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# all2 - Math 5125 Monday October 3 Monday October 3 Exercise...

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Math 5125 Monday, October 3 Monday, October 3 Exercise 6.3.7 on page 220 Prove that the following is a presentation for the quaternion group of order 8: Q 8 = a , b | a 2 = b 2 , a - 1 ba = b - 1 . Q 8 is a group with 8 elements, namely { 1 , - 1 , ± i , ± j , ± k } with identity 1, center { 1 , - 1 } , i 2 = j 2 = k 2 = - 1, i j = k , jk = i , ki = j . Let F denote the free group on { a , b } . Let us map F to Q 8 by sending a to i and b to j . Since Q 8 is generated by { i , j } , we obtain an epimorphism θ : F Q 8 such that θ a = i and θ b = j . Let K = ker θ . Now θ ( a 2 ) = i 2 = - 1 = j 2 = θ ( b 2 ) and θ ( a - 1 ba ) = i - 1 ji = - i ji = - ki = - j = j - 1 = θ ( b - 1 ) , so if R is the normal subgroup generated by a 2 b - 2 , a - 1 bab , we see that R K . Since F / K = Q 8 , we see that | F / K | = 8, so we will be done if we can show that | F / R | < 16 (because by Lagrange | F / R | has to be a multiple of | F / K | = 8, and | F / R | = 8 if and only if R = K ). Since a - 1 ba = b - 1 mod R , we see that a - 1 b 2 a = b - 2 mod R and hence a - 1 a 2 a = b - 2 = a - 2 mod R . We deduce that a 4 = b 4 = 1 mod R . Using ba = ab - 1 , we now see that every element of F can be written in the form a i b j mod R where 0 i , j 3. But a 2 b 2 = 1 mod R , consequently | F / R | ≤ 15. Therefore K is equal to the normal subgroup of F generated by { a 2 b - 2 , a - 1 bab } , and the result follows. Exercise 7.3.7 on page 248 Let R = { a b c d | a , b , d Z } be the subring of M 2 ( Z ) of upper triangular matrices. Prove that the map φ : R Z × Z defined by φ : a b 0 d ( a , d ) is a surjective homomorphism and describe its kernel. Let x 1 = a 1 b 1 0 c 1 , x 2 = a 2 b 2 0 c 2 . Then φ ( x 1 + x 2 ) = ( a 1 + a 2 , d 1 + d 2 ) and φ ( x 1 )+ φ ( x 2 ) = ( a 1 , d 1 )+( a 2 , d 2 ) , so φ ( x 1 + x 2 ) = φ ( x 1 ) + φ ( x 2 ) which shows that φ respects addition. Also φ ( x 1 x 2 ) = ( a 1 a 2 , d 1 , d 2 ) and φ ( x 1 ) φ ( x 2 ) = ( a 1 , d 1 )( a 2 , d 2 ) , so φ ( x 1 x 2 ) = φ ( x 1 ) φ ( x 2 ) and thus φ also respects multipli- cation. This establishes that φ is a ring homomorphism. It is clearly onto, and its kernel consists of matrices of the form 0 b 0 0 where b is an arbitrary element of Z . Exercise 7.3.34 on page 250 Let I and J be ideals of the ring R with a 1.

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(a) Prove that I + J is the smallest ideal of R containing both I and J . (b) Prove that IJ is an ideal contained in I J . (c) Give an example where IJ = I J . (d) Prove that if R is commutative and if I + J = R , then IJ = I J . (a) Obviously I + J is an ideal containing both I and J . Suppose K is an ideal containing both I and J . Since K is closed under addition, it must contain I + J . Therefore I + J is the smallest ideal containing I and J . (b) Since I R , we have IJ IR I . Similarly IJ J and it follows that IJ I J . (c) Let R = Z and I = J = 2 Z . Then IJ = 4 Z and I J = 2 Z . (d) In view of (b), we need to prove that if x I J , then x IJ . Since R = I + J , we may write 1 = i + j where i I and j J . Then x = ix + x j IJ + IJ = IJ as required. Exercise 7.4.4 on page 256 Assume that R is commutative. Prove that R is a field if and only if 0 is a maximal ideal.
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