THE EXISTENCE AND UNIQUENESS OF A SIMPLE GROUP OF ORDER 168


 Poppy Fletcher
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1 THE EXISTENCE AND UNIQUENESS OF A SIMPLE GROUP OF ORDER 168 TIMOTHY L. VIS Abstract. The study of simple groups has been, and continues to be, the most significant area of work in group theory. In this paper, we give a proof of the existence and uniqueness of a simple group of order 168. We begin by presenting (without proof) the major results from group theory upon which this proof draws. We then deduce many of the properties a simple group of order 168 must satisfy. We use these properties to show that such a group, if it exists, must be unique. Finally, we give a concrete realization of the group as the collineation group of the Fano plane from finite geometry. 1. Preliminaries The following results will play an important role in our discussion of the properties a simple group of order 168 would necessarily satisfy. Since we are concerned primarily with the proof of the uniqueness and existence of the simple group of order 168, we present these results without proof. However, the proofs of these results are, in general, instructive in the study of simple groups. Theorem 1.1 (Index Theorem). Let G be a group with subgroup H of index G : H = k. If G k!, then G contains a normal subgroup K. In particular, G is not simple. Definition 1.2. If G is a group of order p a m, where p is a prime not dividing m, then a subgroup H of G of order H = p a is called a Sylow psubgroup of G. Notation 1.3. The number of Sylow psubgroups of G will be denoted n p. Theorem 1.4 (Sylow s Theorem). Let G be a group of order p a m, where p is a prime not dividing m. Then the following hold. (1) G has a Sylow psubgroup. (2) If P and Q are distinct Sylow psubgroups of G, then there exists some g G such that Q = gpg 1. That is, any two Sylow psubgroups of G are conjugate in G. (3) The number of Sylow psubgroups of G, n p satisfies n p 1 mod p. Further, n p is the index in G of the normalizer N G (P) of any Sylow psubgroup P, and as such, n p m. Corollary 1.5. n p 1 for any simple group G. Theorem 1.6. If n p 1 mod p k there exist distinct Sylow psubgroups P and R of G such that P R is of index p k 1 in both P and R. Date: February 26,
2 2 TIMOTHY L. VIS Theorem 1.7. If G has more than one Sylow psubgroup, and P and Q are Sylow psubgroups with P Q is maximal, then N G (P Q) has more than one Sylow psubgroup and any two distinct Sylow psubgroups of N G (P Q) intersect in P Q. Theorem 1.8. If G is a group with subgroup H, then N G (H)/C G (H) is isomorphic to some subgroup of Aut (H). Theorem 1.9. If G is a group of order 12, then either G has a unique Sylow 3subgroup or G = A 4. Theorem The following properties hold for the group S 4. (1) S 4 has a unique normal 2subgroup isomorphic to the Klein 4group and three other conjugate Klein 4groups. (2) Any 2 of the nonnormal Klein 4groups in S 4 generate S 4. (3) The normal Klein 4group and any other Klein 4group in S 4 generate a copy of D 8. (4) Sylow 2subgroups of S 4 are isomorphic to the dihedral group D 8, and there are 3 such subgroups. (5) The normalizer of a Sylow 3subgroup of S 4 is isomorphic to S 3. Theorem The number of conjugates of a subset S in a group G is the index of the normalizer of S, G : N G (S). In particular, the number of conjugates of an element s of G is the index of the centralizer of s, G : C G (s). Theorem If U and W are normal subgroups of a Sylow psubgroup P of G then U is conjugate to W in G if and only if U is conjugate to W in N G (P). Theorem 1.13 (First Isomorphism Theorem). If φ is a homomorphism from a group G into a group H, then kerφ G, and G/kerφ = φ(g). Theorem 1.14 (Second Isomorphism Theorem). If G is a group with subgroups A and B such that A N G (B). Then AB G, B AB, A B A, and AB/B = A/A B. Theorem If V is an elementary Abelian group of order p e, then Aut(V ) = GL e (GF (p)). This group has order (p e 1)(p e p) (p e p e 1). Theorem The following are true of characteristic subgroups. (1) Characteristic subgroups are normal. (2) If H is the unique subgroup of G of a given order, then H is characteristic in G. (3) If K is a characteristic subgroup of H and H G, then K G. 2. Properties of a Simple Group of Order 168 Suppose that G is a simple group, G = 168 = In this section we determine conditions which G must necessarily satisfy in order to be simple. What follows requires no knowledge of G other than the assumptions that G is simple and that G has order 168. The arguments that follow are adapted, in large part, from [1]. For easy reference, we state each individual fact as a separate proposition. Proposition 2.1. G has no proper subgroup of index less than 7. Proof. The index theorem (1.1) implies that if H is a proper subgroup of G of index k, 168 k!. Now if k < 7, k! 6!, which implies that G 6!. But 168 does not divide 720 = 6!. Thus, if H is a proper subgroup of G, H has index at least 7.
3 SIMPLE GROUP OF ORDER Proposition 2.2. G has exactly 8 Sylow 7subgroups; i.e. n 7 = 8 Proof. By Sylow s Theorem (1.4), the number of Sylow 7subgroups of G, n 7 satisfies n 7 1 mod 7 and n = 24. To satisfy n 7 24, n7 {1, 2, 3, 4, 6, 8, 12, 24}. Given these options the requirement that n 7 1 mod 7 restricts n 7 to {1, 8}. The simplicity of G then requires that n 7 = 8. Corollary 2.3. G has 48 elements of order 7. Proof. Since, by Proposition 2.2 G has 8 Sylow 7subgroups, and since each of these contains 6 elements of order 7 different from those contained in any other Sylow 7subgroup, there are at least 48 elements of order 7. Since every element of order 7 in G generates some Sylow 7subgroup of G, there are at most 48 elements of order 7 in G. That is, G has 48 elements of order 7. Corollary 2.4. If P is a Sylow 7subgroup of G, then N G (P) = 21. Proof. By Sylow s Theorem (1.4), n 7 = 8 is the index in G of N G (P). But since G : N G (P) N G (P) = 168, this implies that N G (P) = 21. Corollary 2.5. If P is a Sylow 7subgroup of G, no element of order 2 normalizes P. Proof. If x normalizes P, then, by definition, x N G (P). But then x 21. Since 2 21, x 2. Proposition 2.6. G has no elements of order 14. Proof. Suppose x G, x = 14. Then x 2 P, where P is some Sylow 7subgroup of G and x 7 = 2. More specifically, P = x 2. But then if p P, p = x 2m and vice versa. So x 7 px 7 = x 7 x 2m x 7 = x 2m = p, and x 7 N G (P), in violation of Corollary 2.5. Thus, G has no element of order 14. Proposition 2.7. G has no elements of order 21. Proof. Suppose x G, x = 21. Then x 7 = 3 and x 7 P, where P is some Sylow 3subgroup of G. But then P = x 7. Thus, if p P, p = x 7m and vice versa. So x 3 px 3 = x 3 x 7m x 3 = x 7m = p, and x 3 N G (P). Since x 3 = 7, 7 NG (P). By Sylow s Theorem (1.4), then 7 n3. Since n 3 56, it follows that n 3 8. Since Sylow s Theorem restricts n 3 to 3k + 1 for integer values of k, it follows that n 3 = 4. But then, G : N G (P) = 4, in violation of Proposition 2.1. Thus, G has no element of order 21. Corollary 2.8. If P is a Sylow 7subgroup of G, then C G (H) = 7. Proof. Let p be a generator of P. Then C G (p) = C G (P). Since P = 7 and N G (P) = 21, we have either C G (P) = 7 or C G (P) = 21. In the latter case, P C G (P) with index 3, so that the centralizer would be Abelian. But then C G (P) = Z 21, which is impossible since G has no element of order 21. Corollary 2.9. An element of G of order 7 has exactly 24 conjugates. Proof. By Theorem 1.11, the number of conjugates is the index of the centralizer. Since the centralizer of an element of order 7 has order 7, the element has = 24 conjugates.
4 4 TIMOTHY L. VIS Corollary G contains 2 conjugacy classes of elements of order 7. Proof. By Corollary 2.3, G contains 48 element of order 7. By Corollary 2.9, an element of order 7 has 24 conjugates. It follows then that G contains = 2 conjugacy classes of elements of order 7. Proposition n 3 = 28. Proof. By Slogs Theorem (1.4), n 3 satisfies n 3 1 mod 3 and n = 56. To satisfy n 3 56, n3 {1, 2, 4, 7, 8, 14, 28, 56}. Given these options, the requirement that n 3 1 mod 3 restricts n 3 to {1, 7, 28}. The simplicity of G eliminates n 3 = 1. Let R be a Sylow 7subgroup of G. By Proposition 2.4, N G (R) = 21. But then there is some x N G (R) with x = 3. This x must be in some Sylow 3subgroup Q of G. Then Q normalizes R in G. By Sylow s Theorem (1.4) all Sylow 3subgroups of G are conjugate. Since conjugation is an isomorphism, it follows that any Sylow 3subgroup normalizes some Sylow 7subgroup in G. Specifically, if y maps Q to Q by conjugation, then Q normalizes yry 1. Suppose now that n 3 = 7. Then N G (P) = 24 = Let P be some Sylow 3subgroup of G. So P normalizes a Sylow 7subgroup S of G. Let T be a Sylow 2subgroup of N G (P). For every t T, tpt 1 normalizes tst 1 by the preceding argument, so that P normalizes tst 1 for all t T. Now consider the subgroup T acting by conjugation on the set of Sylow 7 subgroups of G. By Corollary 2.5, no element of order 2 normalizes a Sylow 7 subgroup of G. As such, since all nonidentity elements of T have a power of order 2, no nonidentity element of T normalizes any Sylow 7subgroup of G. Now let R be, as before, a Sylow 7subgroup of G and let t 1, t 2 T be distinct nonidentity elements of G. Then if t 1 R = t 2 R, we have t 1 1 t 2 R = R, so that t 1 1 t 2 normalizes R. But only the identity normalizes R, so this is a contradiction. Thus, each t T maps R to a distinct Sylow 7subgroup of G. Since n 7 = 8 = T, it follows that T acts transitively on the set of Sylow 7subgroups of G. Thus, every Sylow 7 subgroup of G is of the form tst 1, so that P normalizes all Sylow 7subgroups of G. Thus, P is a subgroup of the intersection of the normalizers of all Sylow 7 subgroups, and thus a proper subgroup of G. Further, suppose that x is in this intersection. If R is a Sylow 7subgroup of G, then xrx 1 = R. Consider then gxg 1 for some g G. gxg 1 R ( gxg 1) 1 = gxg 1 Rgx 1 g 1. But this is just gxr x 1 g 1 for some other Sylow 7subgroup R. Since x normalizes all Sylow 7subgroups, this is gr g 1. But note that g 1 Rg = R, so that this is simply R. Thus, this intersection is a normal subgroup of G. Thus, the simplicity of G requires that this intersection be trivial, a contradiction, and n 3 = 28. Corollary G has 56 elements of order 3. Proof. Since n 3 = 28, and each Sylow 3subgroup of G contains 2 elements of order 3 distinct from those contained in any other Sylow 3subgroup of G, G contains at least 56 elements of order 3. Since any element of order 3 generates a Sylow 3subgroup of G, there are at most 56 element of order 3. Thus, G contains exactly 56 elements of order 3. Corollary If P is a Sylow 3subgroup of G, N G (P) = 6.
5 SIMPLE GROUP OF ORDER Proof. By Sylow s Theorem (1.4), n 3 = 28 is the index in G of N G (P). But since G : N G (P) N G (P) = 168, this implies that N G (P) = 6. Lemma There is a pair of distinct Sylow 2subgroups of G with a nontrivial intersection. Proof. By Sylow s Theorem (1.4) n 2 satisfies n 2 1 mod 2 and n = 84. To satisfy n 2 84, n2 {1, 2, 3, 4, 6, 7, 12, 14, 21, 28, 42, 84}. Given these options, the requirement that n 2 1 mod 2 restricts n 2 to {1, 3, 7, 21}. Simplicity of G eliminates n 2 = 1. If n 2 = 3, then the normalizer of a Sylow 2subgroup has index 3, in contradiction to Proposition 2.1. Thus, n 2 is either 7 or 21. But then n 2 1 mod 2 3, so by Theorem 1.6 there exists a pair of Sylow 2subgroups whose intersection has index at most 2 2 in each. Since the order of each is 2 3, the intersection is nontrivial. Proposition The normalizer of the largest intersection of a pair of Sylow 2subgroups is isomorphic to S 4. Proof. Over all pairs of distinct Sylow 2subgroups with nontrivial intersection, let T 1 and T 2 be chosen so that U = T 1 T 2 has maximum order. Now let N = N G (U). Since the order of a Sylow 2subgroup is 8, the order of U is either 2 or 4. Then U is either Z 2 or Z 4. U is clearly a normal subgroup of N, so by Theorem 1.8, the group N/C N (U) is isomorphic to a subgroup of Aut (U). The restriction on the order of U guarantees that Aut (U) {{1},Z 2, S 3 }. In particular, the factor group N/C N cannot have order divisible by 7. If then N has a subgroup of order 7, this subgroup must lie within C N (U). Thus, this subgroup commutes with all elements of U. In particular, this subgroup is a Sylow 7subgroup of G which is normalized by an element of order 2 in U, in contradiction of Corollary 2.5. Thus, the order of N is not divisible by 7, and N { 2, 2 2, 2 3, 2 3, 2 2 3, }. By Theorem 1.7 N has more than one Sylow 2subgroup. Thus, the order of N cannot be a power of 2, otherwise N would be the unique Sylow 2subgroup of N. Further, by Theorem 1.7 any two distinct Sylow 2subgroups of N intersect in the subgroup U, so that the order of N is divisible by 2 2. Thus N { 2 2 3, }. Now suppose that P is a Sylow 3subgroup of N. Then P must also be a Sylow 3subgroup of G, and its normalizer in G, N G (P) has order 6. Since N N (P) N G (P), it follows that N N (P) {3, 6}. Since N {12, 24}, N : N N (P) {2, 4, 8}. However, Sylow s Theorem (1.4) guarantees that the index is the number of Sylow 3subgroups in N and restricts that number to {1, 4}. Thus, N : N N (P) = 4, and N has 4 Sylow 3subgroups, which must, by Sylow s Theorem be permuted transitively by elements of N under conjugation. Now N {12, 24}. If N = 12, then the existence of 4 Sylow 3subgroups implies that N = A 4 by Theorem 1.9. Since A 4 has a unique Sylow 2subgroup and N does not, it follows that N 12, and thus, that N = 24 (and N N (P) = 6). Consider N acting by conjugation on its Sylow 3subgroups, and let K be the associated kernel. Then K is the intersection of the normalizers of each of the Sylow 3subgroups of N. Suppose that K {1}. Then, since K N N (P), K 6. Additionally, since P does not normalize another Sylow 3subgroup of N, P K. Since P = 3, K = 2. Since K is the kernel of a homomorphism (associated to the group action), the group N/K exists and has order 12. But N/K has multiple
6 6 TIMOTHY L. VIS Sylow 2subgroups, corresponding to the Sylow 2subgroups of N factored out by K and four Sylow 3subgroups, corresponding to the four Sylow 3subgroups of N. This is a contradiction to Theorem 1.9, so that K = 1, and thus K = {1}. But then N permutes the four Sylow 3subgroups in all possible ways, so that N = S 4. Corollary The largest intersection of a pair of Sylow 2subgroups is a Klein 4group. Proof. Using all notation as in the proof of Proposition 2.15, note that S 4 has the Klein 4group as its unique normal 2subgroup (Theorem Since U is a normal 2subgroup of N and N = S 4, U must be a Klein 4group. Corollary Sylow 2subgroups of G are isomorphic to D 8. Proof. Again, using all notation as in the proof of Proposition 2.15, note that a Sylow 2subgroup of N has order 8, and is thus a Sylow 2subgroup of G. Since Sylow 2subgroups of S 4 are isomorphic to the dihedral group D 8 (by Theorem 1.10 and all Sylow 2subgroups of G are isomorphic (by conjugation), then the Sylow 2subgroups of G are isomorphic to D 8 as desired. Corollary G has 42 elements of order 4. Proof. Each Sylow 2subgroup is isomorphic to D 8 and thus contains 2 elements of order 4, for a total of at most 42 elements of order 4. On the other hand, no intersection of two Sylow 2subgroups of G contains an element of order 4, so that there are exactly 42 elements of order 4. Corollary G has a unique conjugacy class of elements of order 4. Proof. Any two elements of order 4 in the same Sylow 2subgroup of G are conjugate within D 8, the structure of that Sylow 2subgroup of G. Further, by Sylow s Theorem (1.4), any 2 Sylow 2subgroups are conjugate. That is, an element of order 4 in one Sylow 2subgroup is conjugate to some element of order 4 in any other Sylow 2subgroup, and thus, to every other element of order 4 in G. Corollary The normalizer in G of a Sylow 3subgroup is isomorphic to S 3. Proof. Again, using all notation as in the proof of Proposition 2.15, note (by Theorem 1.10 that the normalizer of a Sylow 3subgroup of S 4 is isomorphic to S 3. Since the normalizer of a Sylow 3subgroup in N has order 6 and the normalizer of a Sylow 3subgroup in G has order 6, and since a Sylow 3subgroup of N is a Sylow 3subgroup of G, it follows that the normalizer of any Sylow 3subgroup of G is isomorphic to S 3. Corollary All elements of order 3 in G are conjugate. Proof. Any element of order 3 is conjugate to its inverse in S 3. Since the normalizer of a Sylow 3subgroup in G is isomorphic to S 3, any element of order 3 is conjugate to the other element of order 3 in that Sylow 3subgroup of G. Further, Sylow s Theorem (1.4) states that all Sylow 3subgroups of G are conjugate. Thus, an element of order 3 is conjugate to some element of order 3 in any other Sylow 3subgroup of G. But then all elements of order 3 in G are conjugate. Corollary G has no elements of order 6.
7 SIMPLE GROUP OF ORDER Proof. Suppose g is an element of order 6 in G. Then g 2 is a Sylow 3subgroup of G and g N G ( g 2 ). But then N G ( g 2 ) contains an element of order 6. But N G ( g 2 ) = S3, and S 3 has no element of order 6. Thus, G has no element of order 6. Proposition The centralizer of the generator of the center of a Sylow 2 subgroup of G is that Sylow 2subgroup of G. Proof. By Corollary 2.5, no element of order 2 in G commutes with an element of order 7. Suppose an element of order 2 commutes with an element of order 3. It would follow that their product has order 6, which violates Corollary Thus, no element of order 2 commutes with any element of odd order (since the only admissible odd orders are 3, 7, and 21, and no element of order 21 exists by 2.7). Let T be a Sylow 2subgroup of G. Then, by Corollary 2.17, T = D 8, and Z (T) = z, where z is a particular element of order 2. Consider then C G (z). Since z has order 2, no element of odd order is in C G (z), and C G (z) {2, 4, 8}. But T C G (z), and T = 8, so that C G (z) = T. Corollary The normalizer of a Sylow 2subgroup of G is that Sylow 2 subgroup of G Proof. Clearly a Sylow 2subgroup T of G normalizes itself. Further, any element normalizing T must normalize its center. But this implies that such an element would commute with the generator of the center, so that the normalizer is a subgroup of the centralizer of the generator of the center. By Proposition 2.23, this centralizer is simply T. That is, N G (T) T N G (T), so that N G (T) = T. Corollary n 2 = 21. Proof. With all notation as in the proof of Proposition 2.23 Suppose g normalizes T. Then g normalizes Z (T); that is, g commutes with z. But then g T and N G (T) = T, so that N G (T) = 8. By Sylow s Theorem (1.4) n 2 = G : N G (T). But this is simply = 21. Proposition G has a unique conjugacy class of elements of order 2. Proof. Since, by Proposition 2.23, there is an element z of order 2 in G with C G (z) a Sylow 2subgroup of G, so that C G (z) = 8. But then z has = 21 distinct conjugates in G by Theorem By Corollary 2.18 there are 42 elements of order 4 in G. By Corollary 2.3 there are 48 elements of order 7 in G. Finally, by Corollary 2.12, there are 56 elements of order 3 in G. Thus, 146 of the nonidentity elements of G have order other than 2, so that at most 21 elements of G have order 2, and the 21 conjugate elements discussed are the only elements of order 2. Thus, the elements of G of order 2 form a unique conjugacy class. Proposition The two Klein 4groups contained in a Sylow 2subgroup of G are not conjugate in G. Proof. By Theorem 1.12, it suffices to show that these two Klein 4groups, which we denote by U and W, are not conjugate in N G (T), where T is a Sylow 2subgroup of G. But recall from Corollary 2.24 that N G (T) = T = D 8. Since each Klein 4group has index 2 in D 8, it follows that each is normal and conjugate only to
8 8 TIMOTHY L. VIS itself, so that U and W are not conjugate in T. Thus, U and W are not conjugate in G. Proposition If T is a Sylow 2subgroup of G containing a Klein 4group U as the intersection with another Sylow psubgroup, then the normalizer of the other Klein 4group in T is isomorphic to S 4. Proof. Let W be the other Klein 4group in T, and let W = z, w, where A(T) = z. By Proposition 2.26 w is conjugate in G to z. Then C G (w) = C G (z) and is a Sylow 2subgroup of G containing W, distinct from U. But then W is the maximal intersection of these two Sylow 2subgroups and, by Corollary 2.15 N G (W) = S 4 as desired. Corollary If U is a Klein 4group contained in a Sylow 2subgroup of G, then U has 7 conjugates. Proof. By Corollary 2.15 and Proposition 2.15, N G (U) = S 4. Then, by Theorem 1.11, U has = 7 conjugates. Corollary G is isomorphic to a subgroup of A 7. Proof. Since G has a subgroup isomorphic to S 4, and thus of index 7, there exists a homomorphism from G into S k with kernel contained in S 4. Then, by the First Isomorphism Theorem (1.13) G is isomorphic to a subgroup of S 7. However, if G is not also isomorphic to a subgroup of A 7, the intersection of G and A 7 within S 7 would be a normal subgroup of G. Hence, G is isomorphic to a subgroup of A 7. Proposition Sylow 3subgroups of G and Sylow 7subgroups of G are cyclic. Proof. Sylow 3subgroups and Sylow 7subgroups of G have order 3 and 7 respectively. The only groups of those orders are cyclic. 3. Relations of the Klein 4Groups in G With these characteristics of a simple group of order 168 known, we now proceed to prove that if such a group exists, it is unique. Recall from Proposition 2.27, that the two Klein 4groups in a given Sylow 2subgroup are not conjugate. Fix a Sylow 2subgroup P and let U and W be its Klein 4groups. Notation 3.1. Denote the 7 conjugates of U by U i and the 7 conjugates of W by W i. Proposition 3.2. Each W i is normalized by exactly 3 of the U j. Proof. Consider N G (W a ). By Corollary 2.15, N G (W a ) = S 4. Further, W a must be the unique normal Klein 4group in S 4. But this S 4 contains three further Klein 4groups by Theorem Since the product of W a with any of these 3 is a subgroup of order 8, and thus a Sylow 2subgroup, it follows that these three are among the U j and W i. But none of these are conjugate to W a in S 4, and thus cannot be conjugate to W a in G (by Theorem 1.12). That is, these are among the U j. Further there are no other Klein 4groups in S 4, so that these are the only subgroups U j normalizing W a. Corollary 3.3. Each U i normalizes exactly 3 of the W j.
9 SIMPLE GROUP OF ORDER Proof. Suppose that U a normalizes W b. Then U a W b = D8, so that W b U a = D8 and W b normalizes U a. By Proposition 3.2 and the symmetry of the U i and W j established in the proof of Proposition 2.28, each U i is normalized by exactly 3 of the W j. But then those 3 W j are each normalized by U i. Further, if any other W k were normalized by U i, U i would be normalized by 4 of the W j, a contradiction of Proposition 3.2. Proposition 3.4. For any U i U j, there exists at most one W k that is normalized by both U i and U j. Proof. Suppose that both U a and U b normalize W c. So U a and U b are nonnormal Klein 4groups in the S 4 that is N G (W c ). But then U a and U b generate N G (W c ) by Theorem Suppose that U a and U b also both normalize W d. Then U a and U b generate N G (W d ) so that N G (W c ) = N G (W d ). Since W c is the unique normal Klein 4group in N G (W c ) and W d is the unique normal Klein 4group in N G (W d ), it follows that W c = W d. That is, at most one W k is normalized by both U i and U j for distinct U i, U j. Corollary 3.5. For any W i W j, there exists at most one U k that normalizes both W i and W j. Proof. Since U a normalizes W b if and only if W b normalizes U a (as in the proof of Corollary 3.3), and since the symmetry of the U i and W j, along with Proposition 3.4, guarantee that at most one U k is normalized by both W i and W j, at most one U k normalizes both W i and W j. Proposition 3.6. For any U i U j, there exists exactly one W k that is normalized by both U i and U j. Proof. Suppose that no W j is normalized by both U a and U b. There are 7 W j, and are normalized by each of U a and U b. Thus, a unique W c is normalized by neither. But then the 3 U i that normalize W c must each normalize exactly one of the W j normalized by each of U a and U b. 5 of the U i are thus accounted for. The remaining U i must each normalize 3 of the W j which are among those normalized by U a and U b. But in order for this to happen, some pair (W d, W e ) must be normalized by two of the U i, a contradiction of Corollary 3.5. Corollary 3.7. For any W i W j, there exists exactly one U k that normalizes both W i and W j. Proof. By Proposition 3.6 and symmetry of the U i with the W j, there exists exactly one U k that is normalized by both W i and W j. Since U i normalizes W j if and only if W j normalizes U i, there exists exactly one U k that normalizes both W i and W j. Each element g G acts by conjugation on the U i and on the W j. Since conjugation is an inner automorphism of G, this action preserves the relation of normalizing. That is, if U a normalizes W b, gu a g 1 normalizes gw b g 1. Further, by the definition of the U i and W j, the action of conjugation maps each of the U i to some U i and each of the W i to some W j.
10 10 TIMOTHY L. VIS 4. The Projective Plane of Order 2 In the preceding section, we determined a great deal about the relationship between the Klein 4groups of G. In this section we describe a geometry realized by the Klein 4groups of G. Definition 4.1. A projective plane is a pointline incidence geometry satisfying the following axioms. (1) Any 2 points are incident with a unique line. (2) Any 2 lines are incident with a unique point. (3) There exist at least 4 points such that no 3 are incident with a common line. The following classic result in finite geometry gives other conditions on the existence of a projective plane. Theorem 4.2. For any projective plane π having a finite number of points, there exists some integer n such that each line in π is incident with exactly n points, each point is incident with exactly n lines, and π has exactly n 2 + n + 1 points. Definition 4.3. A finite projective plane having n points incident with each line is said to have order n. From the definition of a projective plane (Definition 4.1) and Theorem 4.2 it is clear that the smallest possible order of a projective plane is 2. In fact, such a plane exists and is unique. Definition 4.4. The unique projective plane of order 2, is called the Fano plane. The Fano plane has an elegant visual presentation which follows. We note the following properties of the Fano plane. (1) Every line is incident with exactly 3 points and every point is incident with exactly 3 lines. (2) Every pair of points is incident with a unique line, and every pair of lines is incident with a unique point. We also note the following definition and theorem regarding automorphisms of projective planes.
11 SIMPLE GROUP OF ORDER Definition 4.5. An automorphism of a projective plane is a permutation of the points and lines that preserves the incidence relation. That is, if φ is an automorphism of the plane, then the point P is incident with the line l if and only if the point φ(p) is incident with the line φ(l). Theorem 4.6. Any automorphism of the Fano plane is determined by its action on 3 noncollinear points. Finally, we note the following useful characterization of the Fano plane. Theorem 4.7. The geometry induced by taking the 1dimensional vector subspaces of GF (2) 3 as points and the 2dimensional vector subspaces as lines and incidence as subspace inclusion and containment is the Fano plane. 5. Uniqueness of a Simple Group of Order 168 In the preceding two sections, we presented properties of the relation between Klein 4subgroups of G and of the Fano plane. Notice, however, that if we replace the U i by points, the W i with lines, and the relation is normalized by with the relation is incident with in the discussion of the Klein 4groups of G that we obtain the properties of the Fano plane. We can, thus, dispense with the cumbersome terminology from G and instead use their geometrical analogues from the Fano plane in the remaining discussion. Note that this analogue extends to the discussion of G acting by conjugation on the U i and the W j. The analogue of this is that of G acting on the points and lines of the Fano plane. Further, this action preserves the incidence relation in the Fano plane, since conjugation preserves the normalizing relation between the Klein 4groups of G. Proposition 5.1. The kernel of this action by G on the Fano plane is the identity. Proof. Any element g of G normalizing each of the U i would necessarily be in the intersection of their normalizers. But this intersection cannot have elements of order 4 (a consequence of Corollary 2.16). Thus, such g = 2 and, as g N G (U i ), g V, where V is a Klein 4group of N G (U i ). Since g normalizes each of the U i, the structure of S 4 ensures that g U i. But then g cannot normalize all of the W j. By way of the correspondance established, g is not in the kernel of this action. Corollary 5.2. G is isomorphic to some subgroup of the automorphism group of the Fano plane. Proof. The group action described is a homomorphism from G into the automorphism group of the Fano plane. Since the kernel is trivial, the First Isomorphism Theorem (1.13) guarantees that G is isomorphic to the range of this action. Proposition 5.3. The order of the automorphism group of the Fano plane is at most 168. Proof. Any automorphism of the Fano plane is determined by its action on a triple of noncollinear points by Theorem 4.6. Given a triple of noncollinear points, there are 7 choices for the image of the first point, 6 for the second, and 4 for the third, for a total of 168 possible choices. That is, there are at most 168 automorphisms of the Fano plane.
12 12 TIMOTHY L. VIS Theorem 5.4. If G exists, it is isomorphic to the automorphism group of the Fano plane and is thus unique. Proof. By Corollary 5.2, G must be isomorphic to a subgroup of the automorphism group of the Fano plane. By Proposition 5.3, this automorphism group has order at most 168. Thus, if G exists, it cannot be isomorphic to a proper subgroup of the automorphism group and must be isomorphic to the automorphism group itself. Since the automorphism group is unique, G is unique if it exists. 6. Existence of a Simple Group of Order 168 In order to establish the existence of a simple group of order 168, 2 things remain to be proved: that the automorphism group of the Fano plane indeed has order 168, and that this group is indeed simple. Proposition 6.1. The automorphism group of the Fano plane has order 168. Proof. Consider the construction of the Fano plane given by Theorem 4.7. By ( Theorem 1.15, the automorphism group of GF (2) 3 is GL 3 (GF (2)) and has order ) ( )( ) = = 168. Consider the action of GL 3 (GF (2)) on the points and lines of the Fano plane. Suppose that some element of GL 3 (GF (2)) fixes all points and lines of the Fano plane. Such an element fixes all 1dimensional subspaces of GF (2) 3. But these 1dimensional subspaces each consist of a single vector and its scalar multiples. The only scalar multiple of a vector over GF (2) other than that vector is the zero vector. Since elements of GL 3 (GF (2)) are invertible, the generator of a 1 dimensional vector subspace must be mapped to itself by an element fixing that vector subspace. But then an element fixing all 1dimensional vector subspaces fixes all vectors. Thus, the only element of GL 3 (GF (2)) acting as the identity on the Fano plane is the identity element. But then GL 3 (GF (2)) is isomorphic to the automorphism group of the Fano plane. Since this group has order 168, the automorphism group of the Fano plane must also have order 168. Proposition 6.2. The group GL 3 (GF (2)) is simple. Proof. For sake of contradiction, suppose H is a proper nontrivial normal subgroup of GL 3 (GF (2)). Let P be a point in the Fano plane and let N be its stabilizer in GL 3 (GF (2)). Since 168 is the maximum number of triples of noncollinear points, a triple of noncollinear points (P, P, P ) is mapped to every other triple of noncollinear points in the plane by some element of GL 3 (GF (2)). In particular, the action of GL 3 (GF (2)) on the points of the Fano plane is transitive. As such, GL 3 (GF (2)) : N = 7. Consider the intersection of all conjugates of N. This intersection must fix all points of the Fano plane, and, as such, is trivial. Then H N. It follows then that N < HN. But since N has index 7, this implies that HN = GL 3 (GF (2)). By the Second Isomorphism Theorem (1.14), H : H N = HN : N, so that 7 H. Since GL3 (GF (2)) has a subgroup of index 7, it is isomorphic to a subgroup of S 7. Now the normalizers of a Sylow 7subgroup of S 7 have order 42, so that GL 3 (GF (2)) has no normal Sylow 7subgroup, as such would require order 168. Then, by the same counting arguments as in the proof of Proposition 2.2, it follows that for GL 3 (GF (2)), n 7 = 8. Now if H had a normal Sylow 7subgroup,
13 SIMPLE GROUP OF ORDER this subgroup would be characteristic in H and thus normal in GL 3 (GF (2)) by Theorem 1.16, a contradiction. Thus, for H, n 7 1. Then n 7 must be 8 in H, so that H is divisible by 8. Then 56 H. Since H is proper, H = 56. Consider then the number of elements in H. Since n 7 = 8 in H, there are 48 elements in H of order 7. There remain only 8 other elements of H, so that a Sylow 2subgroup of H, which must exist, must be unique, and thus characteristic. Then this is normal in GL 3 (GF (2)), so that GL 3 (GF (2)) has a unique Sylow 2subgroup. Consider the subset of GL 3 (GF (2)) consisting of lower triangular matrices. In order to be invertible, all that is required is that the diagonal entries be nonzero, and thus be 1. The 3 entries below the main diagonal can each be 1 or 0, so that there are 8 such matrices. But multiplication of lower triangular matrices produces lower triangular matrices, so that this subset is in fact a subgroup of order 8, that is, a Sylow 2subgroup of GL 3 (GF (2)). The same argument holds for the upper triangular matrices. But the upper and lower triangular matrices intersect only trivially, so that we have produced distinct Sylow 2subgroups of GL 3 (GF (2)). This is a contradiction, and no proper nontrivial normal subgroup H can exist. That is, GL 3 (GF (2)) is simple. Theorem 6.3. A simple group of order 168 exists. Proof. Proposition 6.1 establishes that the automorphism group of the Fano plane has order 168 and is isomorphic to GL 3 (GF (2)). Proposition 6.2 establishes that GL 3 (GF (2)) is simple. Thus, GL 3 (GF (2)) is a simple group of order 168. Theorem 6.4. There exists a unique (up to isomorphism) simple group of order 168. Proof. Theorem 6.3 gives the existence of such a simple group and Theorem 5.4 establishes its uniqueness. References [1] D. Dummit and R. Foote. Abstract Algebra. John Wiley and Sons, 2004.
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