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## Advanced Linear Algebra

by: Florence Blanda

21

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# Advanced Linear Algebra MATH 481

Florence Blanda
NMSU
GPA 3.93

Patrick Morandi

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Patrick Morandi
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## Popular in Mathematics (M)

This 0 page Class Notes was uploaded by Florence Blanda on Sunday November 1, 2015. The Class Notes belongs to MATH 481 at New Mexico State University taught by Patrick Morandi in Fall. Since its upload, it has received 21 views. For similar materials see /class/233197/math-481-new-mexico-state-university in Mathematics (M) at New Mexico State University.

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Date Created: 11/01/15
The Alternating Group A is simple if n 2 5 Mathematics 481525 Fall 2004 1n this note we will prove that An is simple if n 2 5 by rst proving that A5 is simple and then giving an induction argument for An for n 2 5 This fact is a key step in Galois determination of when there is a formula for the roots of a polynomial The proof we give of the simplicity of A5 uses the idea of conjugacy classes The idea of the proof is that given a normal subgroup of A5 the subgroup is a union of some of the conjugacy classes of A5 from the normality assumption 1f we can show that no union of conjugacy classes other than 6 and A5 can be a subgroup of A5 we will have proved that A5 contains no nontrivial normal subgroup We review now some of the notions behind the idea of conjugacy 1f G is a group then G acts on itself by conjugation 1f 90 E G then the orbit of 90 under this action is the conjugacy class 0 9x971 g E We also have the isotropy group Gm g E G 990971 Note that this group is also the centralizer C90 of x it consists of the elements of G that commute with x Recall a basic fact of group actions of a nite group if G acts on a set X we can de ne the orbit 0 of an x E X and the isotropy group Gm g E G gx x and their sizes are related by the equation 1051 G So in this example of conjugation we have 1051 G for all x E G Now consider S5 1t is known that in Sn for any n the conjugacy class of an element is the set of all elements of the same cycle structure The possible cycle structures of elements of S5 are listed in the following table cycle structure representative element number of elements 5 cycle 12345 24 4 cycle 1234 30 3 cycle 123 20 2 cycle 12 10 1 cycle e 1 3 cycle 2 cycle 12345 20 2 cycle 2 cycle 1234 15 We are interested in the conjugacy classes of A5 1f 90 6 A5 then the conjugacy class of so is 99094 g 6 A5 while the conjugacy class of so in S5 is 99094 g 6 S5 So the conjugacy class of x in A5 could conceivably be smaller than its conjugacy class in S5 We now determine the conjugacy classes for A5 1n order to determine the sizes of these classes we need to determine the order of the centralizer in A5 of an element x 6 A5 We do this by making use of the corresponding information for S5 To distinguish between conjugacy classes in S5 and in A5 we will write 0555 for the conjugacy class of x in S5 and O for the conjugacy class in A5 Similarly G55 will denote the centralizer of x in S5 while we denote by C90 the centralizer of x in A5 First note that A5 consists of the 5 cycles the 3 cycles the product of 2 disjoint 2 cycles and 6 1t is clear that the conjugacy class of e is e For the product of disjoint 2 cycles consider at 1234 1n S5 we have 05551 15 120G55xl so G55l 8 We can determine G55 by producing elements that commute with x and when we have 8 we will have all of G55 By a little trial and error we obtain G55 1324 1324 Moreover C90 G55ac A5 6 1234 1324 1423 So the centralizer of x in A5 has order 4 Therefore 05 A5 605 15 Thus the entire conjugacy class of x in S5 is the conjugacy class of x in A5 Now consider at 123 We have 05551 20 so G55 6 Again by trial and error we see that G55ac 123 45 So C90 123 So 05 603 20 and again this implies that O 0555 Finally for x 12345 we have G55l S5 0555 12024 5 This forces G55ac 12345 and so C90 12345 Thus 05 605 12 Therefore the conjugacy class of 12345 consists of just 12 of the 24 5 cycles Since this argument works for any 5 cycle if we let x be any 5 cycle not in the conjugacy class of 12345 then the conjugacy class of x will consist of the other 12 5 cycles Thus the conjugacy class of a 5 cycle in S5 is the union of two conjugacy classes in A5 We have shown that there are 5 conjugacy classes of A5 and their sizes are 1 12 12 15 and 20 We can now prove the main result of this note Theorem 1 The group A5 is simple Proof Let N be a normal subgroup of A5 Then it is a union of some of the conjugacy classes of A5 However the order of N must divide 60 A short calculation will show that no union of some of these conjugacy classes that includes 6 has order a divisor of 60 unless the union is e or A5 Thus A5 is simple D From this result and induction we can prove that A is simple for each n 2 5 Theorem 2 Ifn 2 5 then A is a simple group Proof We prove this by induction on n the case n 5 is already done So suppose that n 2 6 and that An1 is simple For each i S n let G 039 E An 01 Then G is a subgroup of An isomorphic to A11 So G is simple by induction Moreover if 039 E A is any element with 039j i possible since A is a transitive subgroup of S then G 039Gj03971 Thus any two of the G are conjugate in A Let N be a normal subgroup of A Then N G e or N G G 1f N G G for some i then in fact N Gj G for all j since G is conjugate to G for each j So since N is normal if N contains G for 2 some i it contains all of them But since n 2 6 any product of two transpositions is in G for some i and any element of A is a product of such permutations So N A On the other hand if N N G e for each i then each element of N xes no integer Consequently if 039 739 E N with 0i for some i then 047039 i so 0 17 6 N G 6 This forces 039 739 So distinct elements of N never agree at any integer Consider some 039 E N and write 039 as a product of disjoint cycles say 039 01 ct with c an Ti cycle and 7 1 2 7 2 2 2 n if 7 1 2 3 say 01 i1 Let p igjk with jk i1i2i3 it is possible to do this since n 2 6 Let 739 pap 1 E N Both 039 and 739 send i1 to 1392 but 039 a 739 since 0i2 i3 and 7i2 j a contradiction So any 039 E N is a product of transpositions Now suppose 039 E N Let p lpq with pq ij k I again this is possible since n 2 6 Then if 739 pap l both 039 and 739 send 139 to j but 0k I while 7k p a contradiction So in this case we must have N 6 Thus we have proven that either N A or N e That is A is simple D

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