prob 2 part dprovide handwritten answer for part d We are now ready to prove Theorem 1:Proof of Theorem 1. We know that an element E Zn belongs to Z, if and onlyif k and 2" are coprime, that is, if and only if k is odd. In particular, the elementsa = 2n-1I and b:= 2"-1+ 1belong to Z, and we have(2"-I - T) = 2x2n-2a? =2 + T= I,and62 = (2"-1 + T)? = 2" x2"-2 +2n + T T.Thus, 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 a is non-trivial, f(x) cannot have order 1."Properly justify this statement.(d) Show that Theorem 1 does not hold for n 1 and n 2. That is, show thatthe multiplicative groups Z and Z} are cyclic.(c) 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?%3D Problem 2. Read the following mathematical text and answer the associated comprehensionquestions at the end.The goal of this text is to prove the following theorem:Theorem 1. For n 3, the multiplicative group Z is not cyclic.We start by proving several intermediate results:Proposition 2. For m 1, the additive group Z, contains at most one elementof order 2.Proof. The proposition is true if m is odd. Indeed, a group of odd order doesnot contain any element of order 2 by Lagrange's Theorem. Thus, it is enough toconsider the case where the order of Zm is of the form m = 2l for some integere1.Let ā e Ze be an element of order 2 with a an integer such that 0

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prob 2 part dprovide handwritten answer for part d We are now ready to prove Theorem 1:Proof of Theorem 1. We know that an element E Zn belongs to Z, if and onlyif k and 2" are coprime, that is, if and only if k is odd. In particular, the elementsa = 2n-1I and b:= 2"-1+ 1belong to Z, and we have(2"-I - T) = 2x2n-2a? =2 + T= I,and62 = (2"-1 + T)? = 2" x2"-2 +2n + T T.Thus, 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 a is non-trivial, f(x) cannot have order 1."Properly justify this statement.(d) Show that Theorem 1 does not hold for n 1 and n 2. That is, show thatthe multiplicative groups Z and Z} are cyclic.(c) 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?%3D Problem 2. Read the following mathematical text and answer the associated comprehensionquestions at the end.The goal of this text is to prove the following theorem:Theorem 1. For n> 3, the multiplicative group Z is not cyclic.We start by proving several intermediate results:Proposition 2. For m > 1, the additive group Z, contains at most one elementof order 2.Proof. The proposition is true if m is odd. Indeed, a group of odd order doesnot contain any element of order 2 by Lagrange's Theorem. Thus, it is enough toconsider the case where the order of Zm is of the form m = 2l for some integere>1.Let ā e Ze be an element of order 2 with a an integer such that 0 < a < 2l – 1.We thus have a +ā = 0 in Z, i.e. 2a is divisible by 20. This implies that a isdivisible by l. Since 0 < a < 2l - 1, it follows that we either have a = 0 or a = e.Thus, we either have a = 0, which is the identity element and has order 1, ora = l, which is an element of order 2. Thus, the only element of order 2 of Zeis 7.Proposition 3. Let G be a finite cyclic group. Then G contains at most oneelement of order 2.Proof. Let m be the order of G. We know from the course that a cyclic groupof order m is isomorphic to the additive group Zm, so let us consider a groupisomorphism f: G - Zm. If G does not contain any element of order 2, thereis nothing further to do, so let us assume that G contains an element r of order2. We have a? = e in G. We thus have f(a2) = f(e) in Zm, and by properties ofgroup homomorphisms, this implies thatf(r) + f(x) = 0.Thus, f(r) is an element of Z, of order at most 2. However, since r is non-trivial,f(ax) cannot have order 1. Thus, it has order exactly 2. By Proposition 2, it thusfollows that f(r) is the unique element of Zm of order 2.Let y e G be another element of order 2 of G. The same reasoning shows thatf(y) is also the unique element of order 2 of Zm. Thus, f(x) f(y), and byinjectivity of f, it follows that r = y. Thus, G contains exactly one element oforder 2.

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(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.