THE EXISTENCE AND UNIQUENESS OF A SIMPLE GROUP OF ORDER 168

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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 p-subgroup of G. Notation 1.3. The number of Sylow p-subgroups 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 p-subgroup. (2) If P and Q are distinct Sylow p-subgroups of G, then there exists some g G such that Q = gpg 1. That is, any two Sylow p-subgroups of G are conjugate in G. (3) The number of Sylow p-subgroups 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 p-subgroup 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 p-subgroups 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 p-subgroup, and P and Q are Sylow p-subgroups with P Q is maximal, then N G (P Q) has more than one Sylow p-subgroup and any two distinct Sylow p-subgroups 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 3-subgroup or G = A 4. Theorem The following properties hold for the group S 4. (1) S 4 has a unique normal 2-subgroup isomorphic to the Klein 4-group and three other conjugate Klein 4-groups. (2) Any 2 of the non-normal Klein 4-groups in S 4 generate S 4. (3) The normal Klein 4-group and any other Klein 4-group in S 4 generate a copy of D 8. (4) Sylow 2-subgroups of S 4 are isomorphic to the dihedral group D 8, and there are 3 such subgroups. (5) The normalizer of a Sylow 3-subgroup 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 p-subgroup 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 7-subgroups; i.e. n 7 = 8 Proof. By Sylow s Theorem (1.4), the number of Sylow 7-subgroups 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 7-subgroups, and since each of these contains 6 elements of order 7 different from those contained in any other Sylow 7-subgroup, there are at least 48 elements of order 7. Since every element of order 7 in G generates some Sylow 7-subgroup 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 7-subgroup 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 7-subgroup 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 7-subgroup 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 3-subgroup 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 7-subgroup 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 7-subgroup 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 3-subgroup Q of G. Then Q normalizes R in G. By Sylow s Theorem (1.4) all Sylow 3-subgroups of G are conjugate. Since conjugation is an isomorphism, it follows that any Sylow 3-subgroup normalizes some Sylow 7-subgroup 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 3-subgroup of G. So P normalizes a Sylow 7-subgroup S of G. Let T be a Sylow 2-subgroup 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 non-identity elements of T have a power of order 2, no non-identity element of T normalizes any Sylow 7-subgroup of G. Now let R be, as before, a Sylow 7-subgroup of G and let t 1, t 2 T be distinct non-identity 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 7-subgroup of G. Since n 7 = 8 = T, it follows that T acts transitively on the set of Sylow 7-subgroups of G. Thus, every Sylow 7- subgroup of G is of the form tst 1, so that P normalizes all Sylow 7-subgroups 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 7-subgroup 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 7-subgroup R. Since x normalizes all Sylow 7-subgroups, 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 3-subgroup of G contains 2 elements of order 3 distinct from those contained in any other Sylow 3-subgroup of G, G contains at least 56 elements of order 3. Since any element of order 3 generates a Sylow 3-subgroup 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 3-subgroup 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 2-subgroups of G with a non-trivial 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 2-subgroup 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 2-subgroups whose intersection has index at most 2 2 in each. Since the order of each is 2 3, the intersection is non-trivial. Proposition The normalizer of the largest intersection of a pair of Sylow 2-subgroups is isomorphic to S 4. Proof. Over all pairs of distinct Sylow 2-subgroups 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 2-subgroup 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 7-subgroup 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 2-subgroup. Thus, the order of N cannot be a power of 2, otherwise N would be the unique Sylow 2-subgroup of N. Further, by Theorem 1.7 any two distinct Sylow 2-subgroups 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 3-subgroup of N. Then P must also be a Sylow 3-subgroup 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 3-subgroups in N and restricts that number to {1, 4}. Thus, N : N N (P) = 4, and N has 4 Sylow 3-subgroups, 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 3-subgroups implies that N = A 4 by Theorem 1.9. Since A 4 has a unique Sylow 2-subgroup 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 3-subgroups, and let K be the associated kernel. Then K is the intersection of the normalizers of each of the Sylow 3-subgroups of N. Suppose that K {1}. Then, since K N N (P), K 6. Additionally, since P does not normalize another Sylow 3-subgroup 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 2-subgroups, corresponding to the Sylow 2-subgroups of N factored out by K and four Sylow 3-subgroups, corresponding to the four Sylow 3-subgroups 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 3-subgroups in all possible ways, so that N = S 4. Corollary The largest intersection of a pair of Sylow 2-subgroups is a Klein 4-group. Proof. Using all notation as in the proof of Proposition 2.15, note that S 4 has the Klein 4-group as its unique normal 2-subgroup (Theorem Since U is a normal 2-subgroup of N and N = S 4, U must be a Klein 4-group. Corollary Sylow 2-subgroups of G are isomorphic to D 8. Proof. Again, using all notation as in the proof of Proposition 2.15, note that a Sylow 2-subgroup of N has order 8, and is thus a Sylow 2-subgroup of G. Since Sylow 2-subgroups of S 4 are isomorphic to the dihedral group D 8 (by Theorem 1.10 and all Sylow 2-subgroups of G are isomorphic (by conjugation), then the Sylow 2-subgroups of G are isomorphic to D 8 as desired. Corollary G has 42 elements of order 4. Proof. Each Sylow 2-subgroup 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 2-subgroups 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 2-subgroup of G are conjugate within D 8, the structure of that Sylow 2-subgroup of G. Further, by Sylow s Theorem (1.4), any 2 Sylow 2-subgroups are conjugate. That is, an element of order 4 in one Sylow 2-subgroup is conjugate to some element of order 4 in any other Sylow 2-subgroup, and thus, to every other element of order 4 in G. Corollary The normalizer in G of a Sylow 3-subgroup 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 3-subgroup of S 4 is isomorphic to S 3. Since the normalizer of a Sylow 3-subgroup in N has order 6 and the normalizer of a Sylow 3-subgroup in G has order 6, and since a Sylow 3-subgroup of N is a Sylow 3-subgroup of G, it follows that the normalizer of any Sylow 3-subgroup 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 3-subgroup in G is isomorphic to S 3, any element of order 3 is conjugate to the other element of order 3 in that Sylow 3-subgroup of G. Further, Sylow s Theorem (1.4) states that all Sylow 3-subgroups of G are conjugate. Thus, an element of order 3 is conjugate to some element of order 3 in any other Sylow 3-subgroup 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 3-subgroup 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 2-subgroup 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 2-subgroup 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 2-subgroup of G is that Sylow 2- subgroup of G Proof. Clearly a Sylow 2-subgroup 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 2-subgroup 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 4-groups contained in a Sylow 2-subgroup of G are not conjugate in G. Proof. By Theorem 1.12, it suffices to show that these two Klein 4-groups, which we denote by U and W, are not conjugate in N G (T), where T is a Sylow 2-subgroup of G. But recall from Corollary 2.24 that N G (T) = T = D 8. Since each Klein 4-group 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 2-subgroup of G containing a Klein 4-group U as the intersection with another Sylow p-subgroup, then the normalizer of the other Klein 4-group in T is isomorphic to S 4. Proof. Let W be the other Klein 4-group 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 2-subgroup of G containing W, distinct from U. But then W is the maximal intersection of these two Sylow 2-subgroups and, by Corollary 2.15 N G (W) = S 4 as desired. Corollary If U is a Klein 4-group contained in a Sylow 2-subgroup 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 3-subgroups of G and Sylow 7-subgroups of G are cyclic. Proof. Sylow 3-subgroups and Sylow 7-subgroups of G have order 3 and 7 respectively. The only groups of those orders are cyclic. 3. Relations of the Klein 4-Groups 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 4-groups in a given Sylow 2-subgroup are not conjugate. Fix a Sylow 2-subgroup P and let U and W be its Klein 4-groups. 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 4-group in S 4. But this S 4 contains three further Klein 4-groups by Theorem Since the product of W a with any of these 3 is a subgroup of order 8, and thus a Sylow 2-subgroup, 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 4-groups 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 non-normal Klein 4-groups 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 4-group in N G (W c ) and W d is the unique normal Klein 4-group 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 4-groups of G. In this section we describe a geometry realized by the Klein 4-groups of G. Definition 4.1. A projective plane is a point-line 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 non-collinear points. Finally, we note the following useful characterization of the Fano plane. Theorem 4.7. The geometry induced by taking the 1-dimensional vector subspaces of GF (2) 3 as points and the 2-dimensional 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 4-subgroups 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 4-groups 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 4-groups 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 4-group 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 non-collinear points by Theorem 4.6. Given a triple of non-collinear 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 1-dimensional subspaces of GF (2) 3. But these 1-dimensional 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 1-dimensional 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 non-collinear points, a triple of non-collinear points (P, P, P ) is mapped to every other triple of non-collinear 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 7-subgroup of S 7 have order 42, so that GL 3 (GF (2)) has no normal Sylow 7-subgroup, 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 7-subgroup,

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 2-subgroup 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 2-subgroup. 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 2-subgroup 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 2-subgroups 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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