please send handwritten solution for part d , e only handwritten solution accepted We are now ready to prove Theorem 1:Proof of Theorem 1. We know that an element kEZn belongs to Z if and onlyif k and 2" are coprime, that is, if and only ifk is odd. In particular, the elementsa := 2"-1 –1 and b := 2"-1+ 1belong to Z, and we havea² = (2"-I - T)² :2n x2n-2+ T = T,%3D%3D2nand(2"-1+1)2:+ ī = T.2n x2n-:2nThus, Z contains at least two distinct elements of order 2. Since cyclic groupscontain at most one element of order 2 by Proposition 2, this implies that Z isnot cyclic.Comprehension questions:(a) Following the same approach as in the proof of Proposition 2, show thatfor an integer m 1, the additive group Zm either contains no element oforder 3 or contains exactly two elements of order 3.(b) Explain how Proposition 3 can be used to show that the multiplicativegroup Z is not cyclic.(c) In the proof of Proposition 3, the following statement is not justified:"However, since r is non-trivial, f(x) cannot have order 1."Properly justify this statement.(d) Show that Theorem 1 does not hold for = 1 and n 2. That is, show thatthe multiplicative groups Z and Z are cyclic.(e) The proof of Theorem 1 is thus currently incomplete, as it does not explicitlyuse the hypothesis that n 3. In what parts of the proof is that assumptionimplicitly used? We are now ready to prove Theorem 1:Proof of Theorem 1. We know that an element kEZn belongs to Z if and onlyif k and 2" are coprime, that is, if and only ifk is odd. In particular, the elementsa := 2"-1 –1 and b := 2"-1+ 1belong to Z, and we havea² = (2"-I - T)² :2n x2n-2+ T = T,%3D%3D2nand(2"-1+1)2:+ ī = T.2n x2n-:2nThus, Z contains at least two distinct elements of order 2. Since cyclic groupscontain at most one element of order 2 by Proposition 2, this implies that Z isnot cyclic.Comprehension questions:(a) Following the same approach as in the proof of Proposition 2, show thatfor an integer m 1, the additive group Zm either contains no element oforder 3 or contains exactly two elements of order 3.(b) Explain how Proposition 3 can be used to show that the multiplicativegroup Z is not cyclic.(c) In the proof of Proposition 3, the following statement is not justified:"However, since r is non-trivial, f(x) cannot have order 1."Properly justify this statement.(d) Show that Theorem 1 does not hold for = 1 and n 2. That is, show thatthe multiplicative groups Z and Z are cyclic.(e) The proof of Theorem 1 is thus currently incomplete, as it does not explicitlyuse the hypothesis that n 3. In what parts of the proof is that assumptionimplicitly used?

Question

please send handwritten solution for part d , e only handwritten solution accepted We are now ready to prove Theorem 1:Proof of Theorem 1. We know that an element kEZn belongs to Z if and onlyif k and 2&#34; are coprime, that is, if and only ifk is odd. In particular, the elementsa := 2&#34;-1 –1 and b := 2&#34;-1+ 1belong to Z, and we havea² = (2&#34;-I - T)² :2n x2n-2+ T = T,%3D%3D2nand(2&#34;-1+1)2:+ ī = T.2n x2n-:2nThus, Z contains at least two distinct elements of order 2. Since cyclic groupscontain at most one element of order 2 by Proposition 2, this implies that Z isnot cyclic.Comprehension questions:(a) Following the same approach as in the proof of Proposition 2, show thatfor an integer m > 1, the additive group Zm either contains no element oforder 3 or contains exactly two elements of order 3.(b) Explain how Proposition 3 can be used to show that the multiplicativegroup Z is not cyclic.(c) In the proof of Proposition 3, the following statement is not justified:&#34;However, since r is non-trivial, f(x) cannot have order 1.&#34;Properly justify this statement.(d) Show that Theorem 1 does not hold for = 1 and n 2. That is, show thatthe multiplicative groups Z and Z are cyclic.(e) The proof of Theorem 1 is thus currently incomplete, as it does not explicitlyuse the hypothesis that n > 3. In what parts of the proof is that assumptionimplicitly used? We are now ready to prove Theorem 1:Proof of Theorem 1. We know that an element kEZn belongs to Z if and onlyif k and 2&#34; are coprime, that is, if and only ifk is odd. In particular, the elementsa := 2&#34;-1 –1 and b := 2&#34;-1+ 1belong to Z, and we havea² = (2&#34;-I - T)² :2n x2n-2+ T = T,%3D%3D2nand(2&#34;-1+1)2:+ ī = T.2n x2n-:2nThus, Z contains at least two distinct elements of order 2. Since cyclic groupscontain at most one element of order 2 by Proposition 2, this implies that Z isnot cyclic.Comprehension questions:(a) Following the same approach as in the proof of Proposition 2, show thatfor an integer m > 1, the additive group Zm either contains no element oforder 3 or contains exactly two elements of order 3.(b) Explain how Proposition 3 can be used to show that the multiplicativegroup Z is not cyclic.(c) In the proof of Proposition 3, the following statement is not justified:&#34;However, since r is non-trivial, f(x) cannot have order 1.&#34;Properly justify this statement.(d) Show that Theorem 1 does not hold for = 1 and n 2. That is, show thatthe multiplicative groups Z and Z are cyclic.(e) The proof of Theorem 1 is thus currently incomplete, as it does not explicitlyuse the hypothesis that n > 3. In what parts of the proof is that assumptionimplicitly used?

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Answered: Image /qna-images/answer/08cef5be-5394-4af7-8dda-8925877afa39.jpg

(d) Show that Theorem 1 does not hold for n 1 and n = 2. That is, show that the multiplicative groups Z and Z are cyclic. (e) The proof of Theorem 1 is thus currently incomplete, as it does not explicitly use the hypothesis that n > 3. In what parts of the proof is that assumption implicitly used?