order 2 follows from an inspection of all the subgroups of Sx. Since we have no real need for such proof here, we leave the details to the reader. The upshot of these considerations is that we have produced 4 representations of G of degree 3. 7.3. Find representations of degree 10 and 15 respectively of the cyclic group of order 5. Solution: Let G be cyclic of order 5. Then we can find a E G such that G = {I, a, a'll, a 3, a 4}. Let x = {1,2,..., 10} Y = {1,2,..., 15} and let "'1: 1 ~ 2, 2 ~ 3, 3 ~ 4, 4 ~ 5, 5 ~ 1, 6 ~ 7, 7 ~ 8, 8 ~ 9, 9 ~ 10, 10 ~ 6 "'2: 1 ~ 2, 2 ~ 3, 3 |
~ 4, 4 ~ 5, 5 ~ 1, j ~ j, for j > 5 It follows from a direct calculation that both "'1 and "'2 are of order 5. Then are subgroups of order 5 of Sx and Sy respectively. Thus and 112 : 1 ~ t, a ~ "'2' are both representations of G. The first is of degree 10 and the second is of degree 15. h. Faithful representations Definition: A representation is termed faithful if it is one-to-one. Both faithful and non-faithful representations are useful, as we shall see later. Problems 7.4. Are the representations in Example 1, page 216, faithful? Solution: (i) p is faithful. (ii) The identity isomorphism is one-to-one, so this representation is also faithful. Notice that (i) and (ii) provide examples of faithful representations of the same group which are of different degrees. (iii) Il is faithful. (iv) Il is faithful. (v) The representation is faithful. 7.5. Inspect the representations of (a) Problem 7.2 and (b) Problem 7.3 for faithfulness. Solution: (a) T is not faithful since a # 1 and aT = t, i.e. T is not one-to-one. However Ill' 1l2' 113 are faithful. (b) III and 112 are both faithful. 7.4 PERMUTATIONAL REPRESENTATIONS ON COSETS Definition: Suppose that G is a group and that H is a subgroup of G. Then the right co sets of H in G are HX1, HX2,... (7.1) Sec. 7.4] PERMUTATIONAL REPRESENTATIONS ON CO SETS 219 If H consists of the identity element alone, then (7.1) is simply an enumeration of the ele ments of G. Here we will show how to obtain a permutational representation of G using the cosets (7.1) which coincides with the regular representation when H = {1}. To describe our representation, let us choose a complete system X of representatives of the right co sets Hg of H in G, with 1 the representative of H. In other words, we select in each coset Hg an element which we term the representative of the coset, with 1 the rep resentative of H. X is then simply the set of chosen representatives. |
We call such a set X a right transversal of H in G. Given a right transversal X of H in G, we denote the representative of the coset Hg by g. Thus g is an element of X. Since two right cosets are either identical or disjoint, it since follows that Hg = Hg because g E Hg. Notice that if hE H, Hg = Hg = Hhg = Hhg. then hg = g For example, if G is cyclic of order 4, say G = {1, a, a2 is a sub group of order 2 of G, then {1, a} is a right transversal of H in G. We take X = {1, a} and note that I = 1, a = a, a2 = 1, a3 = a. }, and if H = {1, a2, a3 } With each element gin G we associate a mapping Yg of X into X, where Yg is defined by (7.2) In fact Yg is a permutation of X. To prove this we first show that if gl, g2 E G, then glg2 = glg2. For glg2 is the representative of the coset Hg 1 g2, while glg2 is the representa tive of the coset Hg 1 g2. But Hg 1 = Hg 1, and so H{hg2 = Hg 1g2 = Hg 1g2 Thus (7.3) We use (7.3) to prove that the mapping Yg is a permutation. First we prove Yg is one-to-one. Assume xYg = YYg (x, Y E X). Then xg = yg, and so xgg- 1 = ygg-l. Using (7.3), we find that xgg- 1 = xgg- 1 = X = x and similarly ygg-l = y, from which x = y. Finally we prove Yg is onto. Suppose x E X; then xg- 1 E X and Hence every element of X is an image. Thus Yg is a permutation. We now define a mapping 7r of G into Sx. 7r assigns to g in G the mapping Yg' so that g7r is the permutation of X defined by (7. |
2). The aim of the discussion in this section is to prove that this mapping 7r is a permutational representation of G on X. We have only to then (glg2)7r = (gl7r)(g27r). Note that verify that it is a homomorphism, i.e. if gl' g2 E G, (glg2)7r = (gl7r) (g27r) we must show that (glg2)7r is a permutation of X. To prove the effect of the mapping (glg2)7r is the same as the effect of the mapping (gl7r) (g27r)· that (gl7r)(g27r) is the product of two mappings, i.e. the result of first performing the mapping (gl7r) and then (g27r).) If x E X, using (7.3) we find (Note ~ X((glg2)7r) = X(glg2) = xg 1g2 = (X(gl7r))(g27r) = X((gl7r)(g27r)) Hence (glg2)7r = (gl7r)(g27r) as claimed. We shall refer to 7r as a coset representation of G (with respect to H). Of course 7r In a sense 7r is independent of the choice of the transversal X depends on H, G and X. (see Problem 7.10). 220 Problems PERMUTATIONAL REPRESENTATIONS [CHAP. 7 7.6. Choose right transversals for center Z in the quaternion group!f{ of order 8 (see Table 5.1, page 151). (a) the center Z in the dihedral group D of degree 4, and (b) the Solution: (a) The dihedral group D = {1, aI' az, a3' a4, as, as, a7} table: is most easily described by its multiplication 1 1 1 al a2 al a2 az a3 a4 as as a7 a3 1 a7 a4 as as a3 1 al as a7 a4 as a3 1 al az as as a7 a4 a4 as as a7 1 al a2 a3 as as a7 a4 a3 1 al az as a7 a4 as |
a2 a3 1 al a7 a4 as as al a2 a3 1 (Here al corresponds to a rotation of 90 0, while a4 corresponds to a reflection.) The center Z of G is given by Z = {1, a2}' This can be checked by verifying that a2 commutes with every element of D (and no element other than 1 and a2 has this property). Finally Z,Zal,Za4,ZaS are the co sets of Z in G. Thus is a right transversal of Z in G. X = {1, aI' a4, as} (b) The quaternion group!f{ of order 8, with elements 1, aI' az, a3, a4, as, as, a7, is given by the fol lowing multiplication table: 1 1 1 al a2 a3 a4 as al a2 a2 a3 1 a7 a4 a3 1 al as a7 a4 as a3 1 al az as as a7 a4 a4 as as a7 a2 a3 1 al as as a7 a4 al az a3 1 as a7 a4 as 1 al az a7 a4 as as a3 1 al a3 az as as a7 a6 (Comparing with Table 5.1, we have the following correspondence: 1 ~ 1, a ~ aI, a2 ~ az, a3 ~ a3. b ~ a4' ab ~ as, a2b ~ a6, a3b ~ ~.) The center Z of!f{ is given by Z = {1, az}. One has only to check this from the multiplication table. The co sets of Z in!f{ are Z, Zav Za4, Zas. This again can be checked directly from the multiplication table. Thus is a right transversal of Z in!f{. Sec. 7.4] PERMUTATIONAL REPRESENTATIONS ON COSETS 221 7.7. Find a coset representation of 7.6(b) with respect to Z. (a) D of Problem 7.6(a) with respect to Z, and (b) 3£ of Problem Solution: (a) We work out a coset representation l' using the right transversal X given in Problem 7.6. Thus we must find the permutations of X that l' assigns to each |
element of D. We will calculate in l(al1') = al' detail the permutation a l 1'. Now X = {1, ai' a 4, a 5}. Then l'al = a v alai = 1 since Zai = Z = Zl, and so a l (al1') = 1. a4a l = a5 since Za4al = Za7 = Za5, and so a4(al1') = a5' Finally a5al = a4 since Z(a5al) = Za4' and so a5(al1') = a4' Therefore and so Similarly the other permutations are 11' : a21' : a3'iT : a41' : a51' : a61' : a71' : 1 ~ 1, al --> ai' a4 --> a4, a 5 -? a5 1 -? 1, al -? ai' a4 -? a4, a5 ~ a5 1 ~ ai' al -> 1, a4 4 a5, a5 -? a4 1 -> a4, al ~ a5' a4 ~ 1, a5 ~ al 1 ~ a5' al -> a4' a4 -> ai' a5 -> 1 1 ~ a4, al ~ a5' a4 -> 1, a 5 -> a l 1 ~ a5' al -> a4' a4 ~ ab a 5 -> 1 It is instructive to check directly that this mapping rr is a homomorphism of D into the per mutation group on {1, ai' a4' a5}' (b) Again we must find the permutations that l' assigns to each element of 3{. The argument follows closely that of (a) above, using the set X = {1, ai' a4, a5}' Here we give only the result which the reader is urged to check. 11' : 1 ~ 1, al ~ av a4 ~ a4, a5 -> a5 al1' : a31' : a21' : 1 -> av al ~ 1, a4 --> a5, a 5 -> a4 1 -> 1, al -> ai' a4 4 a4, as 4 a5 1 ~ ai' al ~ 1, a4 -> a5, a5 ~ a4 1 -> a4, al ~ a5 |
' a4 -> 1, a5 -> al 1 -> a5' al..... a4' a4 --> ai, a 5 -> 1 1..... a4' al ~ a 5, a4 -> 1, a5 ~ al a61' : a71' : 1 -> a5, al -> a 4, a4 -> ai' a 5 -> 1 a51' : a41' : 7.8. Consider the permutation group S{L2}' Its elements are <PI = (~ ~) and <Pz = (~ ~). Con sider now the permutation group S{a,b}' Its elements are >¥l = (: :), >¥z = (:!). >¥l and <pv >¥2 and <P2 are essentially the same except for the elements they act on. Give a definition which will make this idea of "essentially the same" precise. Solution: Let F be a permutation group on a set X and let G be a permutation group on a set Y. We say that F and G are isomorphic as permutation groups if there exists a one-to-one onto correspondence fJ: F -> G such that for all x in X and IE F, (xf)a = (xa)(lfJ). a: X..... Y and an onto mapping (In this problem a: 1 --> a, 2 --> band fJ: <PI..... >¥l, <P2 ~ >¥z.) 7.9. Prove that if F and G are isomorphic as permutation groups, then they are isomorphic as groups. (Hard.) Solution: The fJ of the solution of Problem 7.8 provides the isomorphism. First we show that it is a homomorphism. Let 11,12 E F. For any x E X, (x(fd2»a = «XII)12)a «xll)a)(12fJ) = «xa)(llfJ»(fZfJ) = (xa)«(flfJ)(lzfJ» (by the definition of composition of mappings) 22.2 PERMUTATIONAL REPRESENTATIONS [CHAP. 7 Since a is a mapping onto Y, it follows that as x ranges over X, Xa ranges over Y. Hence as (ld2)() and (118)(128 |
) are permutations, 11()12() = (l1()(/2(). Thus 8 is a homomorphism. Next we must show that () is one-to-one. Suppose that II() = 12(); (xa)(l28). Thus then if x E X, (xa)(lI() = Since a is one-to-one, xii = x/2 from which II = 12, Hence () is an isomorphism. 7.10. Let G be any group and H a subgroup of G. Let Xi be a transversal for H in G and 7Ti the cor i = 1,2. Prove that G7TI and G7T2 are isomorphic as permutation responding coset representation, groups. (Hard.) Solution: Since XI and X 2 contain one and only one element in each coset of H in G, we define a: XI..... X 2 if HXI = HX2' a is then a one-to-one correspondence. We define by sending xI E XI to X2 E X 2 /1: G7TI..... G7T2 by (g7TI)/1 = g7T2' It is easy to check that /1 is a mapping. Let g7TI = f3, g7T2 = y. We need only verify that (xf3)a = (xa)y for each x E X. Now by definition of f3 and a, Hxg = H(xf3) = H((xf3)a) Also, Hxg = H(xa)g = H((xa)y) Hence (xa)y = (xf3)a as they are elements of X 2 that belong to the same coset, i.e. Hxg. The result follows. 7.5 FROBENIUS' VARIATION OF CAYLEY'S THEOREM a. The kernel of a coset representation Is a coset representation 7r of a group G with respect to a subgroup H ever faithful? We know that the answer is yes if H = {l}. The object now is to find the kernel of 7r. Theorem 7.2: Let G be a group and H a subgroup of G. Let 7r be a coset representation of G with respect to H. Then the kernel of 7r is the largest normal subgroup of G contained in H, i. |
e. if N <1 G and H"dN, then N C Ker7r. Before proving Theorem 7.2, it should be noted that from this theorem it follows that 7r is faithful if the only normal subgroup of G which is contained in H is the identity subgroup. This implies, for example, that if G is a simple group and H 0/= G, 7r is automatically faith ful. For then, by definition, the only normal subgroups of G are G and the identity subgroup. This observation has been useful in the theory of finite groups. Proof: Let X be the right transversal of H in G from which 7r was defined. First we prove that if K is the kernel of 7r, i.e. the set of all elements g of G such that g7r = t, the identity mapping of X onto itself, then K is contained in H. then a7r = t, If a E K and x E X, In particular on putting x = 1, 1 = ii. Hence a E H. This means that K is contained in H. Of course K, as the kernel of the homomorphism 7r, is a normal subgroup of G. To complete the proof of the theorem, we must show that K is the largest normal subgroup of G contained in H. To do this it is sufficient to prove that if N is any normal subgroup of G contained in H, then N7r = {t}. i.e. x = x(a7r) = xa. Suppose that a EN. If x E X, then Hxa = Hxax-Ix Since N is normal, xax- 1 EN. But N is contained in H, and so xax- I E H. Accordingly Hxa = Hx. Thus xa = x which means Hence a7r = Land N C K as required. x(a7r) = x for all x EX Sec. 7.5] FROBENIUS' VARIATION OF CAYLEY'S THEOREM 223 Problems 7.11. Let G be the symmetric group on {I, 2, 3}. Let u: 1 --+ 2, 2 --+ 3, 3 --+ 1, and let T: 1 --+ 2, 2 --+ 1, (b) H = 3 --+ 3. Then the elements of G are I, u, u2 |
, T, UT, U2T. Let gp(T) = {I, T}. In each case find a coset representation of G with respect to H. Find also the kernels of both representations. (a) H = gp(u) = {I, u, u 2} and Solution: (a) A right transversal,of H in G is {I, T}. Notice that T- 1UT = u- 1 = u2 • So H is normal in G. Hence by Theorem 7.2, every coset representation has kernel precisely H. The coset representa tion 1r associated with the right transversal {I, T} is given by t7T: t ~ t,., ~ 7" 711': L -+ 7', T -+ t (111': t. -+ L, T -+ T (UT)1r: 1--+ T, T --+ I (cr 2T)1r: 1--+ T, T --+ I Clearly Ker 1r = {I, u, u2 }. (b) Let X = {1,u,u 2 }. Since G = HuHuuHu2, X is a right transversal of H in G. The associated permutational representation 1r is 1.11': t. -+ L, u -+ u, u2 --+ u 2 U1r : I. -+ (1, U -+ 0'2, 0'2 -+ L U21r : L -+ 0'2, (1 -+ L, u 2 --+ u It follows immediately that 1r is faithful, since the only element mapped to the identity permuta tion of X is I. Hence the kernel of 1r = {I}. 7.12. Let G be the alternating group of degree 4 and let H be the subgroup consisting of the permutations 1 2 3 4) (1 2 3 41 2 3 4) ( I, Find an associated coset representation of G with respect to H. Is this representation faithful? Solution: Let 1 2 3 231 ( T1 = G ~ : :), T2 = G ~ ~ :), ~), G consists of the elements Then if u - t, (1', 0'2, 7"1' 710', 71(12, 72' T2u, 'T2U2, 73, 'TaU, 7"30'2 |
It follows immediately that a right transversal of H in G is X = {I, u, u2 }. Let 1r be the associated coset representation. It is easy to check directly that H is normal in G. Then Ker 1r = H by Theorem 7.2, and so 1r is not faithful. Finally we list the permutations g1r with g in G: 1.11' : t -+ L, u -+ u, u 2 --+ u 2 T21r : L -+ t., a -+ u, u2 --+ u 2 (111': L -+ U, u --+ u2, u 2 --+ I (T2U)1r : L -+ u, (1 -+ 0'2, u 2 --+ I U21r : L -+ 0'2, U -+ L, u 2 --+ u (T2U2)1r : L -+ 0'2, u -+ t, u2 --+ u T11r : L -+ L, u -+ u, u2 --+ u2 TS1r : t -+ t, U -+ a, u 2 --+ u 2 (T1 U)1r : L -+ u, u --+ u 2, u 2 --+ I (TsU)1r : L -+ u, u --+ u2, u2 --+ I (T1U2)1r : L -+ (12, (1 -+ L, u2 --+ u (TSU2)1r : L -+ 0'2, u -+ t, u2 --+ u b. Frobenius' theorem Let H be a subgroup of G, X a transversal and 71' the associated coset representation. Let p denote the right regular representation of G. Our idea is to express p in terms of 71'. If x E X and g E G, we have Of course xg need not lie in X. x(gp) = xg 224 PERMUTATIONAL REPRESENTATIONS [CHAP. 7 We know, however, that xg belongs to the same coset as xg. Hence xg = axg where a E H. Now a is clearly dependent on x and g, and we denote it by aX,g. Substituting aX,g for a, we have xg = |
aX,gxg (7.4) or (7.5) Now any element of G can be expressed in the form hx for hE H and x EX. Therefore (hx)(gp) = hxg = h(x(gp)) = hax,g X(g7T), i.e. (hx)(gp) = (ha X,g)(x(g7T» This equation suggests that the effect of gp on an element hx of G can be explained by what happens to h (it goes to the element hax,g of H) and what happens to x (it goes to X(g7T)). (7.6) We express this formally in the following theorem which is due (essentially) to Frobenius. Theorem 7.3: Let G be a group and H a subgroup of G. Let X be a right transversal of H in G. Then there is a faithful representation 0 of G as a group of per mutations of H x X (the cartesian product of H and X) defined by (h, x)(gO) = (hax,g, X(g7T» Proof: The proof is an adaptation of the discussion of the last few paragraphs. First let 7T be the coset representation with respect to H with right-transversal X. For each g E G the permutation g7T gives rise to a permutation gA of H x X which is defined as follows: (h, x)(gA) = (h, X(g7T», (h E H, x E X) (7.7) Note that gA is a permutation of H x X. For if then (h, x)(gA) (h', X')(gA) (h, X(g7T» (h', X'(g7T» Since g7T is a permutation of X, it follows from X(g7T) = X'(g7T) that x = x'. Hence we have proved that gA is one-to-one. But gA is also onto since g7T is onto X. Clearly gA is then a permutation of H x X. Now as we saw in (7.5), if g E G, then For each g E G define (h, X)ga = (hax,g, x |
) xg = aX,g(x(g7T», (aX,g E H) We verify that ga is a permutation of H x X. Suppose (h, x) E H x X. Then (ha;'~, X)ga = (ha;'~ax,g, x) = (h, x) Thus ga is a mapping of H x X onto H x X. It remains to verify that ga is one-to-one. Sup pose that (h, X)ga = (h', X')ga. This means that (hax,g, x) = (h'ax·,g, x') and therefore we find x = x' and hax,g = h'ax,g from which h = h' and (h, x) = (h', x'). Hence ga is a permutation of H x X. Finally we compute gagA. (h, X)(gagA) = ((h, X)ga)gA = (hax,g, x)gA Thus (ga)(gA) = gO (7.8) (7.9) Sec. 7.5] FROBENIUS' VARIATION OF CAYLEY'S THEOREM 225 As both gCI and gA are permutations of H X X, g(} is a permutation of H xX. Let gl' g2 E G. We must show that (gl(})(g2(}) = (glg2)(} (Note that the left-hand side of (7.10) is the product of two permutations.) To facilitate the proof of (7.10) we introduce the following notation: (7.10) where g is an element of G. Equations (7.8), (7.9) and (7.11) yield Note that as 17 is a homomorphism, (g117)(g217) = (glg2)17 so that (h, x)g() = (hax,g, Xa) Applying (7.12) twice and (7.13) once, we have (haX,Ql' X(1)g2() = (haX,Q1 a X"1,Q2' (Xa1)a2) (ha X, Q1 aX"1,Q2' X(a |
1a2)) = (haX,Q1 aX"1,Q2' X(3) On the other hand, again on using (7.12), we have (h, X)(glg2)(} = (ha X,Q1Q2' X(3 ) (7.11) (7.12) (7.13) (7.14) (7.15) To prove that () is a homomorphism, we must show that the right-hand side of (7.14) is equal to the right-hand side of (7.15), i.e. we must show that To accomplish this we use equations (7.5) and (7.6) and obtain X(glg2) = aX,glg2(Xa3) Also from (7.13). This means (7.16) (7.17) (7.18) Since X(glg2) = (xg1)g2' follows and therefore () is a homomorphism. the right-hand sides of (7.17) and (7.18) are equal. Thus (7.16) It only remains to show that () is one-to-one. Assume gl() = g2(}' Then (h, X)g2() for all pairs (h, x) E H X X; and in particular, if (h, x) = (1,1), (h, X)gl() = (a1, gl' l(g 117)) = (al,g2' l(g 217)) and 1 (g117) = 1 (g217) (7.19) from which Using equation (7.5) with x = 1, we see that gl = l(glP) = al,gl (1(g117)), aLQ1 = al,g2 g2 = 1 (g2P) = al,g2(1(g217)) Using (7.19) we conclude that gl = g2' Thus () is one-to-one. This completes the proof of Theorem 7.3. We call the homomorphism () of Theorem 7.3 a Frobenius representation (with respect to H). Of course () depends on X as well, but it can be shown that in a sense this dependence does not matter. 226 Problems PERMUTATIONAL REPRESENT |
ATIONS [CHAP. 7 7.13. Describe in detail a Frobenius representation for the symmetric group on {I, 2, 3} relative to the sub groups given in Problem 7.11(a) and (b). Solution: (a) The set X = {" T} resentation of G on H X X described above. The elements of G are is a right transversal of H = {" cr, u2} in G. Let fJ be the faithful rep We use the formula xu = ax,gxu to calculate ax,g. Note that L = " a = I, ;;2 = " T = T, UT = T, U 2 T = T. As an illustration of the procedure we calculate aT,CT. Now TU = aT,CTTU. Since Tcr = cr2T, we have ro = T and aT,CT = TUT = u2 • Similar calculations lead to a",, cr a" CT2 = u2 aL,a- a£~T =, UL,UT = U a" CT2T = u2 aT,L =, aT,u = u2 aT,U2 = u - aT,T, cr2 ur,a-2r = U ar,UT We use the definition of fJ given in the statement of Theorem 7.3. The effect of 1T on U is given in the solution of Problem 7.11(a). The results, repeated here for convenience, are 1':11" = U1T = (u2)1T': T1T = (UT)1T = (u2T)1T:,--+ T, T --+, L ~ t, T 4 T We can now calculate the effect of UfJ for each U in G. In particular, (h,,)uo = (ha"", '(U1T» = (hu,,) and (h, T)UfJ = (haT.CT, T(U1T» = (hu2, T) for all hE H We list the effect of the permutations UfJ for the elements of G.,fJ : (h, x) --+ (h, x) (hEH, x EX) ufJ : u2fJ : (h,,) --+ (hu,,), (h,, |
) --+ (hu2,,), TfJ : (h,,) --+ (h, T), (UT)fJ : (h,,) --+ (hu, T), (h, T) --+ (hu2, T) (hEH) (h, T) --+ (hu, T) (hEH) (h, T) --+ (h,,) (h, T) --+ (hu2,,) (hEH) (hEH) (u2T)fJ : (h,,) --+ (hu2, T), (h, T) --+ (hcr,,) (hEH) One could check that fJ is a homomorphism by inspecting (UlfJ)(U2fJ) and (U1U2)fJ, where (U1U2 E G). The above description of fJ immediately shows that fJ is one-to-one. (6) Here H = {I, T}; X, a right transversal of H in G, is given by X = {" cr, u2}. The Frobenius representation fJ is then given by,fJ : (h,x) --+ (h,x) (h E H, x E X) ufJ : (h,,) --+ (h, u), (h, u) --+ (h, ( 2), (h, ( 2 ) --+ (h,,) u2fJ : (h,,) --+ (h, ( 2), TfJ : (UT)fJ : (h,,) --+ (hT,,), (h,,) --+ (hT, ( 2), (h, cr) -> (hT, cr), (h, cr2 ) --+ (hT,,) (h, ( 2) --+ (h, u) (h, u) --+ (h,,), (h, u) --+ (hT, ( 2), (h, ( 2) --+ (hT, u) (hEH) (hEH) (hE H) (h E H) (u2T)fJ : (h,,) --+ (hT, u), (h, u) --+ (hT,,), (h, ( 2) |
--+ (hT, cr2 ) (h E H) 7.14. Describe in detail the Frobenius representations for the alternating group of degree 4 relative to the subgroup H given in Problem 7.12. Solution: Here G consists of and H = {" Tl, T2, Ta}. A right transversal of H in G is X = {" u, u2 }. The Frobenius representa tion fJ is then described as follows: Sec. 7.6] APPLICATIONS TO FINITELY GENERATED GROUPS 227,0 : (h, x) --> (h, x), (h E H, x E X) uO: (h,,) --> (h, u), (h, u) --> (h, ( 2), (h, ( 2) --> (h,,) u20 : (h,,) --> (h, ( 2), (h, u) --> (h,,), (h, ( 2) --> (h, u) (h E H) (hEH) 'T10 : (h,,) --> (h'Tl',), (h, u) --> (h'T2' u), (h, ( 2) --> (h'Ta, ( 2) (h E H) hu)o: (h,,) --> (h'Tv u), (h, u) --> (h'T2, ( 2), (h, ( 2) --> (h'Ta,,) (h,,) --> (h'Tv ( 2), (h, u) --> (h2,,),. (h, ( 2) --> (h'Ta, u) (~, ( 2) --> (h'TV ( 2) (h, u) --> (h'Ta, u), (h,,) --> (hz,,), (h E H) (h E H) (h E H) (h E H) (h,,) --> (h2, u), (h,,) --> (h'T2, ( 2), (h, u) --> (h'Ta,,), (h, u) --> (h'Ta, ( 2), (h, ( 2) --> (h'Tl',) (h, ( 2) --> (h'Tv u) (h E H) |
('T1U2)/I : 'T20 : ('T2U)O: ('T2u2)O: 'Tao: (h,,) --> (h'Ta,,), (h, u) --> (h'Tl' u), (h, ( 2) --> (h'T2' ( 2) (hEH) ('Tau)o: ('Tau2)o : (h,,) --> (ha, u), (h,,) --> (h'Ta, ( 2), (h, u) --> (h'Tl,,), (h, u) --> (h'Tl> ( 2), (h, ( 2) --> (h'T2,,) (hEH) (h, ( 2) --> (h'T2' u) (h E H) 7.6 APPLICATIONS TO FINITELY GENERATED GROUPS a. Subgroups of finite index Frobenius' representation, although only a variation of Cayley's, is very useful. Here we shall give one application of this representation. First we recall a definition given in Chapter 4. Definition: A group G is finitely generated if it can be generated by a finite set, i.e. if there is a finite subset S (# \Z» of G such that for each g E G there are elements s1' S2' •••, sn E S and integers € l "'" €n g = SEl ••• sEn (€i = ±1) such that 1 n Theorem 7.4 (0. Schreier): A subgroup of finite index in a finitely generated group is finitely generated. Proof: Let G be a finitely generated group. Let S be a finite set of generators of G, with lSI = m. Suppose H is a subgroup of finite index in G. Choose a right transversal X = {Xl'..., X) of H in G, with Xl the identity. Notice that j < <Xl by assumption. Let () be a Frobenius representation of G with respect to H given in Theorem 7.3. If hE H, then h = 1 and au = h so that we have (1, l)(hO) = (h,l). But h can be written as h = S~l ••• s:n (€j |
s- 1s = X = x by equation (7.3), page 219. Let y = XS-l; then ys = x and a -1 x,s -1 = xS- 1SX- 1 = yS(YS)-1 = a - - ~s i.e. a x,s-1 is the inverse of ay,s. Thus H is actually generated by the elements ax,s, x EX and s E S. The number of these elements is jm. (In fact one can lower this bound and prove that H can be generated by 1 + (m -l)j elements. For a proof of this result see Theorem 8.13, page 264.) Note that in Section 7.6a we have actually proved that H is generated by ax,., x EX, s E S, without the assumption that IXI and lSI are finite. Problems 7.15. Let G be the symmetric group of degree 3 and let H be a subgroup of index 2 in G. Find the set of generators of H described above. Solution: We use the description of G in Problem 7.11(a). Now a subgroup of index 2 in any group is IHI = 3. Hence H = {" 0', 0'2}, normal. Thus if H is of index 2 in G, H is normal in G. Moreover, since this is the only subgroup of order 3 in G. A right transversal of H in G is X = {,,7'}. Clearly G can be generated by 0' and 7', and so H is generated by Thus we find that,,0',0'2 generate H. Of course H is actually generated by 0' alone. 7.16. Let G be any group and let H be a subgroup of G of index 2. Prove that if G can be generated by two elements, then H can be generated by three elements. Solution: Suppose that G is a group generated by e and d and that H is a subgroup of index 2 in G. If both e and d are in H, then H d G and H is not of index 2 as initially assumed. Without loss of generality we may suppose that e Iil H. Then the co sets of H in G are H and He. Thus every ele ment of G can be written in the form he or h (h E H). This means that {I, e} is a |
right transversal of H in G. Therefore G is generated by the elements Hence H is generated by the three elements at, d, ac, c and a c, d' Sec. 7.6] APPLICATIONS TO FINITELY GENERATED GROUPS 229 7.17. Let G be generated by a and b and suppose that N is a normal subgroup of G such that GIN is generated by Na and GIN is infinite cyclic. Find a set of generators for N in terms of a and b. (Hard.) Solution: It is clear that X = {I, a±l, a±2,... } is a right transversal of N in G. If b (/: N then since Nb = (Na)W for some integer w, ba- w E N. Put e = ba- w • On the other hand if bEN, put e = b. Clearly gp(a, c) = G. We therefore take S = {a, e}. The generators of N are the elements If x = ai and s ~ a, ax,s aia(aia)-l = aia(ai+ 1)-1 1 ax • s = XS(X"S)-l (xEX, S E S) If x = ai and 8 == C, ax,s = aie(aie)-l = aiea- i Hence the elements {...,a-lea, e, aea- l,... } are a set of generators for N. Since either e = ba- w or e = b, we can restate this set of generators in terms of a and b thereby obtaining a set of gen erators of G of the desired kind. 7.18. Let G be generated by a and b. Find generators for all possible subgroups of index 2. Hence show G has at most three subgroups of index 2. Solution: Let H be a subgroup of index 2 in G. Then we may have (1) a (/: H, b E H, (2) a E H, b (/: H, (3) a (/: H, b (/: H. In case (1) take X = {I, a} and S = {a, b}. Then generators of H are the ax,s (x E X and s E S). Thus H is generated by b, a2 and aba- l. In case |
(2), proceeding as in (1) with X = {I, b}, H is generated by a, b2 and bab- l. In case (3), ab = e E H. Take X = {I, a}, and S = {a, e}. Then H is generated bye, aea- l and a2• So the possible subgroups of index 2 are: (1) gp(b, aba- l, a2 ), (2) gp(a, bab- l, b2 ), (3) gp(ab,a2ba- l,a2 ) 7.19. Let G = gp(a, b, e) and let N be a normal subgroup of G of index 3 with GIN =: gp(Na). Suppose N contains band e. Find a set of generators for N in terms of a, band e. Solution: Choose S = {a, b, e} and X = {I, a, a2} •. Then the elements ax,s, with x E X and 8 E S, gen erate N. Thus N is generated by a3, b, e, aba-l, aea- l, a2ba- 2, a2ea- 2. 7.20. Prove that if a group G contains a subgroup H of index 2 which is cyclic, then every subgroup of G of index 2 can be generated by two elements. (Hard.) Solution: Let H be generated by b and suppose that a(/: H. It follows that G = gp(a, b). From Problem 7.18 the possible subgroups of index 2 are (1) gp(b, aba- l, a2) = H l, (2) gp(a, bab- l, b2) = H 2, Clearly b (/: H2 or H 3, as then each of them would actually be equal to G. Thus H = H l • Since H is the cyclic group generated by b, aba- l = br and a2 = bs for some integers rand 8. We have (7.20) and (7.21) The generators of H2 are a, bab- l and b2. bab- l = (ba)b- l = a-lbr- l from (7.20), and so H2 is generated by a, br- l, b2• But |
gp(br- l, b2) is cyclic generated by c, say, as it is a subgroup of a cyclic group. Thus H2 can be generated by two elements. The generators of Ha are ab, a2ba- l and a2• Hence ab, a2ba- l (ab) and a2 are generators for Ha. Using (7.21) we conclude that ab, bs+ 2 and bs are generators for H 3• But gp(bs+ 2, bS ) is cyclic generated bye, say; so H3 is generated by two elements and the proof is complete. 230 PERMUTATIONAL REPRESENTATIONS [CHAP. 7 c. Marshall Hall's theorem The second application of permutational representations is due to Marshall Hall. Theorem 7.5: The number of subgroups of finite index j in a finitely generated group is finite. This theorem may be restated as follows: Let G be a group which is generated by aI,..., an where n < 00. Suppose j is a fixed positive integer. Then the number of subgroups of G of index j is finite. Proof: Let Sj be the symmetric group on 1,..., j. For each subgroup H of index j in the finitely generated group G choose a right transversal X H (we emphasize that XII contains the identity of G). To avoid confusion between the number 1 and the identity of G, we shall write the identity (for this proof alone) as e. Thus we have e E Xw Let TrH be the coset representation of G with respect to the transversal X H (see Section 7.4). Then TrH is a homomorphism of G into Sx, Since IX HI = j, it is easy to prove that there exists an iso morphism 1>H: SXH ~ Sj such that e E SXH moves e or leaves it fixed according as to whether e1>H moves 1 or leaves it fixed (see Problems 7.21 and 7.22 below). H Note that 'lr H = 7rIJ1>H : G ~ Sj is a homomorphism of G into Sj since it is the composition of two homomorphisms. Note also that if Hand K are two subgroups of index j and H # K, then 'lrH # 'lrK' for there exists an element g E H but g E K (or vice |
versa). Then e(gTrH) = e for e(g7rH) = eg = e as g E H (see Section 7.4). Accordingly 1'lrH = 1. On the other hand, as g E K, e(gTrK) # e and hence 1 'lr H # 1. Thus 'lrJi # 'lr K if H # K. We have therefore found that the number of subgroups of index j in G is certainly not greater than the number of homomorphisms of G into S.. This is where the fact that G is finitely generated comes in. For suppose G is generated by al'..., an' If 1>, e are homomorphisms of G into Sj such that ai1> = aie for i = 1,..., n, then 1> = e. To prove this, observe that if g E G, then J (i = ±1, i j E {1,...,n} and Since 1> and e agree on every element of G, 1> = e. This means that the number of homo morphisms of G into S is finite since the number of possible images of the generators of G is finite (at most (j!)n). This completes the proof of the theorem. d. One consequence of Theorem 7.5 Let G be a finite group and e a homomorphism of G onto G. It follows that e is an iso morphism, for IGI = IGel = IG/Ker el and hence Ker e = {1}. If G is not finite, is it pos sible to have a homomorphism e of G onto G with e not an isomorphism? For example, if P is a p-Priifer group (see Section 6.2c, page 191), let e: P ~ P be defined by xe = px, x E P. Then Pe = P; but as P has an element of order p, e is not an isomorphism. In the following theorem we prove a result which tells us that for a special class of groups every onto homomorphism is an isomorphism. Theorem 7.6 (A. I. Mal'cev): Let G be any finitely generated group whose subgroups of finite index have intersection 1. Then every homomor phism 1> of G onto G is an automorphism. Sec. 7.6] APPLIC |
ATIONS TO FINITELY GENERATED GROUPS 231 Proof: Let K be the kernel of.p and let L be any subgroup of finite index in G. If L is of index j, then the number of subgroups of G of index j is finite. Let these subgroups be L = Lt, L 2, • • •, Lk Now, by Theorem 4.18, page 117, G = G.p ~ GIK and so the number of subgroups of index j in GIK is precisely k, the number of subgroups of index j in G. Let MtlK,..., MklK be these subgroups of index j in GIK. Then M I, • • •, Mk are k distinct subgroups of index j in G, by Corollary 4.20, page 121. Thus the M;'s are simply a rearrangement of the L i• Therefore every Li is an Mi and so contains K. In particular L;;JK This means that every subgroup of finite index contains K. Hence K is contained in the intersection of the subgroups of finite index. By hypothesis, this intersection is 1. So K = 1. Accordingly.p is one-to-one, and.p is an automorphism. This theorem is important in current research in group theory. Problems 7.21. Prove that there exists an isomorphism fl between S{Xl' x2} and S2 such that if xi(J = Xj, then i«(Jfl) = j. (Hint: see Problems 7.8 and 7.9, page 221.) Solution: Let a: {XI, X2} -> {1,2} be defined by Xja = j, j = 1,2. Let fl be defined by (J E S{xl' x } and 2 Then a, fl define a permutation isomorphism (see Problems 7.8 and 7.9) and therefore fl is an isomor phism with the required effect. 7.22. Prove that in general there exists an isomorphism fl between S{xl'".,X,..., xn} and x;(J = xi' then i«(Jfl) = j. (Use Problems 7.8 and 7.9.) (J E S{X 1 } and Sn such that if n Solution: Let a:{x v |
...,xn}->{1,2,...,n} be defined by x;a=i. If (JES{xl""'X if xi(J = Xj (JflESn to be the mapping that sends i -> j (as (J is a permutation of {xI'..., xn}, (Jfl is a per mutation of {1,2,..., n}). It is clear then that fl is onto Sn' and hence a, fl provide an isomorphism of permutation groups. Thus fl is the required isomorphism. }' define n 7.23. Prove that if Hand K are subgroups of G, then each coset of H intersects a coset of K either in the empty set or in a coset of H n K. Hence prove that if Hand K are of finite index in G, so is H n K. Solution: If a coset of H and a coset of K have an element g in common, then the two co sets are Hg and Kg. X E HgnKg if and only if X = hg = kg, for some hE Hand k E K. But hg = kg if and i.e. hE HnK. Thus X E HgnKg if and only if X E (HnK)g. Then HgnKg = only if h = k, (HnK)g and the two co sets meet in a coset of HnK. If H is of index nand K is of index m, at most nm cosets can be found as intersections of a coset of H with a coset of K. Furthermore these are all the cosets of H n K, for any coset (H n K)g = Hg nKg and so is the intersection of a coset of H by a coset of K. Therefore H nK is of finite index in G if both Hand K are. 232 7.24. PERMUTATIONAL REPRESENTATIONS [CHAP. 7 Let G be finitely generated with a subgroup of index j. Prove that the intersection of all subgroups of index j in G is a normal subgroup of finite index. (Hard.) Solution: By Theorem 7.5, G has only a finite number of subgroups, Mv..., Mn say, of index j. Now if M |
is any subgroup of index j, it is easy to prove that x-IMx = M is also of index j. (If the co sets of Mare Mg v...,Mgj, then the cosets of Mare MX-lgIX, Mx-lgzx,...,)iiIx-lgjx. For if g E G, x-lgx E Mg i, for some i, implies that g E x-IMxx-lgix = Mx-lgiX.) Hence M l nM 2 n··· nMn = K is a normal subgroup of·G, for if g E K and x E G, x-lgx E x-IMlx, x- IM 2x,..., x-IMnx. But x- 1M IX, •.., x- 1M nX are n distinct subgroups of index j. Hence they must be all the subgroups of index j (perhaps in a different order). Thus x-lgx E K and so K <J G. K is of finite index by repeated application of Problem 7.23. 7.25. Let G and H be two groups and suppose that G satisfies the conditions of Theorem 7.6. Let o : G ~ Hand ¢: H ~ G be epimorphisms. Prove that G = H. Solution: O¢ is an epimorphism of G to itself and so by Theorem 7.6, O¢ is an isomorphism. Then if g # 1, g E G, g(o¢) # 1 and thus go # 1. Therefore 0 is one-to-one and 0 is an isomorphism of G to H. 7.26. Let Nand M be unequal normal subgroups of a finitely generated group G, M;;;> N. Suppose the intersection of the subgroups of finite index in GIN is the identity. Prove that GIM is not isomorphic with GIN. Solution: Let 0: GIM -> GIN be such an isomorphism. Let /l: GIN ~ GIM be defined by Ng ~ Mg. It is easy to verify that /l is an epimorphism. Then /l0 is a homomorphism of GIN onto itself. By Theorem 7.6, /l0 is an isomorphism. Now if gEM - N |
, is the identity of GIN. Thus /lO is not an isomorphism. This contradiction yields the required result. (Ng)(/lo) = Mo 7.7 EXTENSIONS a. General extension Suppose G is a group with a normal subgroup H and that G/H "'" K. Then, using the terminology introduced in Chapter 5, G is an extension of H by K. It is convenient to gen eralize this concept and to say that G is an extension of H by K if G has a subgroup ii with ii "'" Hand G/il "'" K. It is our aim to investigate how a group is built as an extension of one group by another. In this section let G be a fixed group and H a normal subgroup of G. Let cp be an iso morphism of G/H onto K. Let X be a left transversal of H in G, i.e. a set of elements of G containing one and only one element from each left coset of H in G with 1 EX. If g E G, g = xh for some x E X and some h E H. It is easy to see that this expres sion for g is unique. Let g E G, x E X; then gx belongs to some coset of H in G, say the coset yH where y E X. Therefore gx = yh for some hE H. Now h is uniquely determined by g and x; we denote h by mg,x. Thus gx = ymg,x The elements mg,x correspond to the elements aX,g introduced in Section 7.5b. (We use mg,x instead ofax,g because here we are dealing with left instead of right cosets. We will explain in Section 7.7c the minor reason why we use left co sets here.) (7.22) Note that cp, the isomorphism of G/H onto K, is a one-to-one mapping of the set of left cosets of H onto K. Therefore we can unambiguously denote the representative of the coset gH by XI, if (gH)cp = k. In particular then, Xl = 1. With this notation, X = {Xk IkE K} Sec. 7.7] EXTENSIONS 233 Notice that as (XkXkB)cp = {(xkH)(XkB)} |
cp = (xdl)cp(xkB)cp = kk' where k, k' E K, resentative of the coset XkxkB is Xkk'. Then, from (7.22) we have the rep We suppress the x's and write mk,k' for mxk.xk,. Thus, Every element g in G can be written uniquely in the form xkh where X/c EX, hE H. To express the product of two elements xkh and xk,h' as the product of an element of X by an element of H, we proceed as follows: • (7.24) (7.23) X~,1 hXk' E H Observe that So Xkk' E X and mk,k'x~,1 hxk-h' E H. The right-hand-side of (7.24) looks less complicated if we introduce the notatIOn is a normal subgroup of G. for Xk' Xk'; then since H -1 h hk'..... (7.25) It appears from equation (7.25) that the extension G of H by K that we have been in specting is determined by the mk,k' and by the images hk' of the elements h obtained by con jugation by the Xk', i.e. by forming x-;;~ hXk'. One may conveniently think of the elements mk,k' in H as the images of a function m of two variables (coming from K) with values in H. In other words, we may think of m as a mapping from the cartesian product K x K into H, where we use mk,k' to denote the image of (k, k') E K x K under this mapping m. Con tinuing with this analysis, let us turn to the elements hk. For each k E K we have a mapping, ka say, of H into H, namely the mapping which sends an element h in H to the element hk. In a way then the group G is made up of two mappings: (1) a mapping m from K x K into H, (2) a mapping a of K into a set of mappings of H into H. (The effect of ka is to map h to h k.) • Indeed m and a determine G up to isomorphism (see Problem 7.27). If we |
add enough con ditions to these mappings, one can reverse the procedure we have been outlining and con struct from H, K and the mappings m and a an extension G of H by K. (See A. G. Kurosh, The Theory of Groups, Vol. II, Chelsea, 1960, translated by K. A. Hirsch, for details.) We will not tackle the general problem but we will consider only a particular case (in Section 7.7c). Problems 7.26. Let G be the dihedral group of degree 3. G is an extension of a cyclic group of order 3 by a cyclic group of order 2. After choosing a suitable left transversal, find the mappings m and a introduced above. Solution: Using the notation of Section 3.4f, page 75, let H = {O"l> 0"2' 0"3}' Then, as can be easily checked, IGI = 6, G/H is of order 2 and thus a cyclic group of order 2. Let {t, T} be a left H <J G. Since transversal for H in G. Then "l: is the identity mapping of H onto H, while ra sends 0"1 to 0"2' 0"2 to 0"1 and 0"3 to 0"3' 234 7.27. PERMUTATIONAL REPRESENTATIONS [CHAP. 7 Let G, G be two groups, both extensions of H by K. Assume that H is actually a subgroup of both G and G. Let X, X be transversals of H in G, G respectively. Let m, in and a, ii be the mappings ob tained above. Prove that if m = in and a = ii, then G == G. Solution: The elements of G are uniquely of the form Xkh, where Xk E X, hE H, while the elements of G are uniquely of the form xkh, where Xk E X. Let 8: G..... G be defined by (Xkh)8 = xkh. Then 8 is a one-to-one onto mapping. To prove 8 is a homomorphism, we consider the product of two elements of G. Thus G == G. b. The splitting extension Suppose G is as in Section 7.7a. Consider the particular case where mx,x' = 1 for all x, x' EX. |
By examining equation (7.22) xx' = x", i.e. the product of two elements in X is again in X. Furthermore, we have 1 EX. Let x E X, and let y E X be such that x-IH = yH; then xy E H. Accordingly xy = 1mx,y where mX,y E H However mX,y = 1, and so xy = 1. Thus x has an inverse in X, and so X is a subgroup of G. Since H <J G, XH is a subgroup (Theorem 4.23, page 125). But every element of G is of the form xh, x EX, hE H. Hence XH = G. Since distinct elements of X lie in distinct co sets of H in G and 1 E H, we have XnH = {1} Since X and H are subgroups of G with XH = G, HnX = {1} and H <J G, G is said to split over H. X is called a complement of H. Note that if G splits over H, we can choose any complement X of H as a transversal for H in G, since two distinct elements of X belong to distinct co sets of H. Now X is a sub group of G, and it follows that if we define mg,x as in Section 7.7a with X as transversal, then mx,x' = 1 for all x, x' in X. If G splits over H and X is a complement of H, then G/H = HX/H == X/HnX == X In other words, G is an extension of H by X; that is, if G splits over H, G/H is isomorphic with any complement of H. It is convenient to introduce the following definition. We say that a group G is a splitting extension of HI by Xl if there exists a normal subgroup H of G isomorphic to HI such that G splits over Hand G/H == Xl. Problems 7.28. Prove that the dihedral group D of order 8 is a splitting extension of a cyclic group of order 4 by a cyclic group of order 2. Solution: We use the multiplication table for D given in Problem 7.6(a), page 220. Let H = {I, at> a2' a3}; then H is cycl |
ic of order 4 since Sec. 7.7] EXTENSIONS 235 Moreover H is a normal subgroup of G. This can be verified either by direct calculation, checking that if dE D and hE H then d-lhd E H, or by noting that H is of index 2 in D. Now letting X = {1, a4}, a! = 1 and so X is a subgroup of D. Since a4 tl H, it follows that the co sets H, a4H are disjoint. Therefore, as there are 8 elements in H U a4H, Finally HnX {1} So D is a splitting extension of H by X as required. D = H U a4H or D = XH 7.29. Prove that neither the dihedral group of order 8 nor the quaternion group of order 8 is a splitting extension of a group of order 2 by a group of order 4. Solution: Let E stand for either the dihedral group of order 8 or the quaternion group of order 8. Suppose E is a splitting extension of a subgroup H of order 2 with complement X of order 4. Then H is a normal subgroup of G. Now suppose H = {1, h}. then e-lhe E H. Since h # 1, then x-lhx = h. Now e-lhe # 1. Thus e-lhe = h In particular if x E X, E = XH. Since X is of order 4, X is abelian (see Problem 5.19, page 140). for all e E E. If e E E, Now suppose e, lEE; then e = x'h', I = x"h" (x', x" E X, h', h" E H) Recall that every element of H commutes with every element of X and that X is abelian; then el = x'h'· x"h" = x'x"h'h" = x"x'h"h' = x"h"x'h' = Ie Thus E is abelian. But neither the dihedral group of order 8 nor the quaternion group of order 8 is abelian. Hence we have a contradiction to the assumption that either of these groups is a splitting extension of the type described. Alternate proof: If E is a splitting extension of H by X of order 2 and |
4 respectively, since H n X = {1}, H <l E and X <l E (as X is of index 2), it follows that E = X X H, the direct product of X and H by Corollary 5.17, page 145. From this it again follows that E is abelian, thus producing a contradiction. 7.30. Prove that the quaternion group of order 8 is an extension of a group of order 4 by a group of order 2, but is not a splitting extension. Solution: We use the multiplication table for!J{, the quaternion group of order 8 given in Problem 7.6(b), page 220. First let K = gp(a3)' Then K is of order 4 and therefore of index 2. Thus K is a normal sub group of!J{, and it follows that!J{ is an extension of K by!J{/K. Clearly!J{/K is cyclic of order 2, and so!J{ is an extension of a group of order 4 by a group of order 2. Now suppose!J{ splits over any subgroup K of order 4. Then!J{/K is of order 2 and hence In abelian. Therefore K contains the commutator subgroup of!J{ by Problem 4.68, page 116. particular, K contains a2 since Now we must check that if x is any element except 1 and a2 of!J{, then x is of order 4. This can be done directly, using the mUltiplication table for!J{. Suppose now, if possible, that!J{ is a splitting extension of K by X. Then the subgroup X is of order 2, say X = {1, x}. But as we saw above, X is of order 4 since x # 1, x # a2' So the subgroup X is not of order 2. This is a contradiction and so the desired result follows. Alt~rnate proof: If!J{ is a splitting extension of a subgroup K of order 4 by a subgroup X of order 2, then KnX = {1} and K <l!J{. But every subgroup of the quaternion group is normal (Problem 5.43, page 158). Therefore X <l!J{. It follows, by Theorem 5.16', page 146, that!J{ = K X X. |
This implies!J{ is abelian, which is a contradiction. 7.31. Is the alternating group of degree 4 a splitting extension of a group of order 6 by a group of order 2? Solution: By Problem 5.1, page 131, the alternating group of degree 4 does not contain a subgroup of order 6. Thus the result follows. 236 PERMUTATIONAL REPRESENTATIONS [CHAP. 7 c. An analysis of splitting extensions Suppose now that G is a splitting extension of H by K with complement X. Then by Section 7.7b, we may use X as a transversal for H in G, and mx,x' = 1 for all x, x' EX. Consequently each element g EGis uniquely expressible in the form and equation (7.25) becomes g = xkh where k E K, h E H xkhx/c, h' = Xkk' hk' h' For each k E K the mapping h-'>hk (hEH) (7.26) (7.27) (7.28) is an automorphism of H, for H <J G and hk = Xk 1 hXk that the mapping h -'> hk (h E H) is an automorphism of H. imply, by Problem 3.57, page 85, Finally we remark that if we let a be the mapping which assigns to each element k E K the automorphism (h E H) of H, then a is itself a homomorphism of K into the group of automorphisms of H. (This explains why we used a left transversal in Section 7.7(a), namely so that the mapping a be a homomorphism.) We have only to prove that (kk')a = kak'a (7.29) To verify (7.29), let us take an arbitrary element hE H and apply the automorphism (kk')a to h: h[(kk')a] X';k: hXkk' = (XkXk,)-lh(XkXk') (Xk,lXk1)h(XkXk') = Xk}(Xk 1 hXk)Xk' since Xkk' = XkXk' by (7.27) Xk,l [h(ka)]Xk' [h(ka)] |
(k'a) h[(ka)(k'a)] by the definition of the product of two automorphisms Thus we have (kk')a = (ka)(k'a) We now replace hk' by h(k'a) in (7.27). Then (7.27) becomes xkh· Xk' h' = Xkk" h(k' a)h' (7.30) What is the situation we have arrived at? We started from a group G which splits over H. Then we chose a subgroup X of G such that G = XH and XnH = {I}. We were given an implicit isomorphism cp of G/H with K and we denoted the element in X which corresponds to k in K by Xk. Then we observed that the elements of G were uniquely expressible in the form xkh with k E K, hE H. The way in which elements of G are multiplied was then computed by making use of the existence of a homomorphism a of K into the automorphism group of H and by applying (7.30). Therefore we may suspect that if we are given (a) a group H, (b) a group K, (c) a homomorphism a of K into the automorphism group of H, then we can create a splitting extension of H by K. Indeed this is the case. All we have to do is to reverse the process we have described. To be precise, starting with the data (a), (b) and (c), we let G be the cartesian product of K and H, i.e. G = KxH = {(k, h) I kEK, hEH} Sec. 7.7] EXTENSIONS We define a binary operation in G according to the formula (k, h)(k', h') = (kk', h(k'a)h') The reader will note the strong similarity between (7.30) and (7.31). 237 (7.31) Before verifying that G is a group and a splitting extension of H by K, we will make the similarity of equations (7.30) and (7.31) even more evident. Let Xk = (k, 1) and h = (1, h), and define ka by h(ka) = (1, h(ka)). Then from (7.31) obtain This equation |
resembles equation (7.30) even more closely than (7.31) does. xkhxk' h' = Xkk' h(k' a)h' We will now prove that G is a group and that it is a splitting extension of H by K. (i) We note first that (7.31) defines a binary operation in G. (ii) The binary operation defined by (7.31) is associative: ((k, h)(k', h'»)(k", h") = (Ide', h(k'a)h')(k", h") = ((lcle')le", ([h(lc'a)h'](le"a))h") (7.32) (( kk')k", {[ (h(le' a) )(le" a)] [h'( k" a)] } h") N ow we work out (k, h)((lc', h')(k", h")) = (k, h)(k'k", h'(k"a)h") = (k( k' le"), [h( k' le")a][ h' (k" a) h"J) By (c) above, a is a homomorphism, and so (k'k")a = le'a °lc"a. Thus (k, h)((k', h')(le", h"» = (k(k'k"), [heCk' a)(k" a»)][h'(le" a)h"J) = ((lek')k", [(h(k' a»)(le" a)][h'(k" a)h"]) (7.33) The associative law immediately yields the equality of (7.32) and (7.33). (7.31) is an associative operation. Therefore (iii) There exists an identity in G: (1,1). Notice, of course, that the left-hand 1 in (1,1) is the identity of K while the right-hand 1 of (1,1) is the identity of H. To check that (1,1) is an identity, let (k, h) E G. Then (1, l)(k, h) = (k,l(ka)h) = (k, h) since ka is an automorphism of |
H, and so maps the 1 of H to itself. Similarly since 1a is the identity automorphism of H, and so leaves H identically fixed. (k, h)(l, 1) = (k, h(la)l) = (k, h) (iv) Finally we must check that every element of G has an inverse. Let (k, h) E G. We claim that is the inverse of (k, h). To prove this, we simply observe that (k, h)(lc-t, h-1(k-1a» = (1, [h(k-1a)][h-1(k-1a)]) = (1, (hh-1)(k-1a» since k-1a is an automorphism of H. Thus (k, h)(k-t, h-1(le-1a» = (1,1) Similarly (k- 1, h-1(k-1a»)(k, h) = (1,1) We have thus verified that (k-t, h-1(k-1a» is the inverse of (k, h). 238 PERMUTATIONAL REPRESENTATIONS [CHAP. 7 (i), (ii), (iii) and (iv) above establish in every detail the proof that G is a group. Next we verify that G is a splitting extension of H by K. To accomplish this, put fi = {(l,h) I hEH} and K = {(le,l) I leEK} Then it is easy to prove that fi and K are subgroups of G and that are isomorphisms of H onto fi and K onto K respectively. h ~ (1, h) and le ~ (le, 1) Now to prove that fi is a normal subgroup of G, observe again that if g E G, then If (1, h') E fi, then g = (le, h) = (le, 1)(1, h) g-1(1, h')g (le, h)-1(1, h')(le, h) ((le, 1)(1, h))-1(1, h')((k, 1)(1, h)) [(1, h)-1(k, 1)-1](1, h')((le, 1)(1, h)) Since (le |
, 1)-1 = (le-l, 1), we have (le, 1)-1(1, h')(k, 1) = (1, h'(lea)). Thus g-1(1, h')g = (1, h- 1)(1, h'(lea))(l, h) E fi since the product of elements of fi belongs to fi. Hence fi is normal in G as claimed. Clearly finK = {(1,1)} and G = Kii as we saw earlier, since (k, h) = (le, 1)(1, h). Consequently we have constructed from the data (a), (b) and (c) a splitting extension G of H with complement K. Therefore G is a split ting extension of H by K. The group G that we have constructed is called the splitting extension of H by K via a. We emphasize the importance of the above discussion and the related problems which follow. Problems 7.32. Construct a non-abelian group of order 6 as a splitting extension of a group of order 3 by a group of order 2. Solution: Recall that if we are given (a) a group H, (b) a group K, (c) a homomorphism a of K into the automorphism group of H, then we can construct a group from this data as follows. Consider the set G of all the pairs (k, h) (k E K, hE H) and define a binary operation in G by (k, h)(k', h') = (kk', h(k'a)h') Then G becomes a group which is a splitting extension of H by K. So in the case at hand, we have the group H (the cyclic group of order 3) and the group K (the cyclic group of order 2). We need the homomorphism a. This means in the first place that we need to know more about the automor phism group of H, the cyclic group of order 3. Now if H is generated by h, then H = {I, h, h2 } The mapping which sends each element of an abelian group into its inverse is an automorphism (check this). So if 1): H -> H is this automorphism, 1): 1->1, h->h- 1 =h2, h2 ->h- 2 =h Sec. 7.7] |
EXTENSIONS 239 712 = t, and so gp(7J) is cyclic of order 2. Thus the groups K and gp(7J) are isomorphic, and Now accordingly we can take a to be the isomorphism of K onto gp(7J). Let G be the splitting extension of H by K via a. To see how some of the elements of G combine, let K = {1, k}. Then (k, h)(k, 1) = (k2, h(ka)l) = (1, h2) Now (k, l)(k, h) = (k2, l(ka)h) = (1, h), and so is an extension of H by K. (k, h)(k, 1) :/= (k, l)(k, h). Thus G is non-abelian and 7.33. Are there non-abelian groups of order 2 X 777? Solution: There are non-abelian groups of order 2 X 777. To produce one such group, let H be cyclic of order 777. Then H has an automorphism r of order 2, namely the mapping r which sends every element of H to its inverse. So there is a homomorphism a of K, the cyclic group on k, of order 2, into the automorphism group of H, namely the one which sends k to r. Let G be the splitting exten sion of H by K via a. Then G is the required group. 7.34. Construct two non-isomorphic non-abelian groups of order 168. Solution: Let Hl = gp(h 1) be cyclic of order 84 and Kl = gp(k 1) cyclic of order 2. Let a be the homo morphism of Kl into the automorphism group of Hl defined by k1a: hi -? h~i. Then G1, the split ting extension of Hl by K 1, is of order 168. The center Zl of G1 is of order 2, consisting of (1,1) and (1, hi2) which can be easily checked by direct calculation. Now we construct a second group of order 168. Here we take H2 = gp(h2) to be of order 42 and K2 = gp(k2) to be cyclic of order 4. As usual, H2 has an automorphism |
of order 2, i.e. the mapping defined by i - i r: h2 -? h 2, i = 0, 1,...,42 Let {3 be the homomorphism of K2 onto gp(r) defined by k~{3 = ri, j = 0,1,2,3 Thus k~{3 = t. Let G2 be the splitting extension of H2 by K2 via {3. Then, as the reader may check by direct calculation, the center Z2 of G2 consists of (1,1), (1, h~l), (k~, 1), (k~, h~l) and is therefore of order 4. Now both G1 and G2 are non-abelian since IZll = 2, IZ21 = 4. Moreover if G1 and G2 are iso morphic groups, they have isomorphic centers. Therefore G1 and G2 are not isomorphic. 7.35. Construct all possible groups of order 30 which are extensions of a cyclic group of order 10 by a cyclic group of order 3. Solution: The solution of this problem requires knowledge of the automorphism group of the cyclic group H = gp(h) of order 10. If r is an automorphism of H, then hr is of order 10. The possibilities for hr are therefore Let r; be the automorphisms defined by the possibilities listed above (i = 1,2,3). Now ri = 1, 'Ti = 1, 'T~ = 1; thus none of the automorphisms of H is of order 3. So the only possible homomorphism a of a cyclic group K of order 3 into the automorphism group of H is the one which sends every ele ment of K onto the identity automorphism. The resultant splitting extension of H by K is then abelian (indeed it is isomorphic to the cyclic group of order 30). 7.36. Construct a non-abelian group of order 222 by using splitting extensions. Solution: Form a splitting extension of a cyclic group H of order 111 by a group K = gp(k) of order 2 via the homomorphism taking k to the automorphism which sends every element of H into its inverse. 240 PERMUTATIONAL REPRESENTATIONS [CHAP. 7 7.37. Construct a non-abelian group of order p3 for each prime p |
by using splitting extensions. (Hard.) Solution: Let H be the direct product of a cyclic group gp(a) of order p with a cyclic group gp(b) of order p. Each element of H is uniquely of the form arbs with 0 "" r < p, 0 "" s < p Let T: arbs.... arbs + r. We will show that T is an automorphism of H of order p. First let tn and n be any integers. We claim that (ambn)T = ambm +n Let tn = tn' + qp, n = n' + sp where 0 "" tn' < p, 0 "" n' < p. Then (ambn)T = am'bm'+n' = ambm'b n' = ambmbn ambm +n To verify that T is a homomorphism, observe that (amlbn1. am2 bn2 )T = aml+m2bml+m2+nl+n2 = amlbml+nl. am2bm2+n2 = (amlbnl)T(am2bn2)T Clearly Ker T = {1}, and so T is one-to·one. It is easy to check that T is also onto. Thus T is an automorphism. Note that aTP = ab p = a, so that TP acts as the identity on a and b, which form a set of generators of H. Hence T is of order p. Now let K = gp(k) be of order p. The mapping cr: k i.... Ti is an isomorphism. Then we form the splitting extension of H by K via cr. This gives a group G of order p3. G is non-abelian since (1, a)(k, 1) = (k, ab), but (k, 1)(1, a) = (k, a). d. Direct product Consider the special case of a splitting extension G of a group H by a group K via a homomorphism ll' in which ll' takes K onto the identity group of automorphisms of H. In this case G consists of the pairs (h, k) (h E H, k E K) with binary operation given by (h, k)(h', k') = (hh', kk'). So G is, in the terminology of Section 5.3a, page 143 |
, simply the external direct product of Hand K. The obvious usefulness of this construction is that we do not require any knowledge of the automorphism group of H to construct the direct product. Notice that if Ii = {(I, h) I hE H} and K = {(k,l) IkE K}, then G is the in ternal direct product of its subgroups Hand K, again in the sense of Chapter 5. We will not pursue this concept of direct product here any further. 7.8 THE TRANSFER a. Definition Suppose G is a group with an abelian subgroup A of finite index. The transfer is a special homomorphism of G into A. The use of such transfer homomorphisms has been important in the theory of finite groups. Here we will examine one application of the transfer and briefly mention another (at the end of Section 7.8d). To define the transfer T of G into A, choose a right transversal X of G in A. We repeat that A is an abelian subgroup of G of finite index n, say. Therefore IXI = n. Recall that if g E G, then g is the element of X in the coset Ag. Let Xl, X2, •••, Xn be the elements of X; then if g E G, we define a mapping T of G into A by gT = XIg(Xlg)-l. X2g(X2g)-1 •. •. • xng(Xng)-1 It is clear that gT E A since, as Xig and Xig belong to the same coset of A, Xig(Xig)-l E A (i = 1,...,n) This mapping T is the homomorphism of G into A mentioned above; it is called the transfer of G into A. There are two items to be verified: (1) T is a homomorphism and (2) T is independent of the choice of the transversal X. Sec. 7.8] THE TRANSFER 241 b. Proof that T is a homomorphism We compute (gh)T, where g, hE G: (gh)T (xlgh(Xlgh)-l) • (x2gh(X2gh)-1) •...• (xngh(xngh)-l) (xlgh(Xlgh)-l) |
• (x2gh(X2gh)-1) •...• (xngh(xngh)-l) (Xlg(Xlg)-l • xlgh(Xlgh)-l) • (X2g(Xzg)-1 • x2gh(xzgh)-1) • • (Xng(Xng)-l • xngh(xngh)-l) (since Xigh = xigh by (7.3), page 219) Now everyone of the elements Xig(Xig)-l, xjgh(xjgh)-l lies in A; since A is abelian, they must commute. So we can rewrite (gh)T in the form (gh)T = [XIg(XIg)-l. X2g(X2g)-1 •...• xng(Xng)-l] • [xIgh(Xlgh)-l • x2gh(X2gh)-I •...• xngh(xngh)-l] (i = 1,..., n) is a permutation of X (Section 7.4a). This means that But observe that Xi ~ Xig (7.34) xlgh(Xlgh)-l. x2gh(X2gh)-1 •...• xngh(xngh)-l simply consists of the n elements xih(Xih)-l (i = 1,..., n) multiplied together in some order. Since A is abelian, the order of such a product is immaterial. Thus xlgh(Xlgh)-l. Xzgh(X2gh)-1 •...• xngh(xngh)-l xlh(xlh)-l. x2h(X2h)-1 •...• xnh(xnh)-l = hT, by definition Then it follows from equation (7.34) that (gh)T for all g, h in G. Thus T is a homomorphism. c. Proof that T is independent of the choice of transversal We now prove that T is independent of the choice of the transversal X. The proof depends on an analysis of the product gT = Xlg(Xlg)-l. X2g(X2g)-1 •...• xng(Xng)-l where again |
{Xl, X2,..., Xn} = X is a right transversal of A in G. We recall from Section 7.4a that the mapping Xi ~ Xig is a permutation of X. Now every permutation of a finite set can be written as a product of disjoint cycles (Theorem 5.26, page 167). So, after relabeling the elements of X if necessary, we can assume that Xlg = X2, X2g = X3,..., Xk-lg = Xk, Xkg = XI Xk+lg = Xk+2, Xk+2g = Xk+3,..., Xk+l-l = Xk+l, Xk+lg = Xk+l Xn-m+lg = Xn- m+2, Xn- m+2g = Xn- m+3,..., Xn-lg = Xn, xng = Xn-m+l (Note that k + l +... + m = n, where n is the index of A in G.) Then gT = (Xlg(Xlg)-l. X2g(X2g)-1 •...• Xkg(Xkg)-l) • (Xk+lg(Xk+lg)-l. Xk+2g(Xk+'2lJ)-1 •...• Xk+lg(~)-I) •... • (Xn-m+lg(Xn-m+lg)-1 • Xn - m +2g(Xn- m+2g)-1 •...• Xn(gxng)-l) (Xlgx;l • X2gx;1 •...• Xkgx;l) -1 • (Xk+lgXk+2· Xk+2gXk+3·...• Xk+lgXk •... - I -1 -1) -1 • Xn- m+lgXn - m+2· Xn-m+2gX n- m+3·...• XngXn-m+l ( - I ) 242 PERMUTATIONAL REPRESENTATIONS [CHAP. 7 and thus (7.35) Note that Xlgkx~l |
= (Xlg(Xlg)-I) • (X2g(X2g)-I) •...• (Xkg(Xkg)-I) E A, since it is the prod uct of factors Xig(Xig)-1 which belong to A. Similarly, Xk+lg1xk"L,..., Xn-m+IXmX;;-:m+l EA. Weare now able to prove Lemma 7.7: Let A be an abelian subgroup of finite index in a group G and let X = in G. {Xl, X2,..., Xn }, Y = {Yl, Y2,..., Yn} be two left transversals of A Furthermore let T and 'T be mappings of G into A defined respectively by gT = Xlg(Xlg)-I. X2g(X2g)-I •...• xng(Xng)-1 (where g E G) ~ g'T = Ylg(Ylg)-I. Y2g(Y2g)-1 •...• Yng(Yng)-1 (where g E G) and where, if hE G, h denotes the element of X in the coset Ah and h denotes the element of Y in the coset Ah. Then ~ r-.J Proof: We may assume on suitably reordering Y that Yi = aiXi (ai E A) T == T Now if Xig = Xj, then AYj = Aajxj = AXj = AXig = Aa.;Xig = AYig. This means that ~ Yig = Yj It follows therefore as in equation (7.35) that if g E G, then g'; = (Ylgky~l). (Yk+ 19lyk"! I) •...• (Yn-m+lgmY-;;:m+l) (7.36) But Yi = aiXi, and we know that the elements Xlgkx~l, Xk+lg1Xk"!1,..., Xn-m+lgmX;;-:m+1 lie in the abelian subgroup A. Therefore Ylgky~l = al(xlgkx~l )a~l = Xlgkx~l Yk |
+lg1y"kL = ak+l(xk+lglx"kL)ak~1 = Xk+lg1Xk!1 Thus it follows from (7.35), (7.36) and the above remarks that gT = g'T. This lemma establishes that the transfer homomorphism T is independent of the choice of transversal. Accordingly we may speak of the transfer of G into A. d. A theorem of Schur Using the transfer, we now prove the following important theorem of I. Schur. Theorem 7.8: Let G be a group whose center A is of finite index. Then the derived group G' of G is finite. Proof: Suppose IG/AI = n. Let X = {Xl,...,Xn} be a transversal of A in G and let T be the transfer of G into A. Now if g E G, then by equation (7.35) But Xlgkx;l E A and A is the center of G. So Xlgkx~l = X~l (Xlgkx~l )Xl = gk Similarly, m Xk+lg Xk+l = g,..., Xn-m+lg Xn-m+l - g m -1 l -1 _ l Sec. 7.8] THE TRANSFER 243 Thus, for each g E G, gT = gn. Notice that the image of Gunder T is a subgroup of A and is therefore abelian. Then by Theorem 4.18, page 117, G/K is abelian, where K is the kernel of T. But by Problem 4.68, page 116, this implies G' C K. Now as G/A is finite, so is G'A/A. This means that G'/G'nA is also finite since by Theorem 4.23, page 125, G'A/A == G'/G'nA (7.37) Since G' /G' n A is finite, G' itself is finite if G' n A is finite. SincegT = gn for every g E G, every element of the kernel has finite order. It follows that the elements of G'nA have finite order. We will show that G' n A is finitely generated. Assuming this true for the moment, |
it follows that G' n A is finite (Problem 6.44, page 198). Thus G' is also finite. This completes the proof of Schur's theorem but for the verification that G' n A is finitely generated. We accomplish this by showing first that G' is finitely generated. If g, hE G, then g = aXi and h = bXj, where a, b E A, Xi, Xj EX. Then since A is the center of G, we have This means that there are at most n 2 distinct commutators in G. Therefore G' is finitely generated since it is generated by commutators. Finally G' n A is finitely generated since it is of finite index in a finitely generated group (Theorem 7.4, page 227). This completes the proof of Schur's theorem. It is worth noting one fact that emerged in this proof: the transfer into the center is simply the mapping that takes each element g to gn where n is the index of the center in G. We end our discussion of the transfer by mentioning that if all the Sylow subgroups of a finite group G are cyclic, then G is metacyclic (i.e. an extension of a cyclic group by a cyclic group). This theorem can be proved by using the transfer. The proof is not too complicated; however, it is lengthy and will not be given here. Reference to a proof may be found in Section 5.2a, page 139. A look back at Chapter 7 We re-proved Cayley's theorem, namely that every group is isomorphic to a group of permutations. The ideas that arose from Cayley's theorem were generalized. In particular we called a homomorphism of a group G into the symmetric group on a set X a permutational representation of G (on X). Permutational representations were roughly classified. We explained an important permutational representation of a group G called a coset representa tion. This representation allowed us to provide a variation of Cayley's theorem due to Frobenius. Then we used Frobenius' theorem and the coset representation to prove three theorems: (1) a subgroup of finite index in a finitely generated group is finitely generated; (2) the number of subgroups of fixed finite index in a finitely generated group is finite; (3) if G is a finitely generated group whose subgroups of |
finite index have only the identity sub group in common, then every homomorphism of G onto itself is also one-to-one, i.e. an automorphism. We called a group G an extension of a group H by K if there is a normal subgroup H of G such that G/H == K and fj == H. An analysis of this situation was made simpler because of our discussion of both coset representations and Frobenius' theorem. This ~nalysis of extensions was specialized to splitting extensions, where we provided a method of constructing a splitting extension of two groups Hand K. Finally we defined a special kind of homomorphism of a group into an abelian subgroup, called the transfer. We then used the transfer to prove that the derived group of a group whose center is of finite index is finite. 244 PERMUTATIONAL REPRESENTATIONS [CHAP. 7 Supplementary Problems PERMUTATION REPRESENTATIONS, COSET REPRESENTATIONS, FROBENIUS THEOREM 7.38. Let Dn be the dihedral group of order 2n and let Cn be its cyclic normal subgroup of order n. Find explicitly (1) the representation of Dn onto itself given by Cayley's theorem, (2) a coset representation of Dn using Cn as the subgroup, the representation provided by the Frobenius theorem. (3) 7.39. Give a permutation representation of An of degree 2n. 7.40. Find a faithful representation of G X H on n + m letters if G C; Sn, H C; Sm. EXTENSIONS 7.41. 7.42. 7.43. Construct a non-abelian group which is an extension of an infinite cyclic group by a group of order 2. Prove that if G is a non-abelian group with an infinite cyclic normal subgroup of index 2, then G is a splitting extension of an infinite cyclic group by a group of order 2. Prove that there are precisely three non-isomorphic extensions of an infinite cyclic group by a group of order 2. 7.44. Prove that an extension of a cyclic group of even order by a group of order 3 splits. 7.45. Construct a non-abelian group of order 36. 7.46. Construct five non-isomorphic groups of order 55 X 3. 7.47. 7.48 |
. Let D be the set of infinite sequences of integers, i.e. D consists of the sequences a =..., a-1,aO,al,'" (aiEZ). If b=...,b_ lI bO,b1,..., define a+b=...,a_l+b_1,ao+bo,a1+bl'.. ·. D is an abelian group under the operation of addition of sequences. For each integer n define a mapping an of D by putting aan =..., b_ l, bo, bl,..., where bi = ai-no Prove that (2) aman = am +n, (1) an is an automorphism of D, (3) A = {ai liE Z} is an infinite cyclic group generated by a1' Let W be the splitting extension of D of Problem 7.47 by the infinite cyclic group C = Up(c) via the mapping that sends c to a1' Prove that W is a non-abelian group. Find a proper subgroup of W which is isomorphic to W. Prove that WIW' is infinite cyclic. (Hard.) TRANSFER. MAL'CEV'S THEOREM, MARSHALL HALL'S THEOREM. 7.49. Prove that if A is an abelian subgroup of finite index in a simple group G, then the transfer of G into A sends G into 1. 7.50. Prove that if G is a finite group whose center Z has order co-prime to its index, then the transfer of G into Z is onto. (Hard.) 7.51. Prove that an infinite group with a subgroup of finite index is not simple. 7.52. A group is said to be residually finite if the intersection of all its normal subgroups of finite index is the identity_ Prove that an extension of a residually finite group by a finite group is residually finite. (Hint: The preceding problem gives a clue.) 7.53. 7.54. 7.55. 7.56. Let G be a cyclic extension of a cyclic group N of order n by a cyclic group of order m. Let N = up(a) and GIN = Up(bN). Prove that b-lab = a3, where j is co-prime to nand |
bm = a k (where j and k are integers). Prove that jk == k modulo n. Prove that a cyclic extension of a finitely generated residually finite group (defined in Problem 7.52) is residually finite. (Hint: Use Marshall Hall's theorem. Then use the fact that the automorphism group of a finite group is finite.) (Hard.) Prove that if G is a group and G l, G 2, ••• are subgroups of G such that G1 ~ G 2, G1 C;G2, G2 ~ G3, GZ C;G3, •••, then uG i is a subgroup of G which is not finitely generated. Let G be a residually finite group (defined in Problem 7.52) and suppose every subgroup of G is finitely generated. Prove that if H is a subgroup of G such that HIN == G for some normal sub group N of G, then N = {I}. (Hint: Use Problem 7.55.) 7.57. G is a finitely generated group, every element of which has only a finite number of conjugates. Prove that IG'I < 00. (Hint: /]1 C(Ui) = Z(G) if U1,..., Un are the generators of G.) 7.58. G is a group in which every element has only a finite number of conjugates. Prove that every element of G' is of finite order. (Hint: Use Problem 7.57.) Chapter 8 Free Groups and Presentations Preview of Chapter 8 We begin with a property of the infinite cyclic group and generalize this property to define free groups. We ask questions similar to those we asked in Chapter 4 concerning cyclic groups: (1) Do free groups exist? (2) When are two free groups isomorphic? (3) What are the homomorphisms of free groups? (4) What are the subgroups of free groups? In answering (3) we will learn that every group is a homomorphic image of a free group. This provides a new way of describing a group, i.e. as a factor group of a free group. Such a description of a group is called a presentation. 8.1 ELEMENTARY NOTIONS a. Definition of a free group Recall that if G is a group and X (#~) a subset of G |
, gp(X) = {X;I ••• x:n I Xi EX, €i = ±1} If X;I... x:n and y~1... y:m are two products with xi' Yi E X and €i = ±1, 'fJi = ±1, then they are said to be identical if n = m, x. = y. and €. =". for i = 1,..., m. Two products are said to be different if they are not identical. "It t 1. t It is easy to see that two different products of the form X;I... x:n can give rise to the same element of G. For example, if X = {x, y}, then xy and xx-1xy are different products of the form X;I... X:n but they give rise to the same element of G, i.e. xy. To avoid re dundancy, we introduce the concept of a reduced product: A product X;I... x:n, where €. = ±1 and x. E X, is said to be a reduced X-product if Xi = X Synonyms for reduced X-product are reduced product (X being understood) and reduced 1 implies €i # -€i+I. + 1 1 i product in X. (Examples of reduced products are easily given. Let X = {x, y}; then xy, x-1yxyx- 1 and x-1yyxy-1 are reduced products. However, yxyxx- I and x-1yxyy-1 are not reduced products.) Lemma 8.1: gp(X) = {w I w = 1 or w = a reduced product in X}. Proof: Let {w I w = 1 or w = a reduced product in X} = R. Clearly R ~ gp(X). If u E gp(X), then u = X:l... X~k where Xi E X and €i = ±1. Assume then that any product x: I.. • X~k E R We proceed to show that u E R. If k = 1, u is a reduced product in X, and so u E R. If u = X:I... X~k is a reduced product in X, then u E R. If u is not a reduced product, there Suppose k = n. for k |
~ n - 1. 245 246 FREE GROUPS AND PRESENTATIONS [CHAP. 8 Suppose n > 2; then we can delete 1 and €i = -€i+I' exists an integer i such that Xi = X i + X:iX:~+/ to obtain u as a product involving n - 2 elements of X. By the inductive hypothesis, u then belongs to R. If n = 2, then u = X~IX;2, where XI = x2 and €I = -€2' from which u = 1 E R. Thus gp(X) <: R and accordingly gp(X) = R. Consider now the infinite cyclic group generated by the element x. The reduced products in {x} are of two kinds: X •.. X = xr or X-lX-I. •. X-I = x- r where r is a positive integer. From what has been said about the infinite cyclic group, we know that if m and n are integers, xm = xn implies that m = n. Thus different reduced {xl-products give rise to different elements. (This is by no means the usual situation. Fo>r example in a cyclic group of order 2 generated by y, we have yyyy = yy.) G is said to be freely generated by {x}. More generally we have the following definition: A group G is said to be freely generated by the set X <: G if X"", \2'>, the empty set, and (i) gp(X) = G; (ii) two different reduced X-products define two different nonunit elements of G. Notice that it follows from (ii) that if x E X, X-I rt X. For if not, there exist x and y, both elements of X, with y-I = X. But then x and y-I are two different reduced X-products which are equal. It also follows from (ii) that 1 rt X. A set X of generators of G satisfying (ii) is often called a free set of generators of G. A group G is free if it is the identity group or if it possesses a free set of generators. G is also said to be free on X. If G is a free group freely generated by X, then the study of G is facilitated by the fact that we know exactly whether two X-products are equal. |
All we need to do is to express each of the products in reduced product form. If the reduced products are distinct the elements are not equal. This process of expressing an element as a reduced X-product can be carried out in a finite number steps. To illustrate, let F be freely gen erated by {x, y}. Are the products f = xyx-ly-Ixxx-ly-Iy3 and g = xyy-2y 3 equal? We convert f to a reduced product by deleting inverse pairs (i.e. two adjacent inverse factors). Thus f = xyx- Iy-I X(xx- I)(y-Iy )y2 = xyx- 1y-Ixy2 = xyx-Iy-Ixyy Similarly g = xyy. Hence as f and g are equal to different reduced products, they are not equal. The fact that we can determine in a finite number of steps whether or not two elements are equal is often expressed by saying that the word problem is decidable for (The interested reader may find more details in, e.g., J. J. Rotman, The free groups. Theory of Groups, Allyn and Bacon, 1965.) Problems 8.1. Let G be freely generated by the set X = {x, y, z}. (b) Is xyz(yz)-l a reduced product? x2y3(y3x2)-ly-3 and (xzy)-l xzy2 as reduced products. (c) Is xyy-1z-lyx equal to xz-Iyzz-1X? (a) Write down three distinct elements of G. (d) Express Solution: (a) (1) x, y, z, (2) x, x2, x3, (3) x, xy, xz In fact any three different reduced products in {x, y, z} are distinct elements of G. (b) No, because it is not written in the required x~l... x:n form, since it involves (YZ)-I. However, even if we replace (YZ)-I by its equivalent z-Iy-I to get xyzz-Iy-I, this, though in the form x;l... x:n, is not a reduced product because of the inverse pair ZZ-I. Sec. 8.1] ELEMENTARY NOTIONS 247 |
(c) Yes, for on expressing each as a reduced product we get xz-Iyx. (d) X2y3(y3X2) -ly-3 x2y3x-2y-3y-3 (xzy)-l xzy 2 y-Iz-Ix-Ixzy2 xxyyyx-ix-ly-ly-ly-ly-ly-Iy-l y-Iz-I zy2 = y-Iy2 = Y 8.2. Prove that if G is a free group, G ¥= {I}, then the infinite cyclic group is a subgroup of G. Solution: Suppose G is freely generated by X. If x E X, consider gp({x}). This is a cyclic group. Now x· x·...• x = x r, with r a positive integer, is a reduced product in X. Hence xr = x" where rand 8 are positive integers implies r = 8. Thus gp({x}) is infinite and is infinite cyclic and the result follows. 8.3. Prove that if G is freely generated by X, where X contains at least two elements, then G is not abelian. Solution: There exist two distinct elements x, y of X. Now xy and yx are two different reduced products, so xy ¥= yx. But this implies that G is not abelian. 8.4. Prove that a finite group G is not free if G ¥= {I}. Solution: In Problem 8.2 we proved that except for the identity, every free group has as a subgroup the infinite cyclic group. Consequently if G is free it must have the infinite cyclic group as a subgroup. This is absurd. 8.5. The direct product of two infinite cyclic groups is not a free group. Solution: The direct product of two infinite cyclic groups is abelian. But we have proved in Problem 8.3 that a free group is not abelian if it is freely generated by a set X which contains at least two elements. Hence if the direct product of two infinite cyclic groups is free, it is freely generated by some set X = {x}, i.e. it is infinite cyclic. But a free abelian group of rank two is not cyclic. This follows immediately either from the uniqueness of the type of an ab |
elian group (Theorem 6.21, page 197) or directly. 8.6. Prove that a free group freely generated by X with of the identity element). (Hard.) IXI:=O 2 has no center (i.e. its center consists Solution: Suppose G has an element z ¥= 1 z = x;l... x:n. Let y E X, Y ¥= Xl. Consider in its center. Let z be expressed as a reduced product This is a reduced X-product. On the other hand yz = yx;l... x:n zy = x;l... x:ny If <n ¥= y-l, then x?... x:ny is a reduced product and so clearly zy ¥= yz as yz begins as a reduced product with y but zy begins with X;l ¥= y. If x:n = y-l then for n> 1 (for otherwise Xl = Y contrary to the choice of y), zy = XE1 ••• X I En - 1 n-l and again it is clear that zy and yz are two different reduced products, so zy ¥= yz. But as z is in the center, we must have zy = yz. Therefore this contradicts the assumption that G has a center. 8.7. Generalize Problem 8.2 by showing that if F is freely generated by {Xl'..., xn} with n > 1, {xv..., x;} freely generates Fi = gp(xv..., Xi) for i = 1,2,..., n. Solution: Put Xi = {xv..., X;}. By definition, Xi generates F i. We have only to prove that two different reduced X;-products define different elements of F i• But a reduced Xcproduct is obviously also a reduced X-product and two different reduced X-products define different elements of F. Hence the result follows immediately. then 248 FREE GROUPS AND PRESENTATIONS [CHAP. 8 h. Length of an element. Alternative description of a free group Suppose F is freely generated by a set X. If f E F, f =F 1, it can be expressed in one and only one way as a reduced product X~l... <n. We define the length of f with respect to this free set of generators X to be n. The |
length of the identity is defined conventionally to be O. For example, if F is freely generated by X = {x, y}, then the length of X 2y 2x - l is 5, because A very useful technique in arguments involving free groups is to prove results by induction on the length of elements (see, for example, Lemma 8.9, page 261). Lemma 8.2: A group F is freely generated by a set X =F \2>, the empty set, if and only if (a) gp(X) = F, and (b) no reduced X-product is equal to the identity element. Proof: Assume first (a) and (b) above. To show F is freely generated by X we must show that two different reduced products in X are not equal and are not the identity. Let X~l... x:n and y~l... y:'m (where f; = ±1, "Ii = ±1, and xi' Yi E X) be two different,reduced products. Suppose they are equal. We can assume without loss of generality that xnn =F y:'m; for if x'n = yTlm X'l... x'n-l and yTll... yTlm-l are two different reduced products Since X'l... x'n = yTll... yTlm and x'n = yTlm it follows that X'l... x'n-l = yTll... yTlm-l m-I n-l m-l If again x:n...=-/ = y;::n--?, we can delete them. But we cannot continue indefinitely this way as X;l... <n and y~l... Y:'m are assumed to be different reduced products. So we may as sume x'n =F yTlm. It follows then that Xl'l... x'ny-Tlm.•• y-Tll is a reduced product equal to n 1. But by (b) it is not 1. Hence we have a contradiction, and so it follows that F is freely generated by X. n m m 1 If on the other hand F is given freely generated by X, do (a) and (b) hold? Of course, since this is entailed by |
the definition of a group freely generated by a set X. Problem 8.8. Let F be freely generated by X = {x, y, z}. Determine the length of (iii) f = x-lyxx-2y2x3z-1. (i) 1, (ii) xzyz-l, and Solution: (i) The length of 1 is taken to be O. (This is just the convention mentioned earlier.) (ii) The length of xzyz-l is 4. (iii) The length of f is calculated by expressing it first in reduced form, that is, Hence the length of f is 9. f = x-lyx-lyyxxxz- l c. Existence of free groups An obvious question is: Do free groups exist? Up to now we have tacitly assumed that they do. Theorem 8.3: Let n be any positive integer. There exists a free group freely generated by a set of n elements. Proof: Since every group is isomorphic to a subgroup of the symmetric group of some set, it is natural to look for a suitable subgroup of some symmetric group in order to find a free group of rank n. Sec. 8.1] EXISTENCE OF FREE GROUPS 249 Our plan is as follows: we shall introduce a set T consisting of certain ordered 1-tuples of integers, ordered 2-tuples of integers, and so on. We then choose a set X = {(h,..., On} of permutations of T such that X freely generates gp(X). The elements of T are the ordered m-tuples (r1, r2,..., rm) with r1 = ° and r2,..., rm i = 1,..., m - 1. Thus (0, 1,2, -3) E T but ri + ri+ 1 =1= ° for nonzero integers with (1,2) il T and (0,1,2, -2) il T. We define now for each integer i = ±1,..., ±n a permutation Oi of T as follows. If (r1,..., rm) E T we define (r1,..., rm)Oi = (r1,..., rm, i) if rm =1= -i. (i) (ii) (r1,..., rm)Oi = |
(r1,..., r m-1) if rm = -i. It is clear that Oi is a well-defined mapping of T into T. Moreover OiO-i = t = O-iOi where t is the identity permutation of T (Problem 8.9). Thus each Oi is a permutation of T (Theorem 2.4, page 36), i.e. OJ E S1'. Let G = gp(X), where X = {Ol,..., On}. it makes sense to talk of gp(X).) We prove that X freely generates G, thereby completing the proof of the theorem. (As S1' is a group and X ~ S1', We have only to verify that every reduced X-product is not t. Let then = 01 1, • • • Om' where',...,m' E {1,...,n} and'j = ± £1 Em ( 1 1) be any reduced X-product (so if r' = (r + 1)" on (0). Now 0i- 1 = O-i, and so 0i±l = O±i. Then 'r + 'r+1 =1= 0). We compute the effect of 1 By the definition of the product of permutations, we have 1 = Of l' Of 2' ••• 0, m' 12m (0)1 = ((0)Of11,)(Of2 2' "'O'mm') = ((°"1 1')0'22,)(0'33' "'O'mm') Now if l' = 2', then 'I + '2 =1= ° or '11' =1= -'22'. Therefore It follows in this way that (0, '11')0'22' = (0, '11', '22') (0)1 = (0, '11', '22',...,'mm') =1= (0) Thus 1 =1= t since it does not leave (0) fixed. This completes the proof of the theorem. Note: It is possible to prove similarly that there exist free groups freely generated by sets of arbitrary cardinality. Since we have not introduced cardinal numbers, we cannot prove this more general theorem here. The reader who has a knowledge of cardinal num bers may read the account in J. J. Rotman's The Theory 01 Groups, Allyn and Bacon |
, 1965. Problems 8.9. Prove that "lJ- i = < and "-j"i = <, where "i and < are as defined above. Solution: Let (r1,...,rm )ET. Suppose first that rm7"=-i; then (r1'".,rm)"i"-i = (r1'".,rm,i)"-i = (r1'".,rm) If rm = -i, Now if rm-1 = -(-i) = i, then (r1'".,rm) is of the form (r1'" does not belong to T, rm -1 7"= -(-i). Therefore.,i, -i). Since such an element 250 FREE GROUPS AND PRESENTATIONS [CHAP. 8 (rl'".,rm-I)II_i = (rl'".,rm-l>-i) = (rl'".,rm ) and so (rl"..,rm)lIill_ i = (r j, • • •,rm ) Thus lIill-i leaves all elements of T unchanged and hence 1I;II_i = t. Similarly II-illi = t. 8.10. Prove that there exists a free group freely generated by a set Y such that there is a one-to-one mapping of Y onto the positive integers. Solution: We proceed exactly as in the proof of Theorem 8.3 except that we define IIi for all nonzero the positive integers, defined by is a one-to-one onto mapping, for as none of the IIi have the same effect on (0), they must integers i and put Y = {Ill' 11 2, •.• }. The mapping p: Y -> Z+, (lIi)P = i be distinct. If H = gp(Y), then H is freely generated by Y. d. Homomorphisms of free groups Now it is a fact (see Problem 8.11 below) that if a group G is generated by a set X, then a homomorphism f) is uniquely determined by its effect on X, because each element of G is a product of elements and inverses of elements from X. Consider conversely what would happen if we had a map f) of X into a group H. Could we find a homomorphism of the whole group G into H whose effect on X was the |
same as that of f)? In general the answer is no (see Problem 8.12). However, we have the fol lowing result for free groups. Theorem 8.4: Let F be freely generated by a set X, let H be any group, and let f) be a mapping of X into H. Then there exists a homomorphism 1J of F into H such that 1J agrees with f) on X. 1J is called an extension of f). Proof: Any nonunit element of F is uniquely expressible as a reduced X-product t = X~l..• x:n where Xi E X, €i = ±1 and Xi = Xi+l implies €i 7'= -€i+l' f A Define to = (Xlf)) I (X2f))f2... (Xnf))'n and 1f) = 1 (the latter 1 being the identity of H). Clearly f) is a mapping of F into H agreeing with f) on X. To conclude the proof we must prove that 1J is a homomorphism. A A To do this we shall show that if t = X~l •.• x:n where Xi E X and €i = ±1, then A to = (Xlf)) I. •. (Xnf)) n whether or not X~l •.• x:n is a reduced product. If n = 1, this is true by the definition of O. Assume it is true for all positive integers n < k and consider t = X~l •.• x:n when n = k. If this is a reduced product then to = (Xlf))f j • • • (Xnf))'n by definition. If it is not a €i = -€i+1' reduced product, then there exists an integer i such that Xi = Xi + 1 and Consequently, f f (Xlf))fl.•• (Xnf))fn (Xlf))'!'.. (Xif))fi (Xi+ 1 f))'i+ 1 ••. (Xn(})fn (Xlf))'1 '" (Xi-I f))fi - I (Xi+2f))'i+2 '" (Xnf))fn 1 by our inductive hypothesis. Therefore (Xfl... X~i-IX~i+2... Xfn) f |
),-1,+2 n A (Xlf))fl... (X f))fn = n (Xfl... Xfi-I(Xf;X.i+I)Xfi+2... Xnn)(} =, - I,,+1,+2 1 € A € A ff) So if t = X~I ••. x:n and g = y~1... y:m, €i = ±1, 'YJi = ±1, Xi' Yi E X, then (fg) 0 = (X~l •.• X:ny~1... y:m) B = (Xlf))'1 :.. (Xnf))fn (Ylf))"f/l... (Ymf))"f/m = [(Xlf))'1 '" (x nf))f n ][(Ylf)f 1 ••• (Ymf)fm] Hence B is a homomorphism and the theorem follows. A A to gf) Sec. 8.1] HOMOMORPHISMS OF FREE GROUPS 251 Corollary 8.5: Let G be any finitely generated group (i.e. there exists a finite set Y such that gp(Y) = G). Then G is a homomorphic image of some free group. Proof: Let G be finitely generated by a set {Yl,..•, Yn}. There exists a free group F freely generated by {Xl,..., xn}, say. Define a map () from {Xl,..., Xn} to G by Xi(} = YI for i = 1,..., n. By Theorem 8.4 there exists a homomorphism () of F into G which agrees with () on each Xi. We know that F(} is a subgroup of G. But as F(} contains Yl, ••., Yn, it contains gp(Yl,..., Yn) = G. Therefore F(} = G, and so G is a homomorphic image of a free group. Note: The same result applies whether G is finitely generated or not. However, to prove the more general result we must use some of the ideas involving cardinal numbers. For this reason we have chosen only to consider the finitely generated case. Our last theorem reveals the importance of free groups |
. As every group is a homo morphic image of a free group, from the knowledge of the properties of a free group we may achieve some understanding of other groups. Problems 8.11. Let /11 and /1 2 be homomorphisms of G.... H. Let G = gp(X) and suppose /I, = /12' Solution: /l11X = /l2I x' Prove that Let g E G. Then g = x~' ••• x:n where Xi E X and 'i = ±1, and g/l, = (x,/lt>" ••• (xn/l,)En = (X,/l2)" Thus /I, and /12 have the same effect on the elements of G, and so '" (Xn/l2)'n = g/l2 /I, = /121' 8.12. Find an example of a group G generated by a set X, a group H and a map there exists no homomorphism 8': G.... H with 81x = /I. Solution: /I: X -> H, such that Let G be cyclic of order 2, G = gp({x}). Put X = {x} and let ~= gp({y}) be infinite cyclic. /I: X.... H by x/l = y. Suppose there exists a homomorphism /I: G.... H, such that x/l = y. Define Then G 8' is a subgroup of H which contains gp(y) and hence G 8 = H. But G8, being the image of a finite group, must be finite. Since H is infinite, this is clearly impossible. 8.13. Let H be the cyclic group of order 2, say H = {1, a}. Let F be freely generated by {x}. The map /I : x.... a gives rise to a homomorphism of F.... H. Describe the effect of this homomorphism on all elements of F. Check directly that it is a homomorphism, and find its kernel. Solution: The free group on a single generator is infinite cyclic. Its elements are uniquely of the form x n, n an integer. Now 8' maps xn.... an. If n is even, say n = 2r, an = 1. Hence 1J maps xn.... 1. If n is odd, say n = 2r |
+ 1, /I maps xn to a. A To check that 8 is a homomorphism, consider whether (xnxm) 8 = xn+m 8. Now x" +m 8 is a if n + m is odd, and is 1 if n + m is even. If n + m is odd, then one of the integers nand m is odd and the other even. Hence xn 8 xm 8 = a, as one of xn 8, xm 8' is a while the other is 1. If n + m is even, then either nand m are both even, or they are both odd; if nand m are both even, xn8xm 8'= 1-1 = 1; if both are odd, then xn 8' xm 8 = a - a = 1. Hence 8' is a homomorphism. The kernel of /I = {x 2T I all integers r}. A 8.14. Find a free group that has as a homomorphic image Sn' n any given positive integer. Solution: Let F be free on X where /I: X.... Sn map,:ach element of X onto a distinct element of Sw Then by Theorem 8.4 there is a homomorphism /I of F onto Sn that agrees with /I. IXI = n!. Let 252 8.15. FREE GROUPS AND PRESENTATIONS [CHAP. 8 Prove that a group F freely generated by n + 1 elements has as a homomorphic image a group G freely generated by n elements, where n is any given positive integer and G is a subgroup of F. Solution: Let F be freely generated by X where IXII = n. Then, as we have shown before, gp(Xl ) = G is freely generated by Xl' (Problem 8.7, page 247.) Let o : X ~ G be defined as follows: If x E X, x ~ a, put xo = x. If x = a, put xo = 1, the identity. Then there exists a homomorphism '0 of F onto G by Theorem 8.4. IXI = n + 1. Let X = Xl U {a} where 8.16. Let F be a free group with free generating set {a, b}. Let G be the direct product of two infinite cyclic groups generated by c and d respectively. The map which |
takes a to c and b to d extends to a homomorphism of F onto G (since G = gp({c,d})). Prove that the kernel of this homomorphism is F', the derived group of F. Solution: First we show that if 0 is the homomorphism F ~ G for which ao = c, bo = d, then F' ~ Ker o. F' is generated by the set of all commutators, so it is sufficient to show that each commutator [/1,12] = /";1/;1/d2 where 11> 12 E F. Then belongs to Ker o. Now a commutator is of the form [11,/2]0 = [11°'/2°] = 1, as G is abelian. Hence F' ~ Ker 0. Now every element of F that does not belong to F' is of the form arbSe where not both rand 8 are zero and e E F', as F/F' is abelian and aF', bF' generate it. Under 0 such an element goes onto crds and crds = 1 only if both rand 8 are O. Thus only the elements of F' belong to Ker o. Therefore Ker ° = F'. 8.17. Prove that a free group freely generated by n elements, n any positive integer, has a subgroup of index m for each positive integer m. Solution: We will exploit the homomorphism property. Let em be the cyclic group of order m generated by an element a, say. Let F be freely generated by X = {Xl'..., x n }. Then there exists a homo morphism ° of F onto em for which XiO = a, i = 1,.... n. Hence by the homomorphism theorem (Theorem 4.18, page 117), F/Kero ~ em Since leml = m, the number of co sets of Ker ° in F is m. Hence F has a subgroup of index m. 8.18. Let F be freely generated by X. Let Y be a subset of F such that gp(Y) = F, and Use Theorem 8.4 to find an epimorphism of F onto itself. IXI = IYI < 00 Solution: Let Xl'..., Xn be the elements of X, and Yl,..•, Yn the elements of Y. Def |
ine a homomorphism i = 1,..., n, using Theorem 8.4. Since Fo contains Y, Fo = F. 0: F ~ F by xiO = Yi where Hence ° is an epimorphism of F onto itself. 8.19. Let ° be an isomorphism of F with G. Let F be freely generated by X. Prove that G is freely generated by Xo. Solution: Let (xlO)€l... (xno)€n be a reduced product in Xo. Then we must show that this reduced product is not the identity of G. Clearly and as F is freely generated by X and ° is an isomorphism, the result follows. (x~l... x:n)o = (XlO)€l... (xno)'n 8.20. Let F be freely generated by X and G freely generated by Y. If 0: X ~ Y spondence, prove that F ~ G. is a one-to-one corre Solution: Let '0 be the homomorphism of F into G which is an extension of o. (Theorem 8.4.) Clearly '0 is onto G. Let cp: Y ~ X be the mapping such that Ocp is the identity mapping on X and cpo is the identity mapping on Y. Let;; be the extension of cp to G ~ H. Then '8;; is the identity on F and ¢ '8 is the identity on G. It follows readily that 8 is an isomorphism of F with G. Sec. 8.2] PRESENTATIONS OF GROUPS 253 8.21. Let F be freely generated by a, b, c. Let N be the normal subgroup generated by c, i.e. the inter section of all normal subgroups of F containing c. Prove that FIN is freely generated by aN and bN. (Hard.) Solution: Let G be free on x, y. Let 8 be the homomorphism of F onto G defined by a8 = x, b8 = Y and co = 1. (Since F is free on a, b, c, such a homomorphism 8 exists by Theorem 8.4.) Let K be the kernel of 8. First we shall prove that K = N, the normal subgroup generated by c. Then we shall prove that aN and bN freely generate FIN. To see that K = |
N first observe that as c8 = 1, N is contained in K (K is a normal subgroup containing c). On the other hand, suppose IE F, I El N. Then I can be expressed in the form where 11 is a reduced {a, b}-product and n1 EN. But then 18 = 110 I = 11 n1 I El K. Thus if and 118 is a reduced {x, y}-product. So 118"# 1, which means 18"# 1 and so I El N, I El K which implies K is contained in N. Hence as we observed earlier that N is contained in K, we find K = N as desired. Now a, b, c generate F. So aN, bN and cN generate FIN. Since eN = N, aN and bN generate (YkN)Ek is a FIN. We want to prove that aN and bN freely generate FIN. Suppose (Y1N)E1 '" reduced {aN, bN}-product. Now 8 gives rise to an isomorphism v of FIK with G, i.e. the mapping defined by (IK)v = 18 (the homomorphism theorem, Theorem 4.18, page 117). Observe that (aK)v = x and (bK)v = Y so that ((Y1N)E1... (YkN)Ek)v = XE1 ••• XEk where (YiN)v = xi E {x, y}. The product X~l ••• <k is a reduced {x, y}-product. But {x, y} freely generates G. Thus X~l ••• x~k "# 1. Consequently 1 k and the proof is complete. (Y1N)'1.,. (YkN)<k "# N Instead of completing the proof this way we could refer to Problem 8.19 using the isomorphism 8.2 PRESENTATIONS OF GROUPS a. Definitions We have shown (Theorem 8.4) that if F is freely generated by X, then for every group G and every mapping (J of X into G there is a homomorphism of F into G which extends (J, i.e. which agrees with (J on X. This fact will enable us to "present" a given group in terms of a free group. This idea of |
a presentation is especially important in topology and analysis where groups arise in just this way, as the "groups of certain presentations". First we need a definition. If S is any subset of a group, then the normal closure of S is defined to be the intersection of all normal subgroups of G containing S. Clearly the normal closure of S is a normal subgroup of G containing S. Thus the normal closure of S is often called the normal subgroup generated by S. It is easy to prove that the normal closure of S is gp(g- l sg I g E G and s E S) (see Problem 8.22 below). A presentation is defined to be a pair (X; R) where X is a free set of generators of a free group F and R is a subset of F. The group of the presentation (X;R) is FIN where N is the normal subgroup of F generated by R; we usually denote the group of a presentation (X;R) by IX;RI. Finally a presentation of a group G consists, by definition, of a presenta tion (X;R) and an isomorphism (J between IX;RI and G. A presentation (X;R) is finite if both X and R are finite, and a group G is termed finitely presented if it has a finite presenta tion. Not all groups are finitely generated (a necessary prerequisite for being finitely presented) and not all finitely generated groups are finitely presented. For a more detailed discussion of these notions the reader may consult R. H. Crowell and R. H. Fox, An Intro duction to Knot Theory, Blaisdell, 1963. 254 FREE GROUPS AND PRESENTATIONS [CHAP. 8 h. Illustrations of presentations We shall work some examples to illustrate the definitions of Section 8.2a. First of all, suppose F is a free group freely generated by X = {a, b}. Then the following (i) ({a, b}; {a, b}) (ii) ({a, b}; {a 2b2, a3b3, a4b4,... }) (iii) ({a, b}; ([a, b]}) are patently presentations (by the very definition). The presentations (i) and (iii) are finite, but the presentation (ii) is not. The obvious question is: what are the groups of these presentations? Clearly the group |
of (i) is a group of order 1., a3b3 What about (ii)? Well, we are interested in FIN where N is the normal subgroup generated by a2b2, a4b4,.... We have, since a2b2 E N, a2N = b- 2N. Furthermore a3N = b- 3N since a3b3 EN. Thus it follows that a3N = a2aN = b- 2aN = b- 3N. Cancel ling b- 2N from both sides yields aN = b-lN. Hence ab EN. Indeed we would like to prove that N is the normal subgroup generated by abo Let K be the normal subgroup gen erated by abo Since ab EN, we have K eN. Now observe that aK = b-lK. Then This implies that aibi E K (i = 2,3,... ). Hence N is contained in K and so we have proved that K = N. Therefore FIN is cyclic (because aN and bN clearly generate FIN; but aN = b-1N). In fact FIN is infinite. To see this suppose G is an infinite cyclic group generated by g. Let B be the mapping of {a, b} into G defined by aB = g, bB = g-l A A A Let B be the homomorphism of F into G defined by B (Theorem 8.4). Clearly B is onto and (ab)B = 1. So if L is the kernel of B, L"dN. But as F/L ~ G, F/L is infinite cyclic. Therefore FIN is also infinite cyclic since, as we have already noted, FIN is cyclic. (Actually, N = L; however, we don't need this fact here.) A Finally we come to (iii). In fact I{a, b}; {[a, b]}l is free abelian of rank 2. The reader may attempt to prove this before we do so in a more general case. At this point we simplify our notation. Instead of using our set-theoretic notation which encloses the ele ments of a given set in braces { }, we shall omit the braces in writing presentations. Thus we write (a,b; [a,b]) for ({a,b}; {[a,b]}), la,b |
; [a,b]1 for I{a,b}; {[a,bDI, etc. Let F be freely generated by al,..., Un. We shall prove that is free abelian of rank n (from which it follows that la, b; [a, b]1 is free abelian of rank 2). To this end let H be the free abelian group of rank n. Then H is the direct product of i = 1,...,n, say. We define a homomorphism infinite cyclic groups generated by hi, B: F ~ H by aiB = hi. Now [a;, ai]B = [alB, aiB] = [hi, hi] = 1. Hence [a;, ail E Ker B. Let N be the normal subgroup generated by the [a;, ail, 1""" i """ j """ n. Clearly N e Ker B. Note that [a;N, aiN] = [a;, ai]N = N. As the generators a;N of FIN commute, the elements f of F are of the form f = where c EN Sec. 8.2] PRESENTATIONS OF GROUPS 255 and if f (l N, at least one of the ri # 0. Hence as cB = 1, B = 1 ••• n. j() # 1. It follows that f (l Ker B, and so N = Ker B. the rj # 0, f h Tl hTn N f ow as one 0 Therefore FIN is the free abelian group of rank n, i.e. I aI,..., an; [ai, ail with 1 ~ i ~ j ~ n I is the free abelian group of rank n. Note that as N is generated as a normal subgroup by commutators, N C F'. But as FIN is abelian, N d F'. Thus N = F' and we therefore conclude that FIF' is free abelian of rank n. We state this fact as Theorem 8.6: Let F be a free group freely generated by a set of n elements (n < 00). Then F I F' is a free abelian group of rank n. Corollary 8.7: Let F be a free group. Suppose X and Y both freely generate F. If IXI is finite, then so |
is IYI and IXI = IYI. Proof: By Theorem 8.6 we know that FIF' is a free abelian group of rank n = IXI. then FIF' is free abelian of rank IYI. again by Theorem 8.6. Consequently If IYI < 00, IXI = IYI since the rank of an abelian group is unique (see Section 6.2d, page 193). It remains only to prove that I YI cannot be infinite. To do this, suppose Y1, Yz,..., Yn+ 1 E Y. Now in a free abelian group of rank n every n + 1 elements are dependent. Hence there exist integers m1,..., m n + 1, not all zero, such that (y l F'tl.., (Yn+lF'tn+l = F' i.e. y l F',..., Yn+ IF' are dependent. Let A be the free abelian group on aI,..., an+ 1 and let B be the homomorphism of F to A defined by YiB=ai (i=1,...,n+1), yB=l if y(l{Yl,...,Yn+t) and yEY The kernel K of B contains F' since FIK is abelian. Then 1 = (y lBtl... (Yn+IBt n +l = a71... a:~il But A is free abelian on aI,..., an + 1. Hence m1 = m2 =... = m n + 1 = 0, a contradiction. Thus I YI < 00 and the corollary has been proved. It follows from this corollary that if F is a finitely generated free group, then every pair of sets which freely generate F have the same number of elements. We define the rank of F to be this common number, i.e. the number of elements in any set which freely generates F. Note that free groups of the same rank are isomorphic (Problem 8.20). We can easily give a presentation of A, a free abelian group on aI,..., an, with the results we now have. Let F be free on Xl,..., Xn and let B be the isomorphism defined by (xjF')B |
= aj (j = 1,..., n). Then (Xl,..., X n ; [Xi, Xj] with 1 ~ i ~ j ~ n) together with B is a presentation of A. In some of the following problems we will often be dealing with factor groups. A simple convention makes the arguments simpler to follow. Let G be a group and N a normal subgroup of G. If we use some phrase such as "let us calculate modulo N," then by G we mean the factor group GIN. We shall mean by g = h that Ng = Nh. If we say let M be a subgroup of G, what we really mean is "let MIN be a subgroup of GIN." In other words, we must remember that we are talking of a factor group and instead of writing the cosets, we will simply write the coset representative. (See Problem 8.24 for an example.) 256 FREE GROUPS AND PRESENTATIONS [CHAP. 8 Problems 8.22. If G is a ~roup, show that the normal closure of a subset R (¥ 0) of G is gp({g-l1'g I g E G and l' E R}). Solution: Clearly N = gp({g-l1'g I g E G and r E R}) is a subgroup of G containing R. Also N is a normal subgroup of G. Finally any normal subgroup containing R must contain N. Thus the result follows. 8.23. Let G be a group with subgroups Hand K and let if HdK, then H = K. Solution: [G: H] = n < 00 and [G: K] "" n. Prove that Let Xl = 1 and let the distinct cosets of H in G be H = Hxv HX2'..., HXn- As H d K, it If hE H, i ¥ 1. Hence follows that KXl,Kx2'".,Kxn are distinct. As then hE KXi for some integer i. If i ¥ 1, HnKxi = 0 since HnHxi = 0 for i = 1 and h E KXl = K. Accordingly H C K and thus H = K. [G: K] "" n, G = KxluKxzu··· uKxn- 8.24. Rewrite the first part |
of the argument of presentation (ii), page 254, using the modulo N convention introduced above. Also calculate modulo K and stop after proving K = N. Solution: We are interested in FIN where N is the normal subgroup generated by aZb2, a3b3, a4b4, •••• Let us calculate modulo N. Since aZbz = 1, aZ = b- z. Furthermore a3 = b- 3 since a3b3 = 1. Thus it follows that a3 = a2a = b- 2a = b- 3 • Cancelling b- 2 from both sides yields a = b- l, and so ab = 1. As we are calculating modulo N, this means that ab E N. Indeed we would like to prove that N is the normal subgroup generated by abo Let K be the normal subgroup generated by abo Since ab E N, we have KeN. Now we calculate modulo K. a = b- l, and so aibi = b-ibi = 1 (i = 2,3,... ). This means as we are calculating modulo K that aibi E K (i = 2,3,... ). Therefore we have proved that K = N. 8.25. Let G = la; a21. Prove that G is cyclic of order 2. Solution: The free group F generated by a is infinite cyclic. Now gp(a2) is normal in F since F is abelian. Hence gp(a2) is the normal subgroup generated by a2. It can be easily seen that gp(a)lgp(a2) is the cyclic group of order 2. Thus G is cyclic of order 2. 8.26. Prove that G = la; ani, where n is any positive integer, is the cyclic group of order n (again with F freely generated by a). Solution: The free group F on a is infinite cyclic and N = gp(an) is the normal subgroup generated by an. Therefore FIN = G is cyclic of order n (Theorem 4.9, page 105). 8.27. Prove that la, b; a2, b2, [a, b]i is the direct product of two cyclic groups of order 2. Solution: Although it has not been stated, a and b are (as usual) free generators of a free group |
F. Let N be the normal subgroup generated by a2, b2 and [a, b]. In FIN, Na and Nb commute as [Na,Nbl = N[a, b] = N, since [a, b] EN. Since FIN is generated by Na and Nb, it is therefore abelian. Also (Na)2 = Nand (Nb)2 = N. Let A = {N,Na} and B = {N,Nb}. As A is a normal subgroup of FIN, AB is a subgroup of FIN which contains both Na and Nb. It follows that AB = FIN. Thus IFINI "" IAIIBI "" 2·2 = 4. On the other hand there is a homomorphism (J of F onto the direct product of two cyclic groups of order 2. Clearly Ker (J contains a2, b2 and [a, b]. Hence Ker (J d N. It follows that N = Ker (J (Problem 8.23). Thus FIN is the direct product of two cyclic groups of order 2. 8.28. Find a presentation for S3' (Hard.) Solution: Let p = (1 2 3) and u = (1 2 3). Then u2 213 231 are six distinct elements and hence the whole of S3' Now (~ ~!). So p,pu,pu2 and 1,u,u2 Sec. 8.2] PRESENTATIONS OF GROUPS 257 2 1 2 3)(1 2 3 1 2 1 : ) = (~ 2 3) 1 2 ~ uZ Thus the equation p-Iupu- Z = 1 holds. Also pZ = 1 and u 3 = 1. We use only these equations and hope that they will give rise to a presentation for S3. Let F be the free group on ai' az and let G = lal' az; ai, a~, al-Ia2ala;zl. Furthermore let N be the normal subgroup of F generated by ai, a~, a;-laZala;z, and let 0: F ~ S3 be defined by alo = P and azo = u. Then N C Ker 0. Now we calculate modulo N. We see that ai = 1 and so a;-l = al. Since a~ = 1, a z- l = a;, al-Ia2ala |
Z-z = 1 from which aza l = ala~. Now F is generated by avaz. Let M = gp(az); then M <I F. It follows from Problem 4.62, page 114, that M{1, al} is a subgroup of F and as it contains al and az, M{1, al} = F. Thus the elements of Fare {l, az, a~}{l, al}, i.e. 1, az, a~, ai, aZal' a~al (although we do not know if they are distinct). It follows that IFINI ~ 6. But WIKer 01 = 6 and N C Ker 6, and so Ker 0 = N 8.23). Then (Problem under the isomorphism ¢: alN ~ p, azN ~ u page 117). (by the homomorphism theorem, Theorem 4.18, 8.29. Prove that la, b; a Z, bn, a-Ibabl is isomorphic to the dihedral group of order 2n. Solution: Let G be the dihedral group of order 2n. Then G is the symmetry group of the regular n-gon S (see Section 3.4f, page 75). Recall that Uz rotates S in a clockwise direction through an angle of 27Tln. It follows that Uz is of order n. Put u = uz. If T is the reflection about AIO, where Al is any vertex of S, and 0 the center of S, then T2 = L. Moreover every element of G can be written in € = 0,1 and 5 = 0,1,..., n - 1 (see Section 3.4f and note that ui = Ui for the form T f U 5, where 1 ~ i ~ n). Let 0 be the homomorphism of the free group F, freely generated by a and b, onto G defined by ao = T and bo = u. Then aZo = 1, bno = 1 and Thus a Z, bn and a-Ibab lie in the kernel K of 0. Then N, the normal subgroup of F generated by aZ, bn and a-Ibab, is contained in K. Moreover since 0 is onto, FIK"", G. If. we can show that N = K, then the proof follows. We calculate ( |
compare Problem 8.28) modulo N. We shall show that IFINI ~ 2n. Since IFIKI = 2n and K::l N, this will establish that K = N by Problem 8.23. Modulo N, a-Iba = b- l. Let M = gp(b). Then M <I F. Thus M{l, a} is a subgroup of F by Problem 4.62, page 114. As it contains both a and b, M{1, a} = F. The elements of Mare 1, b, bZ,..., bn - l. Therefore the elements of Fare a f b5 (€ = 0, 1, 5 = 0, 1,..., n - 1) Since we are calculating modulo N, F really stands for FIN. This implies proof is complete. IFINI ~ 2n. Thus the 8.30. Prove that G = la, b; aZ, a-Ibabl Hint: Show that each dihedral group Dn is a homomorphic image of G.) is infinite. (This group is called the infinite dihedral group. Solution: Let F be the free group on a and b and let on be a homomorphism of F onto Dn (as Dn is a two generator group (see, for example, Problem 8.29), we know such a homomorphism exists). Also Keron::l{aZ,a-Ibab}. Thus Keron::lN, the normal subgroup generated by aZ,a-Ibab. Therefore G = FIN has each Dn as a homomorphic image (D n "'" (FIN)/(Ker 6nIN)). If G were finite of order k, say, we would have a contradiction. For then GO k as a homomorphic image of a group of order k is of order ~ k. But GO k = Dk is of order 2k. This contradiction proves that G is of infinite order. 258 FREE GROUPS AND PRESENTATIONS [CHAP. 8 8.31. Prove that every finitely generated group has a presentation. Solution: In Section 8.1d we proved that every finitely generated group is a factor group of a free group. We will use this fact to prove that every finitely generated group has a presentation. Let G be an arbitrary group. Let X be chosen so that X freely generates a free group |
F and, furthermore, so that there is a mapping 8 of X onto one of the finite sets of generators for G. Let </> be the homo morphism of F onto G such that </> agrees with 8 on X and let N = Ker </>. Then (X; N) together with the isomorphism f.I: IN -> fcf> is a presentation of G. 8.32. Let G be a finite group with elements 1 = Xl' X2,..., Xn- Suppose XiXj = XU,j), where XU,j) E {Xl>..., xn} (in other words (i, j) is an integer between 1 and n). Let F be the free group freely gen erated by al,.. "~' N be the normal subgroup of F generated by aiajaZ2j) (1"" i,.i "" n), and 8 be the homomorphism from to G defined by (akN)8 = xk' Show that together with 8 is a presentation of G. Solution: Since aiajN = a(i,j)N, every product of a's in which there are no negative exponents is equal, l = Xj, then XiXj = x(i,j) = xl> and so aiaj = a l modulo N. But modulo N, to some ak' Now if XixlXl = x(l,l) = Xl' Thus l = aj modulo N. Therefore every product of the ak involving This means that al EN and so a iboth positive and negative exponents can be replaced, modulo N, by a product involving only positive exponents. Accordingly every product of the ak and their inverses is equal, modulo N, to an a,.' It follows that IFINI "" n = IGI Now let </> be the homomorphism from F to G defined by ai</> = Xi (for 1"" i "" n). Then (aiaja(i,lj))</> = XiXjX(i,lj) = 1 and so Ker </> includes aiaja(i,lj). Therefore Ker </> ;d N. Since and IF INI "" n, it follows by Problem 8.23 that N = Ker </>. Thus the mapping IFIKer</>1 = n « |
00) is indeed an isomorphism (Theorem 4.18, page 117). 8: akN --> xk 8.33. Prove that lx, y; x2y2, y21 = lx, y; x2, y21. Solution: We have a free group F freely generated by X and y and two normal subgroups Nand M generated, as normal subgroups, respectively by x2y2 and y2, and X2 and y2. We must prove N = M. Since N contains x2y2 and y2, it contains x 2y 2(y2)-1 = X2. Thus N;d M. But M;d {X2y2} since M;d {x2, y2}. Hence M;d Nand M = N. 8.34. Show that a free group of rank n < 00 cannot be generated by n - 1 elements. Solution: Let G be free of rank n. Then by Theorem 8.6, GIG' is free abelian of rank n. If G can be gen erated by n - 1 elements, GIG' can be generated by n - 1 elements. But this contradicts Problem 6.41(b), page 195. Sec. 8.3) THE SUBGROUP THEOREM FOR FREE GROUPS: AN EXAMPLE 259 8.3 THE SUBGROUP THEOREM FOR FREE GROUPS: AN EXAMPLE The object of the next sections of this chapter is to prove that every subgroup of a free group is a free group. This theorem is one of the more difficult in this book. So in order to give the reader a chance to become accustomed to the ideas involved, we first work an example. Example 1: Let F be the free group freely generated by two elements a and b. Consider the set Y = {a- 1ba,a- 2ba2,... }. Show that H = gp(Y) is freely generated by Y. (Thus we see that a free group freely generated by two elements has a subgroup which is freely generated by infinitely many elements.) Proof: Let a-ibai = Yi (i = 0,1,2,... ). Consider any Y-reduced product, where Y = {Vi I i = 0,1,... }. Say f = Y:~... Y~'.', where 1',..., n' are positive integers, and |
Yi' = Y(j+ll' implies Ej # -Ej+l' (Given, for example, Yay;l y1y2, then l' = 3, 2' = 4, 3' = 1 and 4' = 2; El = 1, E2 = -1, Ea = 1 and E4 = 1.) We will prove that f # 1 by induction on n, showing that H is freely generated by Y. Our inductive hypothesis (on n, the number of Y;'S that go into a given reduced product f) is that f when expressed as a reduced product in {a, b} ends in bEn an'. (For example, Yay;l Y1Y2 = a-abaaoa-4b-la4oa-lbaoa-2ba2 = a-3ba-lb-la3ba-lba2 as a reduced product in {a, b}, and, as asserted, it ends in ba2.) If n = 1 this is certainly true. If it is true for n = k, let n = k + 1. Then Yl' ••• Yk' ends in b a when expressed as a reduced product in {a, b}, i.e. Yl"" Yk' = zb a, where z is a reduced product in a and b such that zbEkak' is a reduced product (i.e. z does not end in b- 1 if Ek = lor, if <k = -1, z does not end in b). If now k' = (k + 1)', then, as f is a reduced product, <k # -<1<+1' Since <k and <k+l are the numbers 1 or -1, <k = <k+l' Thus Ek k' Ek k' Ek El Ek £1 El Yl' "'Y(k+ll' - z a a <k+l bEk k' -(k+l)'b<k+l (k+l)' _ a = zbEkak'a-k'bEk+la(k+ll' = zbEkbEk+la(k+l)' this last expression is a reduced product in {a, b} and f ends in Ek! Since <k + 1 bEk + 1a(k+l)'. If |
however k' # (k + 1)" E'(k+l)' Ek Yl'..• Y(k+ll' = zb kak a- Ek+l El then b +la( + k 1)' = zbEka(k'-(k+ll')bEk+la(k+ll' (k + 1)' # 0, the last expression is a reduced product and so f expressed Since k' + 1 a(k+l)'. It follows therefore that in both situa as a reduced product, ends in b tions f # 1. Thus Y freely generates H. Similar arguments will help to prove in general that a subgroup of a free group is free. Ek Problems 8.35. Verify both the inductive assumption of the preceding example and that f # 1 where f = y;:lYa-1Y1Y1Ya Solution: f = a- 2b- 1a2 0 a- 3b- 1o,3. o,-lbo,. o,-lbo,. o,-abo,3 = o,-2b- 1a- 1b- 1o,2bbo,-2ba3 Clearly f # 1. The inductive assumption of the preceding example is that f when expressed as a reduced product should end in ba3, which it does. 8.36. Given the existence of a free group freely generated by two elements, prove the existence of free groups freely generated by any finite number of elements. Solution: In Example 1 we have proved that a free group freely generated by two elements has a subset Y such that Y is infinite and H = gp(Y) is freely generated by Y. If n is any positive integer let Yi>" "Yn be n distinct elements of Y. Then gp({Yl'".,Yn}) is easily shown to be freely gen erated by Yi>..., Yn' Thus there exist free groups of rank n for each positive integer n. 260 8.37. FREE GROUPS AND PRESENTATIONS [CHAP. 8 Let F be freely generated by a and b. Prove that the subgroup of F generated by aba3 and a2 b is freely generated by aba:l and a 2b. (Hard.) Solution: Let Y = {aba3 • a2b} and let Yl = aba 3, Y2 = a2b. Consider a reduced Y-product Then j' |
= (j + 1)' implies fj ¥= - f j + l' £1 En f = Yl'... Yn' (i' EO {1,2}) We will show by induction on n that if n' = 1, then f ends in a3 if fn = 1 and ends in b- 1a- 1 if fn = -1; while if n' = 2, then f ends in b if En = 1 and ends in b- 1a- 2 if fn = -1. (i) For n = 1 this is certainly true. If it is true for n = k, we must proceed case by case in proving it true for k + 1. k' = 1. (a) fie = 1. Then by the inductive hypothesis Yl'... Yk' ends in a3. If now (k + 1)' = 1, then fk + 1 = 1 and f ends in a3 • aba3 = aaaabaaa. as f is a reduced product we must have Thus f ends in a3 as required. If (k + 1)' = 2 and fk + 1 = 1, then f ends in a3 • a2b, and so f ends in b as required. If Ek+l = -1, then f ends in a3 • b- 1a- 2 and so f ends in b- 1a- 2 as required. fk = -1. Then by the inductive hypothesis Y:~ '" Y:~ ends in b- 1a- 1• If now (k + 1)' = 1, then as f is a reduced product fk + 1 = -1 and f ends in (b) fk £1 b- 1a- 1 • a-3b-1a-1 = b- 1a- 4 b- 1a- 1 Hence f ends in b- 1a- 1 as required. If (k+l)'=2 and Ek+l=l, then f ends in b- 1a- 1a2b=b- 1ab, and so f ends in b as required. If (k+l)'=2 and fk+ l= - I, then f ends in b- 1a- 1 • b- 1a- 2 and so f ends in b- 1a- 2 as required. (ii) k' = 2. (a) (b) If ( |
k + 1)' = 1 and fie = 1. Then by our inductive hypothesis Y:~ '" Y~~ ends in b. fk + 1 = 1, f ends in baba3 and so f ends in a3 as required. If (k + 1)' = 1 and fk+ 1 = -1, then f ends in ba-3b- 1a- 1 and f ends in b- 1a- 1 as required. If (k+ 1)' = 2, then as f is a reduced product fk+ 1 = 1 and f ends in ba2b, and so f ends in b as required. fk = -1. Then by our inductive hypothesis Yl'... Yk' ends in b- 1a- 2. If (k + 1)' = 1 and fk+l = 1, then f ends in b- 1a- 2aba3 = b- 1a- 1ba3, and so f ends in a3 as required. If (k + 1)' = 1 and fk+l = -1, then f ends in b- 1a- 2 • a- 3b- 1a- 1 = b-1a- 5b- 1a- 1, and so f ends in b - la -1 as required. If (k + 1)' = 2, then as f is a reduced product €k+ 1 = -1. Thus f ends in b- 1a- 2 • b- 1a- 2 and so f ends in b- 1a- 2 as required. Ek £1 Thus we have proved by induction that any reduced product always ends in one of a3, b- 1a- 1, b and b- 1a- 2• Hence f ¥= 1 and Y freely generates gp(Y). 8.4 PROOF OF THE SUBGROUP THEOREM FOR FREE GROUPS a. Plan of the proof The subgroup theorem for free groups (due to J. Nielsen and O. Schreier) may be stated as follows. Theorem 8.8: Every subgroup H of a free group F is free. Suppose that F is freely generated by S. We know from Section 7.6b, page 228, that if X is any right transversal of H in F, then the non unit elements a x • s = xS(XS)-1 (x EX, s E S) (where, for |
IE F, 1 is the unique element of X in the coset HI) generate H. Let Y = {ax • s I x EX, s E S, ax• s =F 1} Then we shall prove that H is actually freely generated by Y provided X is chosen ap propriately. Thus there are two main steps in the proof: (i) Choose X appropriately. (ii) Prove that Y freely generates H. Sec. 8.4] PROOF OF THE SUBGROUP THEOREM FOR FREE GROUPS 261 Step (ii) of the proof will be broken into two parts: the first part requires a careful look at the elements ax,s and the second involves looking at the way in which products of these ax,s and their inverses interact. h. Schreier transversals Suppose that X is a transversal for H in F, where F is freely generated by a set S. Every element x (x =1= 1) in X may be expressed uniquely as a reduced S-product where ai is an element of S or the inverse of an element of S. Recall that n is termed the length of x and that the length of 1 is O. We shall call the elements x=ala2"'an (n~l) initial segments of x. Definition: A right transversal X is called a right Schreier transversal if every initial segment of an element in X is also in X. Notice that it follows that if X is a right Schreier transversal, then 1 EX. The main result of this section is the following. Lemma 8.9: Suppose that F is a free group freely generated by S and that H is a subgroup of F. Then we can always find a right Schreier transversal X for H in F. Proof: We say that a right coset Hi is of length n if there is an element in Hi of length n but no element of length less than n. We shall choose X inductively using the lengths of the cosets of H in F. First if Hi is of length 0, then 1 E Hi and so Hi = H. We choose 1 to be the repre sentative of H. Suppose that n> 0 and that for each coset of length less than n, representatives have already been chosen so that every initial segment of a representative is again a representa tive. We choose now representatives for the co sets of length n. |
Let Hi be a coset of length n and let ala2... an be an element in Hi of length n. The element ala2... an-l is of length n - 1. Thus the coset Hala2... an-l is of length at most n - 1 (since ala2'" an-l E Hala2... an-l). This means that the representative of Hala2'" an-l has already been chosen, by our induction assumption. Suppose this representative is b 1b2 ••• bm. Now H(b 1b2... bman) = (Hb 1b2 ··· bm)an = (Hala2'" an-l)an = Hala2'" an We select b1b2... bman to be the representative of the coset Hala2... an. The initial segments of blb2... bman, excluding b 1b2... bman, are 1, b1, •••, b1b2 ••• bm which we know have already been chosen as representatives. suitable representatives for all the co sets of length n. In the same way we select We have therefore verified the induction hypothesis and so we are able, in this way, to complete the choice of a right Schreier transversal for H in F. c. A look at the elements ax,s Suppose that we have chosen a right Schreier transversal X for H in F. Consider a nonunit element ax,s where x E X and s E S: ax,s = xS(XS)-l 262 FREE GROUPS AND PRESENTATIONS [CHAP. 8 Now let the reduced S-product for x be x = al... ak where ai or its inverse belongs to S. We will allow k = 0; this will be interpreted as x = 1. Thus we may write ax,s = ala2'" aks(ala2... aks)-1 Let al... akS = bl... b1 where the right-hand side is a reduced S-product with bi or biE S, i = 1,..., l. We assert that al... akS and bl... bls- l are not elements of X. For suppose al... akS E X, then al'" akS = al... akS and so ax,s = 1; whereas if bl ··· bls- l EX, we utilize equation (7 |
.3), page 219. and conclude that l XSS- 1X- l = XSS-l(~)-1 = XSS-l(XSS-l)-1 bl... bls- l(b l... bls 1)-1 = bl... bls- l(b l... bls- l )-1 = 1 From this it follows that the reduced S-product for ax,s is al'" aksbl-... bi""l. For if not, S cancels either with ak or with bl- l. But as every initial segment of an element of X belongs to X, either al'" akS E X or bl ··· bls- l E X, which is a contradiction. Note we have proved ax,s 7'= 1 implies x does not end in S-1 and xs does not end in s. l Let W = {w / w E S or w- l E S}. Then we have the following Lemma 8.10: Let € = 1 or -1. Then if ax,s 7'= 1, a;,s = Cl'" CmWd;;1... d l- l where the right-hand side is a reduced product and both Cl... Cm and d l... dn are elements of X and both Cl... cmw and dl... dnw- l lie outside X. Proof: The lemma is an immediate consequence of the preceding remarks. We need only note that for the case € = -1. Corollary 8.11: Suppose € = ±1 and d- l ax,s = Cl'" cmw n where the right-hand side is a reduced product, Cl'" Cm E X Cl... CmW e: x. Then d-l •.• I E and if w E S, then € = 1, x = Cl... Cm and S = w, (i) (ii) if we: S, then € = -1, x = d l ••• dn and S = w- l • Proof: It follows immediately from the preceding argument. Corollary 8.12: Let € = ±1, 'fJ = ±1, x, y E X, and s, t E S. If a;,s = Cl'" CmWd;;1... di""1 and a~,t = Cl'" CmWe;1... ell with the right-hand |
)-1, contrary to and wd Zthe assumption that g is a reduced Y -product. If e1... em'/) is removed as a result of deleting inverse pairs in the product di e1... emv, then e1'" emv is an initial segment of d 1 ••• d l and hence an element of X. But this is contrary to Lemma 8.10. Similarly is not removed as a result of deleting inverse pairs in the product wdi..• d l 1 1 1... dl 1 • • • d l 1 1 wdi e1... em. Thus d - 1 W Z ••• d-l 1 e1'" emv n 1-1 •., 1 = W... V n 1-1 1-1 ••. 1 1-1 when expressed as a reduced product by deleting inverse pairs (the... between wand v e1... em left after deleting inverse pairs). Consequently represent the factors d Zg ends in W··· VI;:l..• III and the inductive assertion follows. Therefore g # 1, and the result follows. • • • dl 1 1 e. Subgroups of finite index In Section 7.6b, page 228, we proved that a subgroup of index n in a group generated by r elements is generated by nr elements. We shall now find the rank of a subgroup of index n in a free group. To find the rank of the subgroup we use the result of Section 8.4d, i.e. the nonunit ax,s freely generate the subgroup. Let F be freely generated by Sl, •••, Sr and let X = {Xl, •••, Xn} with Xl = 1 be a Schreier transversal for a subgroup H of index n. Consider the elements aX;,Sj = X;Sj(XiSj)-l i = 1,..., nand j = 1,..., r. The number of such elements is nr. To find the rank of H, we wish to determine how many of the aXi,Sj are unit elements. By line 13, page 262: (1) If Xi ends in S;l, then aXi,Sj = 1. (2) If XiSj ends in Sj, then aXi,Sj = 1. We show that (1) and (2) are mutually exclusive. Suppose that Xi ends |
in S;l. Then Xi = WI ••• W mS;l (where WI, •••, Wm E W) is a reduced product. Consequently Wm # Sj (for otherwise WI.•. W mSj1 is not a reduced product). As Xi is in the Schreier transversal X, WI.•• Wm also belong to the Schreier transversal. Thus XiSj = WI ••• Wm = WI'.• Wm and XiSj does not end in Sj. Therefore (1) and (2) are mutually exclusive. Note that if neither Xi ends in Sjl nor XiSj ends in Sj, then aXi,Sj = XiSj(XiSj)-l # 1 as Sj remains when aXi,Sj is expressed as a reduced product. t ) For fixed j, let a. = number of x. E X for which ax.• s. = 1. Clearly a. = number of Xi E X for which Xi ends in S;l plus the number of Xi for which xis j ends in Sj' As Xi runs through X, xisj runs through X (x ~ XS j' X E X, is a permutation, page 219). Thus the number of x. such that x.s. ends in s. is the number of elements in X which end in sJ" We conclude a. = number of x. that end in s. or s,:-1, So the total number (i.e. with j = 1,..., r) ofaxi,sj = 1 is a 1 +... + ar = number of elements of X that end in Sj or Sjt, j = 1,..., r. But except for Xl = 1, every element of X ends in some Sj or Sj1. Hence a 1 +... + ar = n-1. 1 264 FREE GROUPS AND PRESENTATIONS [CHAP. 8 Thus there are exactly n - 1 unit elements among the axi,sj' and we have proved Theorem 8.13: Let F be a free group of rank r and let H be a subgroup of index n. Then H is a free group of rank n(r -1) + 1. Problems 8.38. Let F be freely generated by 81>..., 8 T • Let H be a subgroup |
of index 2 such that 81 e: H, but 8i E H for i = 2,..., r. Find a set of free generators for H by using the method of Section 8.4d. Verify that the number of free generators agrees with the number given by Theorem 8.13. Solution: We choose {1,81} for the Schreier transversal. Then the nonunit elements among al s. and aS1,Sj (j = 2".,r) freely generate H. Now al,Sj = lsj(18j)-1. If j = 1, al,Sj = 1. If j #- 1, al,Sj = 18j(18j)-l = 8j as 8j E H implies 8j = 1. Now, J a sl' Sj = 818'(818.)-1 J J 818j8;-1 if j #- 1 (for then 818j E H81) 8; if j = 1 (as 8i = 1) Thus the subgroup H is freely generated by 8i,82'".,8T and 81828;-1,...,818rB;-1. Thus H is of rank 2(r - 1) + 1, which agrees with Theorem 8.13. 8.39. Let F be freely generated by x and y. Find a set of free generators for F', the derived group of F. Solution: Let X = {XTyS I rand 8 integers}. Then to show that X is a Schreier transversal we need only show that (1) xTyS, XT1 y sl belong to the same coset of F' only if r = r1> 8 = 81, and (2) X is a set of coset representatives. Both (1) and (2) follow easily on using the fact that F/F' is free abelian with basis xF', yF' (by Theorem 8.6, page 255). The free generators of F' are the elements aXTYS,x and aXTyS,y which are nonunit. Now On the other hand, aXTyS,y = xTysY(XTyS+1)-l = 1 Thus a set of free generators for F' are the elements xTySxy-Sx- T - 1 for all integers r |
and all integers 8 #- O. 8.40. Let F be a free group of rank rand Hand K subgroups of index n. Prove that H 2'" K. Solution: By Theorem 8.13, Hand K are free of the same rank. Thus they are isomorphic. 8.41. Let F be a free group on generators x and y. Suppose that R <l F, y E Rand F/R = gp(xR) is infinite cyclic. Prove that the group R/R' is freely generated as an abelian group by the elements xnyx-nR', where n E Z. Then prove that for no integer n, is xnyx-nR' in the center of F/R'. Solution: The method of Section 8.4d with X = {x n I n E Z} gives for the free generators of R the elements aXn,y = xnyx- n, n E Z. By an argument similar to Theorem 8.6 (which deals only with free groups of finite rank), R/R' is free abelian with basis xnyx-nR' for all integers n. Now as (xR')(xnyx-nR')(xR')-l = x n+1yx-(n+l)R' #- xnyx-nR' xnyx-nR' does not belong to the center of F/R'. Sec. 8.4] PROOF OF THE SUBGROUP THEOREM FOR FREE GROUPS 265 f. Intersection of finitely generated subgroups We prove here that if Hand K are subgroups of finite rank of a free group F, then HnK is also a subgroup of finite rank. (This result is due to Howson.) To simplify the problem we note that we may assume that F is finitely generated. For if not, we could consider gp(H, K), which is certainly finitely generated (as Hand K are), instead of F. However, as we have seen in Example 1, page 259, a free group of rank 2 has a subgroup of infinite rank. Therefore we may as well assume that F is free of rank 2. Let F be freely generated by a and b. We say that a coset C is single-ended if every element of C when expressed as a reduced product has the same last factor. For example, if all |
the elements of C end in a, C is said to be single-ended. (We might have for instance baaa E C.) But if C contains for example the elements ababab and ab-1, then C is not single-ended. A coset which is not single-ended is called double-ended. The following lemma is crucial: Lemma 8.14: A subgroup H of the group freely generated by a and b is of finite rank if and only if it has a finite number of double-ended cosets. Proof: Choose a Schreier transversal X for H in F. Put S = {a, b}. (1) Let H be of finite rank. Then only a finite number of the elements a x • s # 1 (where x E X and s E S). But we proved in Section 8.4d that every element of H ended in the l is either S(XS)-I or S-IX- I, where ax • s # 1. Thus form wdZ I... dil where there are we concluded that all the elements of H end in the form wdZ only a finite number of wd ZI... dil and dl ··· dzw- I fl X although dl ··· dz E X. l where wd Z I... d i I... d i Let x EX. Then if Hx is double-ended, there exists an element hE H such that in hx, x cancels completely. As h ends in some wdz- I... di l and dl ··· dtW- I fl X, it follows that if x cancels completely, x is an initial segment of d l ••• d z (otherwise wdl... dz is an initial segment of x and hence in X). It follows that as the number of d l ••• d z that appear is finite, the number of initial segments is finite, and thus the number of double-ended co sets is finite. (2) Let H be of infinite rank. Ifa x • s = xS(XS)-1 # 1, it follows that x does not end in r l (line 13, page 262) and, of course, xs(xs)-1 E H. The coset Hx contains x and (xS(XS)-I)-IX = xsri. Now r l appears in the reduced product for xsr i |
(line 13, page 262). As x does not end in s-1, Hx is double-ended. Now as there are an infinite number of x such that ax.s # 1, there are an infinite number of double-ended cosets. Theorem 8.15: The intersection of two subgroups Hand K of finite rank is again of finite rank. Proof: By Lemma 8.14, H has only a finite number of double-ended cosets, say HI,..., H n, and K has only a finite number of double-ended cosets, say K I,..., Km. Now the co sets of HnK are intersections of co sets of Hand K (Problem 7.23, page 231). Also the intersection of a single-ended coset of H with any coset of K is single-ended (and vice versa). Thus the double-ended cosets of HnK are among the cosets HinK j, i = 1,...,n and j = 1,..., m. Therefore H n K has at most mn double-ended co sets and accordingly HnK is of finite rank by Lemma 8.14. 266 FREE GROUPS AND PRESENTATIONS [CHAP. 8 A look back at Chapter 8 We defined free groups and gave an alternative definition. The existence of free groups was established and homomorphisms of free groups investigated. The main result was that if F is a free group freely generated by X, then for every group H and every mapping (J of X into H there is a homomorphism cf> of F into H which coincides on X with (J. As a consequence every group is a homomorphic image of a free group. We next discussed presentations of groups. The important notion of rank of a free group then arose naturally. We discussed and proved the subgroup theorem for free groups. Using this, the rank of a subgroup of finite index was calculated. Finally we proved that the intersection of two subgroups of finite rank in a free group is of finite rank. Supplementary Problems ELEMENTARY NOTIONS 8.42. Suppose F is freely generated by a and b. Find elements u, v E F so that u # a and a-Ib-Iab = U-Iv-IUV 8.43. Prove that a free group of finite rank has only a finite number of elements of length "" n for any fixed integer n. 8.44. Pro |
ve that if N <1 G and GIN is free, then G splits over N. 8.45. Let F be a free group freely generated by a and b. Let N a- 2ba2, ••• ). Prove N <1 F and verify that F splits over N. gp(...,a2ba- 2, aba-I, b, a-Iba, 8.46. Using the notation of the preceding problem show that if N' is the derived group of N, then N' <1 F. Verify that FIN' is a splitting extension of NIN' by an infinite cyclic group. Construct an isomorphic copy of FIN'directly as an extension of a free abelian group by an infinite cyclic group. PRESENTATIONS OF GROUPS 8.47. 8.48. 8.49. 8.50. Prove that the group G = la, b; a-Iba = b2 1 is not free. (Hint: Prove that GIG' is cyclic. So if G is free it must be free cyclic. Then show G' # {I} by mapping G into a suitable factor group.) Let G = lx, y; x2, y 21. Show that G is infinite. (Hint: Let F be the free group freely generated by x and y. Let (J: x --> a, y --> ab be a homomorphism onto the group of Problem 8.30, page 257.) Let G = la, b, e, d; [a, el, [a, dj, [b, el, [b, dll. Prove G is the direct product of two free groups each of rank 2. Prove that if G = lXI, x2, ••• ; 2x2 = xl,..., (i + l)xi+ I = Xi' ••• 1, then G is isomorphic to the additive group of rationals. (Hint: Note that G has the additive group of rationals as homomorphic image. Also it is abelian. Use Problem 6.72, page 208.) 8.51. Prove that if p is a prime and G = IXI,x2""; Xl>X~X;-I, •••,xf+1x;l,... 1, to the p-Priifer group. then G is isomorphic CHAP |
. 8] SUPPLEMENTARY PROBLEMS 267 THE SUBGROUP THEOREM FOR FREE GROUPS 8.53. Prove that if F is free on x, y, z then the elements xyz, X 2y 2z 2, x 3y 3z3, x 4y 4z4,.•• freely generate a subgroup of F. PROOF OF THE SUBGROUP THEOREM 8.54. Show that if am = bn where a and b are elements of a free group (mn ~ 0), then a and b generate a cyclic group. 8.55. Let Nand M be non identity normal subgroups of G, a free group of rank greater than 1. Prove that NnM ~ {I}. 8.56. Let F be a free group. Prove that there is no sequence of subgroups Fl C F2 C... of F with (Hint: Let G = uFi. Then G is a free group of infinite rank. Fi ~ F i + 1 and rank of Fi = 2. Consider GIG' which is free abelian of infinite rank. Obtain a contradiction by showing that GIG' is of finite rank.) 8.57. Find generators for F2 where F is a free group of rank two and F2 = gp(X2 I x E F). (Note that FIF2 is the Klein 4-group.) 8.58. Let F be a group which can be generated by two elements with a free subgroup of rank 3 which is of index two. Prove F is free by comparing it with a free group of rank 2. (Hard.) 8.59. Let F, R be as in Problem 8.41, page 264. Prove that FIR' has no element other than 1 in its center. Appendix A Number Theory In this book we assume the reader knows the following: 1. The meaning of a divides b, for which we use the notation a I b. The notation a J b is read as "a does not divide b". 2. The definition of a prime, i.e. an integer not equal to 1 which is divisible only by 1 and itself. 3. If a, b are integers, then there exist an integer q and an integer r such that where 0 ~ r < b. a = bq + r 4. The definitions of greatest common divisor of two numbers a and b (the largest integer which divides |
both a and b) and lowest common multiple of a and b (the smallest integer divisible by both a and b). We write the greatest common divisor of a and b as (a, b), the lowest common multiple of a and b as I.c.m. (a, b). If (a, b) = 1, then a and bare said to be co-prime. 5. Given integers a and b, there exist integers p and q such that (a, b) = pa + qb 6. The fundamental theorem of arithmetic which says every integer is expressible in one and only one way (ignoring order) as the product of primes. 7. a == b modulo n means a - b is divisible by n. Also some of the simpler properties such as a == b and x == y implies a + x == b + Y and ax == by. The reader who does not know this material can consult (a) Niven, I., and H. S. Zuckerman, An Introduction to the Theory of Numbers, Wiley, 1966. (b) Upsensky, J. V., and M. A. Heaslet, Elementary Number Theory, McGraw-Hill, 1939. (c) Birkhoff, G., and S. MacLane, A Survey of Modern Algebra, Macmillan, 1953. 269 Appendix B A Guide to the Literature Note: Numbers in brackets refer to the bibliography on pages 272 and 273. General Books which contain much of the material of this text (and frequently more) are, in the order of complexity, [46], [15], [16]. Useful books in algebra are [47] and [23]. Chapter 1 The theory of sets was established by G. Cantor in the last quarter of the 19th century [1]. A central notion in his work is that of cardinality - two sets have the same cardinality if there is a one-to-one mapping from the one onto the other. This leads to the idea of a transfinite number, and much of Cantor's work was devoted to developing these transfinite numbers (see [2] and [3]). In the study of sets a number of paradoxes arose ([4] and [5]). This made it desirable to put set theory on a firm axiomatic footing. Such axiomatic approaches have led to a number of interesting developments ([4], [5] and [6]). Recently a different approach to |
the foundations of mathematics has been originated by Lawvere, which is based on the notion of category [44] (see [7] for the axioms of a category). Chapter 2 The study of groupoids arose in part from a desire to understand more clearly the axioms for group theory. The theory of groupoids can be subdivided into systems which satisfy various "natural" conditions. Thus, for example, among the various classes of groupoids one has semigroups, loops, groups and quasi-groups. Two major references are [8] and [9]. Chapter 3 It is clear from this chapter that groups arise in a great many mathematical disciplines. Groups arose initially from 19th century algebra, analysis and geometry. It was hoped that much of geometry could be handled by associating a group with each geometrical object. To some extent this aspect of group theory has been discussed in the section on isometry groups (see als,o [10], [11], and [14]). For the applications to topology see [37] and [39], and for knot theory see [40]. Groups arise also in quantum physics, crystallography [12] and chemistry [13]. This omnipresence of groups is part of the reason for its importance. In addition, the study of groups has been carried forward for its own sake (see e.g. [15], [16], [17], [18], [19], [20], [21], [22], [43]). Finally we refer the reader to the study of groups with a topology: [41], [42]. 9.70 APPENDIX B] A GUIDE TO THE LITERATURE 271 Chapter 4 Chapter 4 is concerned mainly with cyclic groups and homomorphisms. As we have dealt exhaustively with cyclic groups, there is not really any further informa tion available except perhaps for finding the automorphism group of a cyclic group. If G is cyclic of order n, then the automorphism group of G is the group of integers co-prime to n, with multiplication modulo n; see e.g. [43]. The concept of homomorphism is basic. One can define homomorphisms for other algebraic concepts, such as rings. Factor rings can be defined and very similar theorems obtained [23], [24], [25]. (Indeed, once the group theoretic ones are known, it is routine to obtain the others.) Some work, prompted by algebra |
ic topology, has been done on more complex interaction of group homomorphisms [26]. Chapter 5 Much of this chapter is "arithmetical" in content, i.e. it deals with the order of a finite group. Thus for example we have the theorem of Lagrange: The order of a sub group of a finite group divides the order of the group. There are a great number of important theorems of this kind, embodying various generalizations of the Sylow theorems (see [27], [17]). A very important result of a slightly different kind is the remarkable theorem of W. Feit and J. Thompson, viz. a finite group of odd order is solvable [45]. The classification of groups of small order has not been too successful. This is due to the extraordinary complexity of these groups. The reader might consult [21], [17] and [28] for a discussion of this classification problem. Today the study of finite groups has proceeded at an extraordinary pace. One of the main aims of this study is the classification of finite simple groups. This classification is by no means complete, but much progress has been made (see [29]). Chapter 6 Our proof of the fundamental theorem of abelian groups is nonconstructive, i.e. we do not have a definite procedure for finding a basis. Such procedures exist, e.g. [30]. There are also other criteria for deciding if an abelian group is a direct sum of cyclic groups, e.g. if every element is of bounded order [31]. A more comprehensive result is that of Kulikov [16]. In this chapter we found invariants for finitely generated abelian groups. The in variants for countable torsion groups are subtler and appear in Ulm's theorem ([31], [32]). Some of the theorems of abelian groups extend to wider classes of algebraic structures, e.g. modules over rings ([31]). There has also been an attempt to prove theorems about classes of groups that are quite close to abelian groups, e.g. solvable and nilpotent groups ([17], [19]). Excellent sources for abelian groups appear in [16] and [31]; a more encyclopedic ac count appears in [32]. Chapter 7 The aim of this chapter is to show how representing groups as permutation groups leads to information about the groups themselves. There are other types of representation. |
The most important of these is the representation of finite groups as n X n matrices. (These n x n matrices constitute a natural generalization of the 2 x 2 matrices we have discussed in Chapter 3.) The reader might consult [33], [34]. 272 A GUIDE TO THE LITERATURE [APPENDIXB In infinite group theory the permutational representatien of Frobenius has led to significant progress (§2 of Chapter 2, etc., of [35]). The transfer has been used a great deal in finite group theory. A good source of in formation is [27]. See also [17]. Chapter 8 Our proof of the existence of free groups is really motivated by Cayley's theorem, as a glance at the usual proof will reveal [16]. The property of Theorem 8.4, i.e. the property of being able to extend a mapping of the free generators to a homomorphism of the free group, is of great importance. Indeed, it is often used as the definition of a free group. With this approach, the definition of free group is analogous to the definition of free abelian group (Chapter 6). There are even further generalizations and we can speak of the free group in a variety, where a variety is a collec tion of groups satisfying certain conditions. A detailed account appears in [35]. One can also regard a free group as being the free product of infinite cyclic groups. The free product of two groups is a way of putting the groups together in, roughly speaking, the freest way ([16], [17]). Groups occur frequently as presentations. This is unfortunate as it is always difficult to say what the group of a presentation is or what its properties are. Indeed, there is not even a general and effective procedure for deciding whether the group of any given presenta tion is of order 1. All this is connected with the word problem, which is, roughly speaking, to determine in a finite number of steps if two products in the generators of a given presentation are equal. The word problem is unsolvable in general (see [15]). An important concept which enables us to change from one presentation to another is that of Tietze transformations [36]. There are many proofs of the subgroup theorem, including topological ones (e.g. [37]). An important one is by Nielsen transformations ([16], [17], [36]). One of the advantages of the method we used in Chapter 8 is that it |
extends readily to provide generators and defining relations for subgroups of a group of a presentation [36]. More properties of free groups, free products and presentations of groups appear in [36]. See also [38] for presentations. BIBLIOGRAPHY [1] [2] [3] [4] [5] [6] [7] [8] [9] Cantor, G., Contributions to the Founding of the Theory of Transfinite Numbers (translated by P. E. B. Jourdain), Open Court, 1915. Kamke, E., Theory of Sets (translated from the 2nd German edition by F. Bagemihl), Dover, 1950. Halmos, P. R., Naive Set Theory, Van Nostrand, 1960. Kleene, S. C., Introduction to Metamathematics, Van Nostrand, 1952. Cohen, P. J., and R. Hersh, "Non-Cantorian Set Theory", Scientific American, December 1967. Suppes, P. C., Axiomatic Set Theory, Van Nostrand, 1960. Freyd, P., Abelian Categories, An Introduction to the Theory of Functors, Harper and Row, 1964. Bruck, R. H., A Survey of Binary Systems, Springer, 1958. Clifford, A. H., and G. B. Preston, "The Algebraic Theory of Semigroups", Mathematical Surveys 7, 1961, American Mathematical Society. APPENDIX B] A GUIDE TO THE LITERATURE 273 [10] Weyl, H., Symmetry, Princeton University Press, 1952. [11] Hilbert, D., and S. Cohn-Vossen, Geometry and the Imagination, translated by P. Nemenyi, Chelsea, 1956. [12] Hammermesh, M., Group Theory and its Application to Physical Problems, Addison-Wesley, 1962. [13] Cotton, F. A., Chemical Applications of Group Theory, Interscience, 1963. [14] Yale, P. B., Geometry and Symmetry, Holden Day, 1968. [15] Rotman, J. J., The Theory of Groups: An Introduction, Allyn and Bacon, 1965. [16] Kurosh, A., The Theory of Groups, (translated by K. A. Hirsch), Chelsea, 1960. [17] Hall, Jr., M., The Theory of Groups, Macmillan |
, 1959. [18] Wielandt, H., Finite Permutation Groups, Academic Press, 1964. [19] Schenkman, E., Group Theory, Van Nostrand, 1965. [20] Scott, W., Group Theory, Prentice-Hall, 1964. [21] Burnside, W., Theory of Groups of Finite Order, Dover, 1955. [22] Huppert, B., Endliche Gruppen, Springer-Verlag, 1967. [23] Herstein, 1., Topics in Algebra, Blaisdell, 1964. [24] Van der Waerden, B. L., Modern Algebm, (translated by F. Blum), Ungar, 1953. [25] Cohen, P. M., Universal Algebra, Harper and Row, 1965. [26] Northcott, D. G., An Introduction to Homological Algebra, Cambridge U. P., 1960. [27] Wielandt, H., and B. Huppert, "Arithmetical and Normal Structure of Finite Groups," 1960 Institute on Finite Groups, Editor, M. Hall, Jr., American Mathematical Society, 1962. [28] Hall, Jr., M., and J. K. Seniour, The Groups of Order 2n (n ::S': 6), Macmillan, 1964. [29] Carter, R. W., "Simple Groups and Simple Lie Algebras", Journal of the London Mathematical Society, April 1965, pp. 193-240. [30] Schreier, 0., and E. Sperner, Introduction to Modern Algebra and Mat1'ix Theory, (translated by M. Davis and M. Haussner), Chelsea, 1959. [31] Kaplansky, 1., Infinite Groups, U. of Michigan Press, 1954. [32] Fuchs, L., Abelian Groups, Akademiai Kiado, 1958. [33] Burrow, M., Representation Theory of Finite Groups, Academic Press, 1965. [34] Curtiss, C. W., and 1. Reiner, Representation Theory of Finite Groups and Associative Algebras, Interscience, 1962. [35] Neumann, H., Varieties of Groups, Springer, 1967. [36] Magnus, W., A. Karass and D. Solitar, Combinatorial Group Theory: Present |
ations of Groups in Terms of Generators and Relations. Interscience, 1966. [37] Massey, W. S., Algebraic Topology; An Introduction, Harcourt, Brace, 1967. [38] Coxeter, H. S. M., and W. O. J. Moser, Generators and Relators for Discrete Groups, Springer-Verlag, 1965. [39] Wallace, A. H., An Introduction to Algebraic Topology, Pergamon, 1957. [40] Crowell, R. H., and R. H. Fox, Introduction to Knot Theory, Ginn, 1963. [41] Pontriagin, L. S., Topological Groups, Gordon and Breach, 1966. [42] Cohn, P., Lie Groups, Cambridge U. P., 1957. [43] Zassenhaus, H., The Theory of Groups, Chelsea, 1949. [44] Lawvere, F. W., "The Category of Categories as a Foundation for Mathematics", Proceedings of the Conference on Categorical Algebra, La Jolla, 1965, Springer-Verlag, 1966. [45] Feit, W., and J. G. Thompson, "Solvability of groups of odd order", Pacific Journal of Mathematics, vol. 3, no. 3, pp. 773-1029, 1963. [46] Lederman, W., Introduction to the Theory of Finite Groups, Oliver and Boyd, 1957. [47] Birkhoff, G., and S. MacLane, A Survey of Modern Algebra, Macmillan, 1953. INDEX Abelian group, 177-212 divisible, 205, 209 finitely generated, 196 free, 186 of type poe = p-Priifer group, 191, 206 primary = p-group, 190 rank of, 193 Abelian groupoid, 29 Additive notation, 19, 178 Alternating group, 62 simplicity, 172 Ascending central series = upper central series, 142 Associative groupoid, 29 Automorphism, group, 84, 87 inner, = mapping Pa' 85 of field, 87 of group = of groupoid, 83 Basis of free abelian group, 186 Basis theorem = fundamental theorem for finitely generated abelian groups, 197 Bijection, 14 Binary operation, 19 Block, R-, 9 Cardinality, 14 Carrier, 26 Cartesian product, 6 Cayley's theorem, 46, 214 Center, 112 |
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