Question

3. Consider the four groups C2 X C3 C5, S3 x C5, C3 x D5, D15. (i) Show that each of them has a unique subgroup of order 3 and a unique subgroup of order 5, and they are both normal. (ii) Identify the normal subgroups and the quotients (without further justifications). (iii) Show that the four groups are not isomorphic. (iv) Give a presentation of each of them.

Answer 1

(2) C₂ X C3 x C 2 All these fourk groupe (6) Sz x C3 ? have order as 30 = 2 X3 X5 (E) C₂ x Dg (d) Dis (1) since I lost = 30 for all four, Let us Ary to find 3-stylow to subgroups and 5-sylow to subgroups. Let n (3) & n (5) respectively denote the number of such sylow subgroupe. now n (3) E l mod 3 t n (3) / 10 ( 34 sylone a Treorem) => n() is either or 10 Suppose a (3) = 10, then we must have 1043 - 1) = 20 elements of order 3 in G which is not the case in any of the given groups above . . Hence n (3) = 1 . There exist only one sylow subgroup of order 3 and hence it must be a normal subgroup too. ( - 2 nd Sylower theorem) Similarly n(s) = 1 mod 5 & n (5) 16 = n (5) is either 1 or 6 but n(5) 76. as it forcer G to have 6 x 4 = 24 elements of order 5 which in not possible in either of a, b, c) or d e

so that n (5)=1 & hence each a), b), c) & d) have an unique subgroup of order 5 which must be normal too by Sylow'e and theorem, 1) (a) normal subgroups are C₂ X {o} x {o}, {o}x x {0}, {0} x {0} XCg and their quotients are o 3x C, XCs, c_x{}x & ex G x {0} resp. (6) normal subgroups are H, x {0} and {0} , (s and their quotients are ₂ x (g and S, X{0} respectively where H, is the unique subgroup of order 3 in Sa and He is a subgroup of order a in S (c) Normal Subgroups are. ₂ x {0} and {o} x Mg and their quotiente are {0}x Dg and C X respectively where H₂ is the unique subgroup of order 5 in Ds and He is a subgroup of order a in Ds (d) Normal Subgroups are (15 and its quotient is į where Cis is the rotation subgroup of order is in Dys and , is any reflection Subgroup in Dis of order 2 Note: Here only basic normal groups are listed we can take further direct product of the normas subgroups listed above which again is a normal sube. for a), b) & c). In d) 2,5 = C, XCg which again individuelle normal subgroup in Dis.

(ii) (a) is an abelian group wherear (b), (c) & (d) G) (a) ir an asbelian and here (c) & (d) ras (a) is not isomorphic to (6) (c) & (d) respectively, (b) has only 3 elemente of order a (2-cycler in S3) but (c) has elements of order 2 . Creflections in lg) and (d) has 15 elements of order a Creflections in Dis) Thue (b), (c) and (d) are not isomorphic to each other (GV) (a) C, X C₂ XC 9 = {(a,b,c) ) a = b & ap} (6) Sz x C₂ = {(x, a), Cy a)/ xy = y, x ² y = a=1} (2) C₂ X D s = {(2,4), (9,8) | r = rit, frares} (d) Dis - {% f) for = er nf f f = r = 1} are the presentations of G.