School of Mathematics and Statistics MT5824 Topics in Groups Problem Sheet V: Direct and semidirect products (Solutions)
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1 CMRD 2010 School of Mathematics and Statistics MT5824 Topics in Groups Problem Sheet V: Direct and semidirect products (Solutions) 1. Give an example of two groups G and H and a subgroup of the direct product G H which does not have the form G 1 H 1 where G 1 G and H 1 H. Solution: Let X be a non-trivial group, and let G = H = X. Let D = { (x, x) x X }. If (x, x), (y, y) D, then (x, x)(y, y) =(xy, xy) D(x, x) 1 =(x 1,x 1 ) D. Hence D is a subgroup of X X. We cannot write D as G 1 H 1 where G 1 G and H 1 H, because such G 1 and H 1 would have to be non-trivial (as D contains elements with non-identity values in both entries), yet D contains no element of the form (g, 1) with g = 1, while G 1 H 1 contains such elements. (Subgroups of the form D are often described as diagonal subgroups, although this refers to a slightly larger class of subgroups.) 2. Let M and N be normal subgroups of a group G. By considering the map g (Mg,Ng), or otherwise, show that G/(M N) isisomorphictoasubgroupofthedirectproductg/m G/N. Solution: Let φ denote the above map, so φ: G G/M G/N. Now (gh)φ =(Mgh,Ngh)=(Mg Mh,Ng Nh)=(Mg,Ng)(Mh,Nh)=(gφ)(hφ) so φ is a homomorphism. Furthermore, ker φ = { g G (Mg,Ng)=(M1,N1) } = { g G Mg = M1 and Ng = N1 } = { g G g M and g N } = M N. The First Isomorphism Theorem implies that G/(M N) = im φ G/M G/N. 3. Using Question 2, or otherwise, show that if m and n are coprime integers, then C m C n = C mn. 1
2 Solution: Let G = x = C mn,so x = mn. Let M = x m and N = x n. Then M has index m in G and N has index n in G. Hence G/M = Mx = C m and G/N = Nx = C n. Now M = n and N = m, som N = 1 (since m and n are coprime). Hence G = G/(M N) (a group of order mn) is isomorphic to a subgroup of G/M G/N (the latter also a group of order mn). Thus C mn = G = G/(M N) = G/M G/N = Cm C n. 4. Let X 1, X 2,...,X n be non-abelian simple groups and let G = X 1 X 2 X n. (In this question we will identify the concepts of internal and external direct products. Thus we speak a subgroup of G containing a direct factor X i instead of it containing the subgroup X i in the notation of the lectures.) Prove that a non-trivial normal subgroup of G necessarily contains one of the direct factors X i. Hence show that every normal subgroup of G has the form X i1 X i2 X ik for some subset {i 1,i 2,...,i k} of {1, 2,...,n}. [Hint: If N is a non-trivial normal subgroup of G, chooseanon-identity(x 1,x 2,...,x n) N. Consider conjugating this element by (1,...,1,g,1,...,1).] Now suppose that X 1, X 2,...,X n are abelian simple groups. Is it still true that every normal subgroup of the direct product has this form? Solution: Let N be a non-trivial normal subgroup of G and consider a non-identity element x =(x 1,x 2,...,x n ). One of these entries must be not the identity, say x i = 1. Consider conjugating by a =(1,...,1,g,1,...,1), where the g occurs in the ith entry: so N contains a 1 xa =(x 1,...,x i 1,g 1 x i g, x i+1,...,x n ) N x 1 a 1 xa =(1,...,1,x 1 i g 1 x i g, 1,...,1). Now X i is a non-abelian simple group, so Z(X i )=1. In particular, x i Z(X i ), so there exists some g X i which does not commute with x i. Thus if we take this g in the above, we deduce that 1 = x 1 a 1 xa N X i,where X i = { (1,...,1,h,1,...,1) h X i } G. Now N X i is a normal subgroup of X i (since N G) and X i = Xi is simple. Since N X i = 1, it follows that N X i = X i.thus X i N. This holds for every index i such that N contains an element of the form x = (x 1,x 2,...,x n )withx i = 1. Let i 1, i 2,...,i k be the indices for which this holds. The above shows X i1, X i2,..., X ik N, so X i1 X i2 X ik = X i1 Xi2... X ik N. (Here we have identified the direct product on the left with the obvious subgroup of G isomorphic to it.) Furthermore, if (x 1,x 2,...,x n ) N, then x i = 1 for i {i 1,i 2,...,i k } by the choice of the indices i j,son is contained in the product X i1 Xi2... X ik. Hence N = X i1 Xi2... X ik = X i1 X i2 X ik. 2
3 On the other hand, suppose that X 1 = X 2 = C p, the cyclic group of prime order. Then G = X 1 X 2 is abelian, so every subgroup is normal in G. In particular, D = { (x, x) x C p } is a normal subgroup of G which does not have the claimed form. Thus the result no longer holds if the X i are abelian. 5. Let p be a prime number. (a) Show that Aut C p = Cp 1. (b) Show that Aut(C p C p) = GL 2(F p)(wheref p = Z/pZ denotes the field of p elements). [For (a), let C p = x. Observe that an automorphism α is given by x x m where m is a representative for a non-zero element of F p. Recall the multiplicative group of a finite field is cyclic. For (b), write C p C p additively and view it as a vector space over F p. Show that automorphisms of the group then correspond to invertible linear maps.] Solution: (a) Let G = C p = x where x = p. First note that we can specify a homomorphism θ : G H, whereh is any group, by specifying the image y of x. This is a homomorphism whenever the order of y divides p. In particular, we can specify a homomorphism φ: G G by specifying the value of xφ to be one of {1,x,x 2,...,x p 1 }.Ifφ is an automorphism, then xφ = x m where m = 1, 2,..., p 1. (For φ is onto and therefore maps x to one of the elements of G of order p. Sincep is prime, this is just saying that xφ = 1.) Thus Aut G = {φ 1,φ 2,...,φ p 1 } where φ m : x x m. We have now parametrised Aut G by the non-zero elements in the field F p = Z/pZ. Let m, n be non-zero elements in F p (i.e., the congruence classes of the integers m and n (mod p) wherep m, n). Then xφ m φ n =(x m )φ n =(xφ n ) m =(x n ) m = x mn = xφ mn and we deduce that the map m φ m is an isomorphism from the multiplicative group of F p to Aut G. Thus Aut G = (F p ) = Cp 1. (We use the fact that the multiplicative group of F p is cyclic of order p 1.) (b) Consider G = C p C p.writeg additively, so that it is generated by two elements x and y and consists of all elements of the form: g = λx + µy where λ, µ {0, 1, 2,...,p 1}. It then makes sense to multiply g by the scalar ν F p = {0, 1, 2,...,p 1}. We simply multiply g by ν (as an integer) and reduce coefficients modulo p. Inthisway,G can be viewed as a vector space over the field F p, and now the generators x and y form a basis for G. Furthermore, since a scalar λ F p can be viewed as λ = , λ times 3
4 any (group) homomorphism φ: G G satisfies (λx)φ =(x + x + + x)φ =(xφ)+(xφ)+ +(xφ) =λ(xφ). Hence a group homomorphism φ from G to itself is the same thing as a linear transformation. (The only difference is the requirement that φ preserves scalar multiplication, and this follows from the fact that φ preserves addition.) Therefore the set of invertible group homomorphisms G G (i.e., automorphisms of G) is the same as the set of invertible linear transformations of G (as a 2-dimensional vector space over F p ) and hence Aut G = GL 2 (F p ). 6. Let G = x be a cyclic group. Show that Aut G is abelian. Solution: Let φ, ψ Aut G. They (and their products) are determined by their effect on the generator x. Suppose that xφ = x r and xψ = x s. Then, using the fact that φ and ψ are homomorphisms, and xφψ =(x r )ψ =(xψ) r =(x s ) r = x rs xψφ =(x s )φ =(xφ) s =(x r ) s = x rs. Hence φψ = ψφ since they have the same effect on x. Therefore Aut G is abelian. 7. Show that the dihedral group D 2n is isomorphic to a semidirect product of a cyclic group of order n by a cyclic group of order 2. What is the associated homomorphism φ : C 2 Aut C n? Solution: Recall that D 2n = α, β where α has order n and β has order 2. Now N = α has index 2 in D 2n,soN D 2n. Let H = β. Since α and β do not commute, β N, soh N is a proper subgroup of H. As H = 2, this forces H N = 1. Now HN = H N / H N =2n. Hence D 2n = HN, N H and H N = 1. ThusD 2n = H N. The associated homomorphism φ: H Aut N is determined by the value of βφ (since H = β) and this automorphism is determined by its effect on α. Now α βφ = β 1 αβ by definition of φ (see the proof that internal and external semidirect products are equivalent) and by calculating in D 2n,weseeβ 1 αβ = α 1. Thus φ maps β to the inversion automorphism (α α 1 ) of N. 8. Show that the quaternion group Q 8 may not be decomposed (in a non-trivial way) as a semidirect product. [Hint: How many elements of order 2 does Q 8 contain?] Solution: Suppose that Q 8 = H N, wheren,h are non-trivial subgroups of Q 8. Then N has order 2 or 4 and H has order 4 or 2. In particular, N and H both contain elements of order 2, so Q 8 contains at least two elements of order 2 (since N H = 1). However, this is not the case since Q 8 has a unique element of order 2. Thus Q 8 does not decompose as a nontrivial semidirect product. 9. Show that the symmetric group S 4 of degree 4 is isomorphic to a semidirect product of the Klein 4-group V 4 by the symmetric group S 3 of degree 3. Show that S 4 is also isomorphic to a semidirect product of the alternating group A 4 by a cyclic group of order 2. 4
5 Solution: Let N = V 4 S 4 and H = S 3 S 4. Only the identity in V 4 fixes the point 4, so N H = 1. Therefore HN =(6 4)/1 = 24 = S 4. We deduce that S 4 = HN = H N = S 3 V 4. Let M = A 4 S 4 and K = (1 2) S 4. Clearly, M K = 1, so KM =(2 12)/1 = 24 = S 4. We deduce that S 4 = KM = K M = C 2 A Let G be a group of order pq, wherep and q are primes with p<q. (a) If p does not divide q 1, show that G = C pq, thecyclicgroupoforderpq. (b) If p does divide q 1, show that there are essentially two different groups of order pq. Solution: Sylow s Theorem tells us that there is a unique Sylow q-subgroup. Thus G has a normal subgroup Q of order q. The number n p of Sylow p-subgroups divides q and is congruent to 1 modulo p. Hence if n p = 1, then n p = q and then q 1(modp), that is, p (q 1). (a) If p (q 1), then there is a normal Sylow p-subgroup P G, withq G, P Q = 1 and PQ = pq = G. ThusG = PQ = P Q = C p C q = Cpq. (b) Now suppose that p (q 1). Let P be any Sylow p-subgroup of G. (In this case, we cannot deduce it is unique.) Then P Q = 1, Q G and PQ = pq = G, so G = PQ. Therefore G = P Q and so the structure of G is determined by the conjugation action of P on Q (since the structures of Q = C q and P = C p are determined). Note that Aut Q = C q 1 by Question 5(a). Consider the homomorphism φ: P Aut Q determined by the conjugation of elements of Q by elements of P. If the image of φ is trivial, then choose a generator x for Q and a generator y for P.Then y 1 xy = x yφ = x (since yφ is the identity). Hence G has presentation x, y x q = y p =1,xy= yx = x x q =1 y y p =1 = C q C p = Cpq. If im φ = 1, then by the First Isomorphism Theorem, im φ = C p. Now Aut Q = C q 1 has a unique subgroup of order p (as p (q 1)) and this must be im φ. Fix a generator α for this subgroup and fix a generator x for Q. Now φ is a bijection from P to im φ = α, so we can choose a generator y for P such that yφ = α. Then we deduce that G has presentation x, y x q = y p =1,y 1 xy = x α. This group exists (it is the semidirect product C p φ C q where φ maps the generator of C p to the automorphism α) and is not isomorphic to C pq as α = 1. Hence if p (q 1), there are two different groups of order pq up to isomorphism. 11. Classify the groups of order 52 up to isomorphism. 5
6 Solution: Let G be a group of order 52. Let n 13 denote the number of Sylow 13-subgroups of G. Sylow s Theorem tells us that n 13 1 (mod 13) and n Hence n 13 = 1, so G has a normal Sylow 13-subgroup P = C 13. Choose x such that P = x. Let T be a Sylow 2-subgroup of G. Then P T = 1 by Lagrange s Theorem, so TP = T P / P T = 52. Hence G = TP, P G and P T = 1, so G = T φ P with respect to some φ: T Aut P. Now Aut P = C 12. Consider the automorphism α: x x 2. We calculate xα 2 =(xα)α =(x 2 )α =(xα) 2 =(x 2 ) 2 = x 4 xα 3 =(xα 2 )α =(x 4 )α =(xα) 4 =(x 2 ) 4 = x 8 xα 4 =(xα 3 )α =(x 8 )α =(xα) 8 =(x 2 ) 8 = x 16 = x 3 xα 6 =(xα 4 )α 2 =(x 3 )α 2 =(xα 2 ) 3 =(x 4 ) 3 = x 12 = x 1 xα 12 =(xα 6 )α 6 =(x 1 )α 6 =(xα 6 ) 1 =(x 1 ) 1 = x. Hence α 12 =id,butα, α 2, α 3, α 4 and α 6 are non-identity. Thus, α is an element of order 12, so Aut P = α. In particular, it follows that α 3 is the unique subgroup of order 4 in Aut P. To determine the possibilities for φ, note that T = C 4 or T = C 2 C 2. Case 1: T = C 4. Now Tφis a subgroup of Aut P and is isomorphic to a quotient of T.Then Tφ = 1, 2 or 4, so Tφ = 1, α 6 or α 3. If Tφ = 1, lety be any generator of T,soyφ = id. Hence G = x, y x 13 = y 4 =1,y 1 xy = x = x, y x 13 = y 4 =1,xy= yx = C 13 C 4 = C 52. (10) If Tφ = α 6,thenkerφ must be the unique subgroup of order 2 in T. Choose y T such that yφ = α 6.Theny must be an element of order 4 (as y ker φ), so T = y. Therefore G = x, y x 13 = y 4 =1,y 1 xy = x 1, (11) since xα 6 = x 1. If Tφ = α 3,thenφ induces an isomorphism between T and α 3. We may pick a generator y for T satisfying yφ = α 3.Then since xα 3 = x 8. G = x, y x 13 = y 4 =1,y 1 xy = x 8, (12) There are therefore at most three groups of order 52 with Sylow 2-subgroup cyclic. Note that group (10) is abelian while (11) and (12) are not. Furthermore in group (11), the centraliser C G (P ) of the unique Sylow 13-subgroup has order 26 (it contains P and ker φ), while in (12) it has order 13 (as ker φ = 1). Hence these three groups are non-isomorphic. Case 2: T = C 2 C 2. Now Tφ is a subgroup of the cyclic group Aut P,somustbecyclic.Sinceg 2 = 1 for all g T, we deduce that either Tφ = 1 or Tφ = C 2,i.e.,T = α 6. 6
7 If Tφ = 1, thenlety and z be two generators for T,so G = x, y, z x 13 = y 2 = z 2 =1,yz= zy, y 1 xy = x, z 1 xz = x = x x 13 =1 y y 2 =1 z z 2 =1 = C 13 C 2 C 2 = C 2 C 26. (13) If Tφ = α 6,then ker φ = 2. Choose an element y T such that yφ = α 6 and a generator z for ker φ. ThenT = y, z since y ker φ. Therefore G = x, y, z x 13 = y 2 = z 2 =1,yz= zy, y 1 xy = x 1,z 1 xz = x = x, y x 13 = y 2 =1,y 1 xy = x 1 z z 2 =1 = D 2 13 C 2 = D2 26. (14) These two groups are not isomorphic since (13) is abelian while (14) is not. Finally, none of these groups in Case 2 are isomorphic to those in Case 1, since the Sylow 2-subgroups are not isomorphic. We conclude that there are five groups of order 52 up to isomorphism. We have not yet justified the claim that D 2 13 C 2 = D2 26. Consider D 2 26 = α, β α 26 = β 2 =1,β 1 αβ = α 1. Now α has order 26, so α = α 13 α 2 (this simply being the result from Q 3 that C 26 = C2 C 13 ). Now α 13 is the unique element of order 2 in α, so while β 1 α 13 β =(β 1 αβ) 13 =(α 1 ) 13 = α 13 = α 13, β 1 α 2 β =(β 1 αβ) 2 =(α 1 ) 2 =(α 2 ) 1. It follows that α 2,β = D 2 13 and α 2,β D 2 26 (as it has order 2), while α 13 D 2 26 (since α 13 commutes with both α and β). Therefore D 2 26 = α, β = α 13 α 2,β = α 13 α 2,β = C 2 D [A similar argument shows that D 2 2m = C2 D 2m whenever m is an odd number, since C 2m = C2 C m.] 12. Show that a group of order 30 is isomorphic to one of C 30, C 3 D 10, D 6 C 5, D 30. Solution: Let G be a group of order 30. Let n 3 and n 5 denote the number of Sylow 3- and Sylow 5-subgroups, respectively. Sylow s Theorem tells us that so n 3 = 1 or 10. Equally n 3 1 (mod 3) and n 3 10, n 5 1 (mod 5) and n 5 6, so n 5 = 1 or 6. If n 3 = 10, then the Sylow 3-subgroups account for 2 10 = 20 elements of order 3, while if n 5 = 6, then the Sylow 5-subgroups account for 4 6 = 24 elements of order 5. Since G = 30, at least one of n 3 and n 5 equals 1. 7
8 Thus G has a normal Sylow p-subgroup P for p = 3 or 5. Consider the quotient group G/P. This has order 10 or 6 (respectively according to p) and Sylow s Theorem applied again shows that G/P has a normal Sylow q-subgroup for q = 5 or 3 (according to the value of p). Hence, by the Correspondence Theorem, G has a normal subgroup K of order 15 (namely the one that corresponds to this normal subgroup of the quotient group). Apply Question 10(a). This shows that K = C 15. Let H be a Sylow 2-subgroup of G, soh = C 2. Now H K = 1 and so HK = ( H K )/ H K = 30. Hence we deduce G = H K and the structure of G is determined by the action of H on K by conjugation. Now K = C 15 = C3 C 5, so we may choose elements x and y in K of orders 3 and 5 respectively such that K = x y = x, y. Now x is the unique Sylow 3-subgroup of G and hence any automorphism of K must map this subgroup back to itself. The same applies to the effect of any automorphism of K on the subgroup y. Hence if α Aut K, then we induce an automorphism α 1 of x by restriction and similarly we induce an automorphism α 2 of y. Let z be the generator for H. (In this case, H = 2, so there is no choice!) If Hφ = 1, the action of z on K by conjugation is now determined: z 1 az = a where a is a generator for K, because zφ is the identity map. Hence in this case G has presentation a, z a 15 = z 2 =1,az= za = a a 15 =1 z z 2 =1 = C 15 C 2 = C30. Suppose then that Hφ = 1. ThenHφ is cyclic of order 2 and the generator z of H is mapped to to an element α of order 2. The induced automorphisms α 1 and α 2 of x and y above then have order 1 or 2. Since α 1 Aut C 3 = C2 and α 2 Aut C 5 = C4, each of these automorphisms are either the identity map or the unique automorphism of order two, namely the inversion map. Hence we have three further possibilities: (a) α 1 is inversion, α 2 is the identity: so z 1 xz = x 1 and z 1 yz = y. Hence G has presentation x, y, z x 3 = y 5 = z 2 =1,xy= yx, z 1 xz = x 1,yz= zy (b) α 1 is the identity, α 2 is inversion: so G has presentation = x, z x 3 = z 2 =1,z 1 xz = x 1 y y 5 =1 = D 2 3 C 5 = S 3 C 5. x, y, z x 3 = y 5 = z 2 =1,xy= yx, xz = zx, z 1 yz = y 1 = y, z y 5 = z 2 =1,z 1 yz = y 1 x x 3 =1 = D 2 5 C 3. (c) α 1 and α 2 are both inversion. In this case, a = xy is an element of order 15 which generates K. Furthermore z 1 az = z 1 xz z 1 yz = x 1 y 1 = a 1,so G has presentation a, z a 15 = z 2 =1,z 1 az = a 1 = D Classify the groups of order 98 up to isomorphism. 8
9 Solution: By Sylow s Theorem, if G = 98, then G has a normal subgroup S of order 49. If T is a Sylow 2-subgroup (so T = 2), then S T = 1 and TS = G, so G = T S. We now divide according to the structure of S. Case 1: S = C 49. Let x be a generator for S. We need to consider φ: T Aut S. IfTφ = 1, thenthe (unique) generator y for T satisfies y 1 xy = x and so G has presentation x, y x 49 = y 2 =1,xy= yx = x x 49 =1 y y 2 =1 = C 49 C 2 = C98. Otherwise, yφ = α where α is an element of Aut S of order 2. Such an α is determined by its effect on x. If xα = x m,thenx = xα 2 =(x m )α =(xα) m =(x m ) m = x m2. Hence x m2 1 = x, som 2 1=(m 1)(m + 1) is divisible by 49. Hence 49 divides m 1 or m + 1. (It is not possible for 7 to divide both!) We may assume that 0 m 48, so m = 1 or m = 48. If m =1thenα is the identity, which does not have order 2. If m = 48 then xα = x 48 = x 1, and we do indeed have an element of order 2. Then y 1 xy = xα = x 1,soG has presentation Case 2: S = C 7 C 7. x, y x 49 = y 2 =1,y 1 xy = x 1 = D 98. We can view S as a vector space of dimension 2 over the field F 7 of order 7. An automorphism of S corresponds to an invertible linear transformation of the vector space. Consider the generator y for T and its image yφ = α GL 2 (F 7 ). Then α is a linear transformation satisfying α 2 = I (the 2 2 identity matrix). Hence X 2 1 is a polynomial satisfied by α, so the minimal polynomial m(x) of α divides X 2 1=(X 1)(X + 1). There are three possibilities for m(x): m(x) =X 1, X +1, or (X 1)(X + 1). If α = I, we may pick any basis (i.e., generating set) for S, sayu and v, and see that uα = u and vα = v. This means that the conjugation action of y on S is given by y 1 uy = u and y 1 vy = v. Hence G has presentation u, v, y u 7 = v 7 = y 2 =1,uv= vu, uy = yu, vy = yv = u u 7 =1 v v 7 =1 y y 2 =1 = C 7 C 7 C 2 = C7 C 14. If m(x) =X + 1, then α = I, so in additive notation, uα = u and vα = v for a basis {u, v} for S. Switching back to multiplicative notation, this means that y 1 uy = u 1 and y 1 vy = v 1. Hence G has presentation u, v, y u 7 = v 7 = y 2 =1,uv= vu, y 1 uy = u 1,y 1 vy = v 1. If m(x) =(X 1)(X +1) then our transformation α is diagonalisable (as the minimal polynomial is a product of distinct linear factors) so we may pick a basis {u, v} for S of eigenvectors. Thus uα = u and vα = v. (The corresponding eigenvalues are 1 and 1.) In multiplicative notation, y 1 uy = u and y 1 vy = v 1. Hence G has presentation u, v, y u 7 = v 7 = y 2 =1,uv= vu, uy = yu, y 1 vy = v 1 = u u 7 =1 v, y y 7 = y 2 =1,y 1 vy = v 1 = C 7 D
10 We to show that the two non-abelian groups we have constructed are not isomorphic. The last group has a subgroup of order 7 lying in the centre of the group (namely, the C 7 direct factor), whereas the previous group has no elements in the Sylow 7-subgroup which commute with y (all elements in the Sylow 7-subgroup are inverted by y), so has no non-trivial central elements. Hence up to isomorphism there are five groups of order 98: C 98 and C 7 C 14 are abelian, while D 2 49, C 7 D 2 7 and C 2 (C 7 C 7 ) are non-abelian. In the last group, C 2 acts by inversion on the (normal) Sylow 7-subgroup. 14. Classify the groups of order 117 up to isomorphism. Solution: Let G be a group of order 117 = By Sylow s Theorem, the number of Sylow 13-subgroups equals 1. Let T be the unique Sylow 13-subgroup, so T G, T = C 13 and Aut T = C 12. Let x be a generator for T. Then any automorphism has the form x x i for some i with 1 i 12. Consider the automorphism α: x x 2. We calculate that and therefore xα 2 =(x 2 )α =(xα) 2 =(x 2 ) 2 = x 4 xα 3 =(x 4 )α =(xα) 4 =(x 2 ) 4 = x 8 xα 4 =(x 8 )α =(xα) 8 =(x 2 ) 8 = x 16 = x 3 xα 6 =(x 3 )α 2 =(xα 2 ) 3 =(x 4 ) 3 = x 12 = x 1 xα 12 =(x 1 )α 6 =(x 1 ) 1 = x. Hence α 12 = 1 and α, α 2, α 3, α 4 and α 6 are not the identity map. This shows that α has order 12 and consequently Aut T = α. Let Θ be a Sylow 3-subgroup of G. ThenT Θ=1, as they have coprime order, so ΘT = Θ T = G and G =ΘT. Therefore G = Θ φ T for some φ: Θ Aut T. We now divide according to the structure of Θ. Case 1: Θ = C 9. If φ: Θ Aut T is a homomorphism, then im φ Aut T and im φ 9. Therefore im φ = 1 or α 4, the latter being the unique subgroup of order 3 in Aut T. If im φ = 1, lety be a generator for Θ. Then yφ = 1, so y 1 xy = x in G. Therefore G has presentation G = x, y x 13 = y 9 =1,xy= yx = C 13 C 9 = C117. If im φ = α 4, then choose y Θ such that yφ = α 4. Necessarily y generates Θ, since it does not lie in the kernel of φ (which is the unique subgroup of Θ of order 3). Then y 1 xy = xα 4 = x 3,soG has presentation G = x, y x 13 = y 9 =1,y 1 xy = x 3. Finally, note this is a non-abelian group, so is not isomorphic to C 117. Case 2: Θ = C 3 C 3. Again there are two possibilities for the image of a homomorphism φ: Θ Aut T, namely im φ = 1 or im φ = α 4. 10
11 If im φ = 1, take y and z to generate Θ. Then G has presentation G = x, y, z x 13 = y 3 = z 3 =1,yz= zy, xy = yx, xz = zx = C 13 C 3 C 3 = C3 C 39. If im φ = α 4, choose y Θ such that yφ = α 4 and choose z to generate ker φ (which is a subgroup of Θ of order 3). Then Θ = y, z and G has presentation G = x, y, z x 13 = y 3 = z 3 =1,yz= zy, y 1 xy = x 3,xz= zx = x, y x 13 = y 3 =1,y 1 xy = x 3 C 3. This last group is non-abelian, so is not isomorphic to C 3 C 39. No group in Case 1 is isomorphic to one in Case 2, as their Sylow 3-subgroups are different. In conclusion, there are four groups of order 117 up to isomorphism. 11
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