Introduction. One of the things most people retain from their early study of mathematics is the

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1 Introduction One of the things most people retain from their early study of mathematics is the quadratic formula that solves for the roots of the quadratic equation: ax 2 +bx+c=0. This equation stands out because of how useful it is in many areas of mathematics. Less famous than their quadratic counterpart, there are also formulae that find the roots of cubic and quartic equations. Finding a formula for the roots of polynomial equations of degree higher than four proved to be difficult. This was the original impetus for group theory. At the age of 19, Niels Abel, a Norwegian mathematician of the early 19 th century, proved that there is in fact no formula, involving ordinary arithmetic and extraction of roots, that works for the roots of polynomials of degree five or higher. This discovery, however, was missing some pieces, including the determination of which equations of degree five could be solved with such a formula. It was the work of another brilliant young mathematician, Évariste Galois, that further developed this theory by providing a criterion, in terms of the symmetry group of root permutations, for determining which equations can be solved with a formula. Galois work was cut short, as he died in a dual at the age of 20. A statement and proof of Galois great theorem can be found in [3]. A loose definition of a group is a set on which we have defined an associative binary operation with the additional stipulations that there be an identity element and that every element have an inverse. Perhaps the most familiar example of a group is the set of ordinary integers with addition as the operation. With the foundation of group theory laid by such young geniuses as Abel and Galois, mathematicians have been kept busy exploring the concept further. It has been 1

2 applied heavily to geometric theory. For example, the group of symmetries of a square consists of the rigid motions that a square can undergo while still keeping its shape, namely four rotations and four reflections, including eight unique orientations. This is called the dihedral group of degree four or D 4. Similar to the symmetries of D 4, there are symmetries of an equilateral triangle and a regular pentagon, denoted by D 3 (of order 6) and D 5 (of order 10), respectively. In fact, there are dihedral groups for all regular polygons with n sides when n 3 such that the order of D n is 2n. Because of their resemblance to geometric shapes, it is easily seen how naturally occurring dihedral groups can be observed in animals such as sand dollars and starfish (D 5 ). Symmetry groups manifest themselves in nature in more subtle ways in the formation of naturally occurring crystals, which often present patterns of D 4 and D 6. The relevance of group theory, however, is not limited solely to the scope of mathematics. Group theory is also useful in cryptography as well as in coding for data compression and error correction and is therefore present in the everyday lives of people through their technology. Applications of group theory outside of mathematics include their use in fields such as physics and chemistry. For example the same concept of dihedral groups is often applied to molecular structures to predict how the physical properties of molecule change when its arrangement changes. Ammonia (NH 3 ), made up of a single nitrogen atom surrounded by three hydrogen atoms in the trigonal pyramidal formation, has symmetry group D 3, as it is generated by a 120 rotation and a reflection. Thus there are six permutations for the molecule. 2

3 The permutations of the arrangement of this molecule present an interesting look at the value of characterizing groups. Because there are six unique ways the molecule can be organized, it is of order six. It is known (and follows in Table 1 and Table 2 below) that a group of order six is isomorphic to, and therefore essentially the same as, one of the two groups D 3 or Z 6 (the rotation group of a regular hexagon). A proof for the fact that every group of order six is isomorphic to one of these two groups involves showing that if it contains an element of order six, it is isomorphic to Z 6 ; otherwise it is isomorphic to D 3. In the case of a molecule of ammonia, one can easily find that it has characteristics of the group D 3. This is a simple example of the utility of characterizing groups of a given order. The significance of studying group theory lies not only in its many uses as mentioned above, but also in its complexity. An illustration of the difficulty of group theory is provided by the classification of all finite simple groups, the building blocks of all finite groups. Finding such a classification was the primary interest of group theorists for much of the 20 th century and one of the great mathematical achievements of this time is a collaborative journal compilation of more than 10,000 pages that provides a complete list of all simple groups! Given the emphasis that the mathematical community has placed on this area of study, especially after considering its afore mentioned uses in multiple scientific disciplines, it is clear that the significance of my thesis exists in its relevance to my understanding, as a budding mathematician, of an important mathematical area of study with many practical applications. My thesis work focused on characterizing all groups of certain small orders. It should be noted that small should not be taken literally, but instead will refer to 3

4 numbers that are products of few primes. For example, there are 14 groups of order 16 (the product of 4 primes), so I will not be considering such groups. As seen below, I have proven theorems similar to the characterization above of groups of order 6. In order to achieve this, I have immersed myself in an in-depth study of group theory, a concept that was new to me at the beginning of the year. In my study I have read through various textbook chapters on groups, learning many theorems and working through exercises to build upon my foundation of knowledge of groups, going so far as to prove many theorems for which proofs are not provided in the books as well as applying those theorems to group related problems, demonstrating my comprehension of the intricacies of groups. Now, at the end of my study of groups, I have a stronger understanding of the nature of group theory, having learned and proven many theorems that are required in order to characterize groups. In this thesis, I will present characterizations of all groups of certain small orders. This will include a characterization of all groups of order pq where p and q are primes. Furthermore, I have included some specific cases, such as groups of order 30 and 12. My work required the use of sophisticated concepts from group theory such as semidirect products and the Sylow theorems. 4

5 Semidirect Products We begin by looking at a group formed by the semidirect product of two groups. Semidirect products are similar to direct products except that the operation in one of the components is defined differently. We begin here because we will make frequent use of semidirect products. Lemma 1: Let G, H, and K be groups. If G H and H K, then G K. Proof: Let f : G H and g : H K be isomorphisms, we will prove that the composite function g f: G K is an isomorphism. We have: g f(xy) = g(f(xy)) = g(f(x)f(y)) = g(f(x))g(f(y)) = (g f)(x)(g f)(y) for each x, y ε G. Thus g f: G K is a homomorphism. Let a ε K. Since G is surjective, we have that there exists b ε H such that g(b) = a. Additionally, since F is surjective, there is a c ε G such that f(c) = b. This gives g f(c) = g(f(c)) = g(b) = a, so G K is surjective. Let g f(x) = g(f(x)) = g(f(y)) = g f(y). We want to show that this implies x = y. g(f(x)) = g(f(y)) allows us to conclude that f(x) = f(y) because g is injective. Furthermore x = y because f is also injective. Therefore G K is injective. Therefore, g f is an isomorphism. Theorem 2: Let G be a group and let Aut G be the set of all automorphisms of G. Then Aut G is a group under the operation of compositions of functions. Proof: Let f, g ε Aut G. Because g f is an isomorphism by Lemma 1, g f ε Aut G. Thus the closure axiom is satisfied. 5

6 Let f, g, h ε Aut G and a ε G, ((f g) h)(a) = (f g)(h(a)) = f (g(h(a))) = (f(g h))(a) = (f (g h))(a). Thus the associative axiom is satisfied. Let f, g be in Aut G. Let id signify the identity function in Aut G. f id(a) = f(id(a)) = f(a) = id(f(a)) = id f(a). Thus the identity axiom is satisfied. Let f ε G. Then f is one-to-one and onto. Therefore, we know that for x ε G there exists exactly one element a in G such that f(a) = x. We set f 1 (x) = a. Now f f 1 (x) = f(f 1 (x)) = f(a) = x. Additionally, we must consider f 1 f(a) = f 1 (f(a)) = f 1 (x) = a. Furthermore, let f(b) = y for b in G and f 1 (y) = b. Then f 1 (xy) = f 1 (f(a)f(b)) = f 1 (f(ab)) = ab = f 1 (x)f 1 (y). Then f 1 is a homomorphism. Therefore the inverse axiom is satisfied. With the axioms of a group satisfied, we have that Aut G is a group under the operation of compositions of functions. Let A, B be groups and let θ: B Aut(A) be a homomorphism. For b ε B, θ(b) is a function from A to A. Therefore, instead of using expressions θ(b)(a) for a ε A, we will denote them as θ b (a) to avoid complicating notation. Next, recall from Theorem 2 that Aut(A) is a group with function composition for the operation. Let us write στ for the composition, so that (στ)(a) = σ(τ (a)) for a ε A. With the conventions above, for b 1, b 2 ε B, we have θ b1 b 2 = θ b1 θ b2 because θ is a homomorphism. Theorem 3: Let us define an operation on the set A B by (a, b)(a, b ) = (aθ b (a ), bb ). This operation satisfies the axioms of a group. 6

7 Proof: θ b1 b 2 means θ(b 1 b 2 ) and θ(b 1 b 2 ) = θ(b 1 )θ(b 2 ) by definition of a homomorphism. For a, a, a ε A and b, b, b ε B, we have (a, b)((a, b )(a, b )) = (a, b)(a θ b (a ), b b ) = (aθ b (a θ b (a )), bb b ) = (aθ b (a )θ b (θ b (a )), bb b ). Additionally: ((a, b)(a, b ))(a, b ) = (aθ b (a ), bb )(a, b ) = (aθ b (a )θ bb (a ), bb b ) = (aθ b (a )θ b (θ b (a )), bb b ) by the above statement on homomorphisms. Thus the associativity axiom is satisfied. (a, b)(e, e) = (aθ b (e), be) = (a, b) and (e, e)(a, b) = (eθ e (a), eb) = (a, b). Thus the identity axiom is satisfied. We will use b = b 1 and a = θ b 1(a 1 ). Then (a, b)(a, b ) = (aθ b (a ), bb ) = (aθ b (θ b 1(a 1 )), bb 1 ) = (aθ bb 1(a 1 ), e) = (aθ e (a 1 ), e) = (aa 1, e) = (e, e). Additionally, (a, b )(a, b) = (a θ b (a), b b) = (θ b 1(a 1 )θ b 1(a), b 1 b) = (θ b 1(a 1 a), e) = (θ b 1(e), e) = (e, e). Thus the inverse axiom is satisfied. Notation: The group in Theorem 3 is denoted by A θ B. Theorem 4: If θ is the trivial homomorphism (or θ(b) is equal to the identity automorphism for all b ε B) then A θ B is the usual direct product of A and B. Proof: If θ is the trivial homomorphism then θ b (a ) = a for all b ε B and all a ε A, so A θ B is defined by the operation (a, b)(a, b ) = (aθ b (a ), bb ) = (aa, bb ) which is the operation that defines the usual direct product of groups A and B. 7

8 Theorem 5: If θ is not the trivial homomorphism, then the semidirect product gives us a group different from the ordinary direct product; moreover, the semidirect product is nonabelian. Proof: Let b be such that θ b is nontrivial and choose a such that θ b (a) a. Then (e, b)(a, e) = (eθ b (a), be) = (eθ b (a), b). Thus (e, b)(a, e) (a, b). On the other hand, (a,e)(e,b)=(aθ e (e),eb) = (a,b). Therefore, the semidirect product of groups A and B give us a different group than their direct products if θ is not the trivial homomorphism, and the semidirect product is nonabelian. We show next that S 3, the symmetric group of order 6, can be realized as a semidirect product. Table 1: Abstract representation of the symmetric group of order 6, S 3, where a =3, b =2, ba=a 2 b: e a a 2 b ab a 2 b e e a a 2 b ab a 2 b a a a 2 e ab a 2 b b a 2 a 2 e a a 2 b b ab b b a 2 b ab e a 2 a ab ab b a 2 b a e a 2 a 2 b a 2 b ab b a 2 a e 8

9 Let C 2 = <b> and C 3 = <a> be the multiplicative cyclic groups of order 2 and 3. Define θ: C 2 C 3 by θ(b i ) = σ i, where σ(a j ) = a 2j. Table 2: C 3 θ C 2 : (e, e) (a, e) (a 2,e) (e, b) (a, b) (a 2,b) (e, e) (e, e) (a, e) (a 2, e) (e, b) (a, b) (a 2, b) (a, e) (a, e) (a 2, e) (e, e) (a, b) (a 2, b) (e, b) (a 2, e) (a 2, e) (e, e) (a, e) (a 2, b) (e, b) (a, b) (e, b) (e, b) (a 2, b) (a, b) (e, e) (a 2, e) (a, e) (a, b) (a, b) (e, b) (a 2, b) (a, e) (e, e) (a 2, e) (a 2, b) (a 2,b) (a, b) (e, b) (a 2, e) (a, e) (e, e) It is clear that the two tables are identical except for notation. Thus the semi-direct product C 3 θ C 2 is isomorphic to the group S 3. 9

10 Groups of order PQ We now look at characterizing general types of groups. Here we look at a group whose order is the product of two primes, p and q, making it of sufficiently small order for our consideration. We begin with a simple counting argument: Lemma 6: If H and K are subgroups of G, then HK denotes the set {hk ε G h ε H, k ε K}. If H K = {e}, then HK = H K. Proof: H = {h 1, h 2,, h m } and K = {k 1, k 2,, k n }. Then HK = {h 1 k 1, h 1 k 2,, h 1 k n, h 2 k 1, h 2 k 2,, h 2 k n,, h m k 1,, h m k n }. Therefore, there can be at most mn elements in HK. To demonstrate that these elements are distinct, we will assume to the contrary that h a k b = h c k d, or h 1 c h a = k d k 1 b. However, H K = {e}. Thus h -1 c h a = k d k -1 b = e, that is, h c = h a and k d = k b. Therefore, HK = mn = H K. Theorem 7: Let G be a group of order pq where p and q are primes, with p < q and p not dividing q 1. Then G has normal subgroups of orders p and q and G Z pq and is therefore cyclic. Proof: By the First Sylow Theorem, G has a Sylow p-subgroup [2, Thm 9.13]. By the Third Sylow Theorem, there must be p, q, pq or 1 Sylow p-subgroups because these are divisors of G [2, Thm 9.17]. However, the Third Sylow Theorem states that the number of Sylow p-subgroups must also be of the form 1+pk for some nonnegative integer k. Let us test these cases: 10

11 Case 1.) There are p Sylow p-subgroups: p = 1 + pk gives p(1 k) = 1, implying that p 1, which is not possible. Case 2.) There are q Sylow p-subgroups: q = 1 + pk gives pk = q 1, implying that p q 1, which contradicts the hypothesis, so this is not possible. Case 3.) There are pq Sylow p-subgroups: pq = 1 + pk gives p(q k) = 1, implying that p 1, which is not possible. Therefore, there is one Sylow p-subgroup, H, in G. G has a normal subgroup of order p if and only if H is the only Sylow p-subgroup in G, thus this subgroup H of order p is normal in G [2, Cor 9.16]. Now we must similarly show that G has a normal subgroup, H, of order q. Let J be a Sylow q-subgroup. By the Third Sylow Theorem, there must be p, q, pq or 1 Sylow q-subgroups because these are divisors of G. However, the number of Sylow q- subgroups must also be of the form 1+qk for some nonnegative integer k. Let us test these cases: Case 1.) There are p Sylow q-subgroups: p = 1 + qk, but p < q, so this is not possible. Case 2.) There are q Sylow q-subgroups: q = 1 + qk gives q(1 k) = 1, implying that q 1, which is not possible Case 3.) There are pq Sylow q-subgroups: pq = 1 + qk gives q(p k) = 1, implying that q 1, which is not possible. Therefore, there is one Sylow q-subgroup, J, in G. G has a normal subgroup of order q if and only if J is the only Sylow q-subgroup in G, thus this subgroup J of order q is normal in G [2, Cor 9.16]. 11

12 We now have that this group G has one subgroup of order p and one of order q. Now we consider H J, which is a subgroup of both H and J and, therefore, must have an order that divides both H = p and J = q by Lagrange s theorem [2, Thm 8.5]. Thus, H J = 1 and H J = {e}. By Lemma 6 we have that this implies HJ = H J, thus G = HJ. G is isomorphic to H J [2, Thm 9.3]. Furthermore, H Z p and J Z q [2, Thm 8.7]. Therefore, G H J Z p Z q. Finally G Z p Z q Z pq [2, Lem 9.8]. Theorem 8: Let H, K be subgroups of a group G. If HK = KH then HK is a subgroup of G. Proof: Assume HK = KH. Then for h 1, h 2 ε H and k 1, k 2 ε K, we have (h 1 k 1 )(h 2 k 2 ) = h 1 (k 1 h 2 )k 2 = h 1 (h 3 k 3 )k 2 = (h 1 h 3 )(k 3 k 2 ) = h 4 k 4 ε HK. Where the second equality is true for some h 3 ε H and k 3 ε K by the hypothesis Thus the closed axiom is satisfied. (h 1 k 1 ) 1 = (k 1 1 h 1 1 ) ε KH, thus (h 1 k 1 ) 1 ε HK. Furthermore, (h 1 k 1 )(h 1 k 1 ) 1 = (h 1 k 1 )(k 1 1 h 1 1 ) = e, and the inverse axiom is satisfied. Therefore, HK is a subgroup of G. Lemma 9: If H is a normal subgroup in G and K is a subgroup of G, then HK is a subgroup. Proof: hk = kk 1 hk. k 1 hk ε H [2, Thm 8.10]. Thus, kk 1 hk ε KH. Hence HK KH. Similarly, KH HK. Therefore, HK = KH. By Theorem 8, HK is a subgroup. 12

13 Lemma 10: Let U(q) denote the group of order q 1 containing the nonzero elements of Z q under multiplication modulo q. U(q) has an element, i, of order p. Proof: Because the order of U(q) is q 1, and p q 1, by Cauchy s Theorem, we have that U(q) has a cyclic subgroup of order p [2, Cor 9.14]. Theorem 11: Let p, q be primes, with p < q and p q 1. Then there is exactly one nonabelian group of order pq, up to isomorphism. Proof: Now, let i generate a cyclic subgroup of U(q) of order p. Each of the elements i, i 2, i 3,, i p-1 also has order p and these elements are distinct [2, Thm 7.15]. Now, the equation x p = 1 has at most p roots and we found them to be 1, i, i 2,, i p 1, thus there are no other elements in U(q) that are of order p [2, Cor 4.17]. Next, suppose that G is a nonabeleian group of order pq and let H = <b> and K = <a> be the Sylow subgroups of order p and q respectively. Because K G (see proof of Theorem 7), bab 1 = a i for some i. Thus a = b p ab p = a ip. a ip = a gives e = a ip a 1 = a ip 1. Thus i p 1 is a killing power for a. However, a is of order q, thus i p 1 0 or i p 1 (mod q). Now, bab 1 = a i, b 2 ab 2 = a i2,, b p 1 ab (p 1) = a ip 1. Let k denote the smallest number among i, i 2,, i p 1 (all reduced modulo q). For some j, we must have b j ab -j = a k. Let c = b j then cac 1 = a k. Then G is generated by a and c with a q = e, c p = e, and ca = a k c. Since this completely determines the Cayley table for G, there is only one nonabelian group of order pq. 13

14 Definition: The centralizer of an element a in a group G is denoted C(a) and consists of the elements in G that commute with a. Definition: The center of G, denoted Z(G), is the subset of elements a in G that commute with every g ε G. We complete our study of groups of order pq by considering the case q = p. Theorem 12: If G is a group of order p 2, with p prime, then G is abelian. Hence G is isomorphic to Z p 2 or Z p Z p. Proof: In the case of group of order p 2, we have Z(G) = p or p 2 [2, Thm 9.27]. Let a be an element of G, and assume that a is not in Z(G). Theorem 9.20 gives that C(a) is a subgroup of G; this implies that C(a) = 1, p or p 2 by Lagrange s theorem. C(a) 1 because a can commute with e and itself, so C(a) 2. Since a is not in Z(G), C(a) G, so that C(a) p 2, so C(a) = p by default. Similarly, Z(G) p 2 because Z(G) is nonabelian, so Z(G) = p. Now, Z(G) C(a) because the elements in Z(G) commute with each element of G, but C(a) also includes a, which poses a contradiction as there are shown to be at least p + 1 elements in C(a). Thus our assumption that C(a) G must be false. Therefore, G is abelian. 14

15 Groups of Order 30 With the knowledge gathered by characterizing groups of small order, we can now consider specific groups of larger order. Considering groups of order 30 presents an interesting case where there are four distinct, nonisomorphic groups, which we will here characterize. Theorem 13: Z 30, Z 5 S 3, Z 3 D 5 and D 15 are exactly four distinct groups of order 30. Proof: Z 30 is the only abelian group among them, so it is not isomorphic to any others. Furthermore, Z 5 S 3 has 3 elements of order 2: {(0,b), (0,ab), (0,a 2 b)}, Z 3 D 5 has 5 elements of order 2: {(0,b), (0,ab), (0,a 2 b), (0,a 3 b), (0,a 4 b)} and D 15 has 15 elements of order 2: {b, ab, a 2 b, a 3 b,, a 14 b}, thus, these groups are pairwise nonisomorphic. Therefore, there are at least four nonisomorphic groups of order 30. Now we must show that these are the only groups of order 30 up to isomorphism. Let G be a group of order 30. By the Third Sylow Theorem, there are 1 or 6 Sylow 5- Subgroups and 1 or 10 Sylow 3-Subgroups. Let us consider the case of the maximum number (6) of Sylow 5-subgroups possible. Because each such subgroup has order 5, each would have 4 non-identity elements. The intersection of any two of these groups is <e>, so there are a total of 24 distinct elements of order 5 in this case. Considering the maximum number (10) of Sylow 3-subgroups possible, we similarly find that there are also 20 distinct elements of order 3 in this case. If there were 6 Sylow 5-subgroups and 10 Sylow 3-subgroups, then there are shown to be at least 45 distinct elements in G, which is a contradiction of G = 30. Therefore, one of 15

16 the assumed cases must not be true, so there is either one Sylow 5-subgroup or one Sylow 3-subgroup. The existence of only one of these groups implies normality [2, Cor 9.16]. Thus we know that G must have a normal subgroup of order 5 or a normal subgroup of order 3. Let H, K be subgroups of G such that H = 5 and K = 3 where at least one of them is normal as seen above. We know that HK is a subgroup of G by Lemma 9 and HK = 15 by Lemma 6. Let HK = J. Then J is a subgroup of G of order 15. Moreover, J is a cyclic subgroup of order 15 generated by a by Theorem 7. By Cauchy s Theorem, there exists b ε G such that b = 2. Therefore b is not included in H. Then G = J Jb = {e, a, a 2,, a 14, b, ab,, a 14 b} with a 15 = b 2 = e. We will find that there are only four possibilities for ba that will give bab 1 = bab: a = b 2 a i2 b 2 = a i2, which implies that a i2 1 = e. So i 2 1 mod(15). Let i = 14: a = a 195 = e Let i = 11: a = a 120 = e Let i = 4: a 42 1 = a 15 = e Let i = 1: a 12 1 = a 0 = e These are the only possible options for i. Thus ba = a 14 b, a 11 b, a 4 b, ab represent the only possible options, meaning that there are at most four nonisomorphic groups of order 30. This gives that there are exactly four nonisomorphic groups of order 30, which are listed above. 16

17 Groups of order 12 We conclude by considering the most difficult case here studied, groups of order 12. Being the product of three primes, or more specifically, of order p 2 q, this characterization is similar to that of groups of order 30, except it will be seen that there are five distinct groups of order 12. One of these groups is an alternating group and another is a previously unnamed group, both of which prove to be difficult. Lemma 14: Consider a group G of order 4 such that G is not cyclic. Then x 2 = e for every x ε G. Such a group is abelian. Proof: Let G = {e, a, b, c}. Each element must have order dividing G = 4. There is no element of order 4 because that would make the group cyclic. Thus every non-identity element has order two. From here, the table fills itself in: Table 3: e a b c e e a b c a a e c b b b c e a c c b a e It is clear that G is abelian. G is isomorphic to the group Z 2 Z 2, and is called the Klein 4-group. 17

18 Theorem 15: If H G and G/H = k, then x k ε H for all x ε G. Proof: G/H denotes the set of all right cosets of H in G. The fact that G/H = k gives that (Hx) k = He for every Hx ε G/H [2, Cor 8.6]. Then Hx k =(Hx) k = He. Furthermore, because H is a subgroup, x = ex k ε Hx k = He = H. Theorem 16: If every nonidentity element of a group G has order 2, then G is abelian. Proof: a = 2 implies that aa = e or a = a 1 for every nonidentity element in G. Let a, b ε G. Then there exists c ε G such that ba = c by the closure axiom of a group. Therefore, because every nonidentity element of a group G has order 2 we have cc = (ba)(ba) = e. This gives b 1 = aba or a 1 b 1 = ba. Because all nonidentity elements are their own inverse, this gives us a 1 b 1 = ab = ba. Therefore, G is abelian. Next, we describe a specific group, T, of order 12: Example: Let G be the group S 3 Z 4 and let a = ((123), 2) and b = ((12), 1). (a) a 2 = ((123), 2)((123), 2) = ((132), 0) a 3 = ((132), 0)((123), 2) = ((1), 2) a 4 = ((1), 2)((123), 2) = ((123), 0) a 5 = ((123), 0)((123), 2) = ((132), 2) a 6 = ((132), 2)((123), 2) = ((1), 0) Thus a = 6 18

19 Furthermore: b 2 = ((12), 1)((12), 1) = ((1), 2) = a 3 from above. Thus b 2 = a 3. a 1 = ((132), 2) as seen in a 6 above. Now: ba = ((12), 1)((123), 2) = ((23), 3) a 1 b = ((132), 2)((12), 1) = ((23), 3) Thus a 1 b = ba. (b) Claim: The set T = {e = a 0, a 1, a 2, a 3, a 4, a 5, b, a 1 b, a 2 b, a 3 b, a 4 b, a 5 b} consists of 12 distinct elements. Proof: b = ((12), 1) ab = ((123), 2)((12), 1) = ((13), 3) a 2 b = ((132), 0)((12), 1) = ((1), 1) a 3 b = ((1), 2) ((12), 1) = ((12), 3) a 4 b = ((123), 0)((12), 1) = ((13), 1) a 5 b = ((132), 2)((12), 1) = ((1), 3) As seen by comparing these, along with what was found in part (a), we see that each of these elements are in fact distinct. (c) The fact that T is a subgroup of G follows from Table 4 below. From part (a), ba = ((23), 3) ((13), 3) = ab, demonstrating that T is nonabelian. 19

20 Table 4: e a a 2 a 3 a 4 a 5 b ab a 2 b a 3 b a 4 b a 5 b e e a a 2 a 3 a 4 a 5 b ab a 2 b a 3 b a 4 b a 5 b a a a 2 a 3 a 4 a 5 e ab a 2 b a 3 b a 4 b a 5 b b a 2 a 2 a 3 a 4 a 5 e a a 2 b a 3 b a 4 b a 5 b b ab a 3 a 3 a 4 a 5 e a a 2 a 3 b a 4 b a 5 b b ab a 2 b a 4 a 4 a 5 e a a 2 a 3 a 4 b a 5 b b ab a 2 b a 3 b a 5 a 5 e a a 2 a 3 a 4 a 5 b b ab a 2 b a 3 b a 4 b b b a 5 b a 4 b a 3 b a 2 b ab a 3 a 2 a e a 5 a 4 ab ab b a 5 b a 4 b a 3 b a 2 b a 4 a 3 a 2 a e a 5 a 2 b a 2 b ab b a 5 b a 4 b a 3 b a 5 a 4 a 3 a 2 a e a 3 b a 3 b a 2 b ab b a 5 b a 4 b e a 5 a 4 a 3 a 2 a a 4 b a 4 b a 3 b a 2 b ab b a 5 b a e a 5 a 4 a 3 a 2 a 5 b a 5 b a 4 b a 3 b a 2 b ab b a 2 a e a 5 a 4 a 3 Theorem 17: The groups Z 12, Z 2 Z 2 Z 3, A 4, D 6 and the above described group T are pairwise nonisomorphic. Moreover, if G is a group of order 12, then G is isomorphic to one of these groups. Hence there are, up to isomorphism, exactly 5 groups of order 12. Proof: Recall from above that T = {e = a 0, a 1, a 2, a 3, a 4, a 5, b, a 1 b, a 2 b, a 3 b, a 4 b, a 5 b} where a 1 b = ba, a 6 = e, b 2 = a 3. We will demonstrate that the groups A 4, D 6 and T are pairwise nonisomorphic by comparing their elements of order 2 by considering their Cayley tables. 20

21 As seen in Table 4, T has 1 element of order 2: {a 3 } Using ba = a 5 b gives the dihedral group D 6 with 7 elements of order 2, as seen in Table 5: {a 3, b, ab, a 2 b, a 3 b,, a 5 b}. Table 5: e a a 2 a 3 a 4 a 5 b ab a 2 b a 3 b a 4 b a 5 b e e a a 2 a 3 a 4 a 5 b ab a 2 b a 3 b a 4 b a 5 b a a a 2 a 3 a 4 a 5 e ab a 2 b a 3 b a 4 b a 5 b b a 2 a 2 a 3 a 4 a 5 e a a 2 b a 3 b a 4 b a 5 b b ab a 3 a 3 a 4 a 5 e a a 2 a 3 b a 4 b a 5 b b ab a 2 b a 4 a 4 a 5 e a a 2 a 3 a 4 b a 5 b b ab a 2 b a 3 b a 5 a 5 e a a 2 a 3 a 4 a 5 b b ab a 2 b a 3 b a 4 b b b a 5 b a 4 b a 3 b a 2 b ab e a 5 a 4 a 3 a 2 a ab ab b a 5 b a 4 b a 3 b a 2 b a e a 5 a 4 a 3 a 2 a 2 b a 2 b ab b a 5 b a 4 b a 3 b a 2 a e a 5 a 4 a 3 a 3 b a 3 b a 2 b ab b a 5 b a 4 b a 3 a 2 a e a 5 a 4 a 4 b a 4 b a 3 b a 2 b ab b a 5 b a 4 a 3 a 2 a e a 5 a 5 b a 5 b a 4 b a 3 b a 2 b ab b a 5 a 4 a 3 a 2 a e The alternating group A 4 : A 4 has 3 elements of order 2 {(12)(34), (13)(24), (14)(23)} [1, pp. 107]. Thus, these groups A 4, D 6 and T are pairwise nonisomorphic 21

22 Additionally, the groups Z 12, Z 2 Z 2 Z 3 are abelian (and nonisomorphic because Z 12 is cyclic while Z 2 Z 2 Z 3 is not). Therefore Z 12, Z 2 Z 2 Z 3, A 4, D 6 and T are pairwise nonisomorphic. Now we must show that an arbitrary group G of order 12 is isomorphic to one of the afore mentioned groups: Z 12, Z 2 Z 2 Z 3, A 4, D 6 or T. Suppose G is abelian. Then G is isomorphic to either Z 12 or Z 2 Z 2 Z 3 by the Fundamental Theorem of Finite Abelian Groups [2, Thm 9.7]. If the group contains an element of order 12 then it is isomorphic to Z 12. If it does not, then it is isomorphic to the group Z 2 Z 2 Z 3. Suppose that G is nonabelian. Because G is a nonabelian group of order 12, the nonidentity elements of G must be of order 2, 3, 4, 6 by Lagrange s Theorem. Furthermore, G cannot be entirely composed of elements of order 2, because this would make it abelian by Theorem 16. Hence all nonidentity elements of G have order 2, 3, 4 or 6. Suppose G does have an element a of order 6; let H = < a > = {e = a 0, a 1, a 2, a 3, a 4, a 5 }. Let b be any element of G not contained in H. Then the 12 elements {e = a 0, a 1, a 2, a 3, a 4, a 5, b, a 1 b, a 2 b, a 3 b, a 4 b, a 5 b} are distinct because a = 6 and a i = a j b implies b = a i j ε H, which is contrary to the choice of b. Thus G = {e = a 0, a 1, a 2, a 3, a 4, a 5, b, a 1 b, a 2 b, a 3 b, a 4 b, a 5 b}. bab 1 is an element of H by normality and, furthermore, bab 1 = 6 by a standard isomorphism theorem, since conjugation is always an automorphism. Hence bab 1 = a or a 5 as these are the elements are of order 6. But G being nonabelian rules out bab 1 = a, so bab 1 = a 5 or, equivalently, 22

23 ba = a 5 b. There are then two cases: there is an element b not in H such that b = 2 or b 2 for all elements b not in H. In the first case, we take b to have order 2, and the table fills itself in as shown above in Table 5. It follows that G must be isomorphic to D 6 in this case. Then suppose that no element outside of the cyclic group H has order 2. Choose b outside of H. We claim that b has order 4. Otherwise b has order 3 or 6, as it must divide G = 12. But b 2 ε H by Theorem 15. If b has order 3, then b 1 = b 2 ε H. Additionally, b ε H because H is closed under the inverse axiom. This presents a contradiction to the assumption that b is not in H. Therefore, b does not have order 3. If b has order 6, then b 3 has order 2. But b 3 is not in H, and we assumed no element outside of H has order 2, thus presenting a contradiction. Hence b = 4. Furthermore, b = 4 implies b 2 = 2, but a 3 is the only element in H of order 2, so b 2 = a 3. The Cayley Table fills itself in using these relations as seen above in Table 4. It follows that G is isomorphic to T in this case. Finally, suppose that G has no element of order 6, thus all nonidentity elements have order 2, 3 or = By the Third Sylow Theorem, there must be 1 or 4 Sylow 2-subroups and 1 or 4 Sylow-3 subgroups. If there is only one Sylow 3-subgroup of order 3 in G, then it is normal [2, Cor 9.16]. If there are 4 Sylow 3-subgroups, then each of them has 2 nonidentity elements, and each nonidentity element has order 3 by [2, Cor 8.6]. Because the intersection of any two of these subgroups is <e>, there are 4 2 = 8 elements of order 3 in G. Every Sylow 2-subgroup of G has order 4 and each element of a Sylow 2-subgroup must have order dividing 4 by [2, Cor 8.6]. Therefore, no such elements can be in the set of 8 elements of 23

24 order 3. With the 8 elements of order 3 and the 4 additional elements in a Sylow 2- subgroup, there are 12 total elements. Thus there is room in G for only one group of order 4. In this case, the Sylow 2-subgroup of order 4 is normal by [2, Cor 9.16]. Therefore, either a Sylow 2-Subgroup or Sylow 3-subgroup must be normal. Let H be a subgroup of order 4 and K = <b> be a subgroup of order 3. Suppose by way of contradiction that K G. Let x ε H, x e. Then xbx 1 = b or b 2. We claim that x = 2. Otherwise x = 4; hence x 2 = 2 and, in particular, x 2 is not in K. If xbx 1 = b, then xb = bx. Hence (xb) 2 = xbxb = x 2 b 2 e (since x 2 is not in K), and (xb) 3 = x 3 b 3 = x 3 e = x 3 e. However, this yields the contradiction that xb 6. Thus H contains no elements of order 4, an H must be the Klein 4-group. Let H = {e, u, v, w}, with uv = w. If, ubu 1 = b, then an argument similar to that just given yields ub = 6, a contradiction. Hence ubu 1 = b and, similarly, vbv 1 = wbw 1 = b 2. However, we then have wbw 1 = (uv)b(uv) 1 = (uv)bv 1 u 1 = u(vbv 1 )u 1 = ub 2 u 1 = ubbu 1 = ub(u 1 u)bu 1 = (ubu 1 )(ubu 1 ) = b 2 b 2 = b, a contradiction. Therefore, we must have a normal Sylow 2-subgroup, which we denote by H. Let K = <b> be one of the Sylow 3-subgroups. Suppose that H = <a> is cyclic. If bab 1 = a, then an argument similar to that just given yields ba 6. Hence bab 1 = a 3. However, this gives b 2 ab 2 = a, which yields the contradiction that b 2 a 6. Hence H is the Klein 4-group where H = {e, a 1, a 2, a 3 }, and we have G = {e, a 1, a 2, a 3, b, a 1 b, a 2 b, a 3 b, b 2, a 1 b 2, a 2 b 2, a 3 b 2 }. If ba i b 1 = a i, then ba i has order 6, a contradiction. Without loss of generality, we then have ba 1 b 1 = a 2. If ba 2 b 1 = a 1, we obtain ba 3 b 1 = ba 1 a 2 b 1 = ba 1 b 1 ba 2 b 1 = a 1 a 2 = a 3, a 24

25 contradiction. Hence ba 2 b 1 = a 3, and ba 3 b 1 = a 1. This then determines b 2 a i b 2 for each i. Thus we know ba i and b 2 a i for each i, and this completely determines the Cayley table for such a G. Thus there is at most one nonabelian group of order 12 with no element of order 6. There is such a group, namely A 4. It follows that G is isomorphic to A 4 in this case. We have seen that there are 5 nonisomorphic groups of order 12 and that there are no more than 5 such groups. Thus there are exactly 5 distinct groups up to isomorphism of order 12. Therefore, a group G of order 12 must be isomorphic to exactly one of Z 12, Z 2 Z 2 Z 3, A 4, D 6 or T. Conclusion We have characterized groups of certain small order, including all groups of order pq where p and q are prime in the general case, and the specific cases of groups of orders 30 and 12. As demonstrated above, this required the understanding of complex mathematical concepts such as the Sylow theorems and semidirect products. Though group theory is a vast area of study, the theorems proven above provide the background and foundation for the study of groups and are essential to the exploration of groups of higher order. 25

26 References [1] J. Gallian, Contemporary Abstract Algebra, 7 th ed., Brooks Cole, Boston [2] T. Hungerford, Abstract Algebra, 3 rd ed., Brooks Cole, Boston, [3] J. Rotman, Galois Theory, 2 nd ed., Springer, New York,

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