model4 - e . 10. Suppose that for every a G , a = a-1 ....

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MODEL ANSWERS TO THE FOURTH HOMEWORK 1. (a) No, this is not a group. The rule for multiplication is not associative. For example, (1 - 0) - 1 = 0 but 1 - (0 - 1) = 2 . Also there is no identity. 0 is a right identity, as a - 0= a , but 0 - 1= - 1 ± = 1, so that 0 is not a left identity. (b) No, 2 does not have an inverse. (c) Yes, this is a group. Associativity, follows from associativity for adding natural numbers. a 0 plays the role of the identity. a 7 - i is the inverse of a i . (d) Yes, this is a group. Indeed it is a non-empty subset of Q , which is closed under mutliplication and inverses. 3. Let a and b G . Expanding ( ab ) 2 = a 2 b 2 , we get abab = aabb. Thus a ( ba ) b = a ( ab ) b. Multiplying on the left by a - 1 and on the right by b - 1 we get ba = ab. 7. e , (1 , 2), (2 , 3) and (1 , 3) satisfy x 2 = e . e , (1 , 2 , 3), (1 , 3 , 2) satisfy y 3 =
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Unformatted text preview: e . 10. Suppose that for every a G , a = a-1 . Then a 2 = e . Let a and b G , and set c = ab . Then e = c 2 = ( ab ) 2 and a 2 b 2 = e e = e. Thus the equation ( ab ) 2 = a 2 b 2 , is satised for somewhat vacuous reasons. Now apply (3). 4. Suppose that ( ab ) i = a i b i and ( ab ) i +1 = a i +1 b i +1 . Expanding the second equality, we get a ( ba ) i b = aa i b i b. 1 Multiplying on the left by a-1 and by b-1 on the right, we get ( ba ) i = a i b i = ( ab ) i . Hence we have c = ( ba ) i = ( ab ) i and ( ba ) i +1 = ( ab ) i +1 . Multiplying both sides of the second equation by c-1 , we get ab = ba, which is what we want. 2...
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model4 - e . 10. Suppose that for every a G , a = a-1 ....

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