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Unformatted text preview: MAT 322: Algebra with Galois Theory Professor Claus Sorensen Problem Set 5 Rik Sengupta October 23, 2009 1. Let F 4 be a field with four elements { , 1 ,x,y } . Write down the tables for addition and multiplication, hence showing F 4 is unique. Solution. The tables are as follows: + 1 x y 1 x y 1 1 y x x x y 1 y y x 1 and 1 x y 1 1 x y x x y 1 y y 1 x Those are the addition and multiplication tables, and they serve an im portant purpose. They show that the operations on the field are fixed if the field axioms are assumed, and so the field is essentially unique. Hence, they prove welldefinedness and uniqueness. 2. Let D n be the dihedral group of order 2 n , where n > 2. Pick generators r and s such that r n = e , s 2 = e , and rs = sr 1 . Write n as 2 m k , where k is odd. For each i m , show that Z i ( D n ) is generated by r n/ 2 i . Moreover, verify that Z i ( D n ) = Z m ( D n ) for every i m . Solution. When i = 0, Z ( D n ) = { e } by definition, and r n = e , hence it trivially follows. We consider the case i = 1. Then, for x D n , write x = r a s b . So now, xs = r a s b +1 and sx = sr a s b = r a s b +1 , so that sx = xs iff r a = r a , which is true iff n  2 a . Now, rx = r a +1 s b , and xr = r a s b r , which equals r a +1 if b = 0 and r a 1 s if b = 1. Hence, rx = xr iff b = 0 or b = 1 and n  2. Since n 3, we have that x Z ( D n ) iff x = r a where n  2 a . 1 Now, n  2 a = n 2  a . But 0 a < n , so that a = n 2 or a = 0, and it follows immediately that Z ( D n ) = { e,r n 2 } = h r n 2 i . Now, suppose we have shown that Z i ( D n ) = h r n 2 i i for some 0 i m 1. Once again, let x = r a s b . Then, we have [ s,x ] = sr a sr a = r 2 a . On the other hand, [ x,s ] = r a s b ss b r a s = r 2 a . Note that [ x,r ] = r a s b rs b r a r 1 , which equals e when b = 0 and r 2 when b = 1. On the other hand, [ r,x ] = rr a s b r 1 s b r a , which equals e when b = 0 and r 2 when b = 1. Hence, we easily get [ x,y ] Z i ( D n ) for all y D n iff n 2 i  2 a and b = 0, or when b = 1 and n 2 i  2. It follows immediately that x Z i +1 ( D n ) iff x = r a where n 2 i +1  a . Thus, we get Z i +1 ( D n ) = h r n 2 i +1 i , and hence, by an inductive argument, the problem is solved. For the second part, note that Z m ( D n ) = h r n 2 m i = h r k i . Once again, setting x = r a s b and using the same commutators from our calculations above, we see that x Z m +1 ( D n ) iff k  2 a and b = 0, or when b = 1 and k  2. Since k > 1 is odd, we have x = r a where k  a . Since 0 a < 2 m k , it follows immediately that Z m +1 ( D n ) = h r k i = Z m ( D n ). As for the trivial final step, since Z m +2 ( D n ) is defined recursively from Z m +1 ( D n ) = Z m ( D n ), it follows that Z m +2 ( D n ) = h r k i = Z m ( D n )....
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 '09
 CLAUSSORENSEN
 Algebra, Addition, Multiplication

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