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# Abstract Algebra I MATH 455

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This 3 page Class Notes was uploaded by John MacGyver on Monday October 26, 2015. The Class Notes belongs to MATH 455 at University of Tennessee - Knoxville taught by Staff in Fall. Since its upload, it has received 29 views. For similar materials see /class/229829/math-455-university-of-tennessee-knoxville in Mathematics (M) at University of Tennessee - Knoxville.

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Date Created: 10/26/15

Groups of Order 18 Math 455 7 Fall 2006 Question Classify all groups of order 18 Solution Let C 18 2 32 and 5 denote the number of Sylow p subgroups Then by the Sylow Third Theorem 53 2 and so 53 6 12 and 53 E 1 mod 3 Thus 53 1 and we have only one subgroup of order 9 say H Since 9 by Corollary 6114 H 2 C9 or H 2 C3 gtlt C3 Now 52 6 139 since 52 9 We split the problem in cases Case 1 Assume that 52 1 Then if K is the only subgroup of order 2 we have that K lt1 G since 52 1 H K 1 since the groups have relatively prime orders and H So by Proposition 286 we have that C E H gtlt K Therefore G203gtltOggtlt02203gtlt06 H g 03 X 03 01 G g 09 X 02 g 018 if H 2 C9 Case 2 Assume that 52 9 This means that there are 9 elements of order 2 one in each Sylow 2 subgroup Since H already has 9 elements this means that every element not in H has order 2 Let y E C 7 H Hence y has order 2 Then G H U Hy since G H 2 or using Proposition 286 again If h E H then by has order 2 since it is not in H ie by2 1 i hyhy 1 a yhy h 1 a yh h ly i If H g C9 and H ltgt then we have that 8 CHU Hy1zx2x8yy2y the order of z is 9 the order of y is 2 and yzi z iy Thus D18 is the dihedral group of order 18 as we7ve seen in class It is completely charac terized by the properties given above If H 2 C3 gtlt C3 we can write H 11 2 12 21 where 1 and 2 have order 3 and commute with each other Hence C H U Hy 1xh i g x12x 2x1x g 2 1117 xfy 9622 9632 96196211 zzy 961x311 zfxiy and for any h E H yh h ly We have not encountered this group before but we can check that these properties indeed give us a group the properties allows us to make a multiplication table and check all the requirements Note for example that since 1 and 2 commute with each other we have ij 27139739 ijiijij712iijiiij zy zyx zyi zw z y 121 2 So every element not in H indeed has order 2 as we knew it should be the case This group is in fact the semi direct product of C3 gtlt C3 with C2 but we havent seen those Case 3 Assume nally that 52 3 We have in this case only 3 elements of order 2 The elements of H can have order 1 the identity 3 or 9 if any Therefore all the elements left G ie not of order 2 and not in H must have order 6 since their orders has to divide 18 and cannot be equal to 1 because 1 E H 2 because we7re excluding those 3 these must be in H 9 these if exist must also be in H or 18 since if we have an element ofo rder 18 C would be cyclic and hence all subgroups would be normal and we would have to have 52 1 not 3 Hence we have 9 elements of order 1 3 or 9 in H 3 elements of order 2 and 6 elements of order 6 Let y be an element of order 2 So y E C 7 H and as before C H U Hy i If H C9 let x be a generator Let7s nd what are the other two elements of order 2 besides lf y which is how every element not in H can be represented has order 2 then as before y2 1 i y y 1 i i yxiy z i yxi x iy But then y ki y i sil fiy sil fiy i simi 72iyk72i 7k71iyi fkiy7 and thus lmy kiy kiy ki ikiyy 1 Hence if y has order 2 so does fly fly etc Since we have only 3 elements of order 2 they have to be y3yz6y We can conclude that xy has order 6 since its not in H and doesnt have order 2 But then my2 has order 3 But the only elements of order 3 which must be in H are x3 and 6 and so 9613 9692961 963w 9642 or 9613 9692961 966w 9672 But my must have order 2 since xy has order 6 so my 6 y3y6y giVing us a contradiction which means that if 52 3 then H 5 C9 ii So assume that H C3 gtlt C3 Then there is some element of Hy besides y that also has order 2 Remember that y has order 2 Let x E H 7 1 be an element such that zy has order 2 Again this means that ym x ly and we can easily check that y also has order 2 Note that since z E H 7 1 it must have order 3 So the subset Sg wwaywywyz

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