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by: Kavon Feest

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Introduction to Abstract Algebra MATH 113

Marketplace > University of California - Berkeley > Mathematics (M) > MATH 113 > Introduction to Abstract Algebra
Kavon Feest

GPA 3.93

K. Poirier

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K. Poirier
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This 3 page Class Notes was uploaded by Kavon Feest on Thursday October 22, 2015. The Class Notes belongs to MATH 113 at University of California - Berkeley taught by K. Poirier in Fall. Since its upload, it has received 20 views. For similar materials see /class/226591/math-113-university-of-california-berkeley in Mathematics (M) at University of California - Berkeley.

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Date Created: 10/22/15
H 2 Math 113 Section 5 Fall 2010 Homework 2 Solutions Due Monday September 13 Determine which of the following sets with 2 to 1 operations form groups For those that are groups prove that they are groups For those that are not groups describe why they are not groups Complete solutions are not given here but sketchesideas are a Z with usual addition 1 Yes b Z with usual multipication No 0 is not invertible c ZnZ where n is any positive integer with multiplication Not a group in general 6 and 7 for 0171 3A 1 are not invertible d ZnZX the set of invertible elements of ZnZ with multiplication Yes Mnxm the set of n x m matrices with entries in R with matrix addition Yes f Mnxm the set of n x m matrices with entries in R with matrix multiplication No multiplication would only be de ned in m n and then a matrix A would be invertible only if detA 7 0 g Mn the set of square nxn matrices with entries in R with matrix multiplication No see above h C a b 1 011 6 Q C R with multiplication No 0 E C but 0 is not invertible i G z E C l 2 1 where n E Zgt0 with multiplication of complex numbers Yes j Q where lowest terms No the identity would necessarily be but Q is given in lowest terms and is de ned by g 1 reduced to Let G be a group a Show that for all g E G g l 1 g 9 1 19 1 e 99 1 5 9 1 19 19 99 19 5 9 1 19 19 99 19 9 1 16 9e 5 9 1 1 b Show that for all g and h e G g h 1 11494 97171 19 1 9017759 1 969 1 99 1 e 9 th 9quot 03 and 7171971 h h lgilgh h leh hilh 5 so gh 1 h lg l c Show that ifg h 9 71 then h h gh gh gt gilgh 97171 gt eh eh gt h 71 Write down the multiplication table for the dihedral group D24 of order 8 Use it to determine the elements of order 2 D24 5 r 72 T3 3 rs e e r r2 r3 3 TS 723 3 r r r2 r3 e TS r23 r33 3 r2 r2 r3 e r 723 r s 3 rs r3 r3 e r r2 r33 3 TS 723 s s r33 723 T3 5 r3 r2 7 rs T3 3 r33 r23 7 e r3 r2 r23 723 T3 3 r33 72 r e r3 r33 r33 723 T3 3 r3 r2 r 5 Elements of order 2 are the non identity g for which the entry on the diagonal is 92 5 Therefore the elements of order 2 are 72 373 s r 3 Let D2 be the dihedral group of order 271 generated by r and s such that T represents rotation by 2quot radians and 3 represents reflection in an appropriate line 6 a Show that rs sr l Label the vertices of the n gon by 12 n Without loss of generality let 6 be the line through the vertex 1 and the center of the n gon and let 3 be reflection in 6 Then write r and s as permutations of the set of vertices r 1 2 3 and s 2n3n71 T1 WT ifnisoddands 2n3n71 T 2 L32 ifn is even You can then check explicitly that the compositions on the left and right are equal b Show that Ms 57quot for all 0 g i S n Hint use associatiVity of composition and proceed by induction using part a Ifi 0 the statement is triVial Part a 13 ST Then rls TTlTl 87717141 80717714 877139 describes the case i 1 Assume rsr l1 Tsr l1 s 771713 a Updated S4 has been changed to 53 Write down the multiplication table for the symmetric group 53 of degree 3 Next page b Write down the cycle decomposition of the element 1432 o 12 o 24 in S4 1432 o 12 o 24 23 Show that 3 is isomorphic to D23 Hint de ne a bijection of sets 4p 3 gt D23 which is a group homomorphism that is 4pzy 4pzltpy for all m y 6 3 In par 12 23 13 123 132 12 23 13 123 132 12 12 e 123 132 23 13 23 23 132 e 123 13 12 13 13 123 132 e 12 23 123 123 13 12 23 132 e 132 132 23 13 12 e 123 ticular7 you will want to let ltpe e and 4pz 1 Apm l Updated hint What you are doing here is showing that the groups are effectively the same77 You should write out the multiplication table for each group to see where to send each element of 53 by p The above hint means that multiplying and then applying p is the same as applying 4p and then multiplying Compare your multiplication table in the previous question with the multiplication table for D23 lt r slra 52 5765 372 gt below D23 5 s 723 rs r 72 e e s 723 rs r r2 s s e r2 723 rs 2 723 r2 e 7 rs 3 rs T3 7 r2 e s 723 r 7 T3 s 723 r2 e r2 r2 723 T3 3 e 7 With this ordering of elements7 it is clear that if we set M123 r and M12 s and let these generate the value of Lp on other elements requiring that 4pzy 4pz4py for all m y 6 53 then the multiplication tables look identical Therefore7 we have a bijection of sets that preserves the group structure

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