please send solution for part b only handwritten solution accepted 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. Forn 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 = 20 for some integere1.Let ā e Zu be an element of order 2 with a an integer such that 0

Question

please send solution for part b only handwritten solution accepted 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. Forn> 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 = 20 for some integere>1.Let ā e Zu be an element of order 2 with a an integer such that 0 < a < 2l – 1.We thus have a +ā = 0 in Z2e, 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 = l.Thus, we either have a =0, which is the identity element and has order 1, ora = 1, which is an element of order 2. Thus, the only element of order 2 of Zuis 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 a2 = e in G. We thus have f(a2) = f(e) in Zm, and by properties ofgroup homomorphisms, this implies thatf(r) + f(x) = 0.%3DThus, f(x) is an element of Z, of order at most 2. However, since r is non-trivial,f(r) cannot have order 1. Thus, it has order exactly 2. By Proposition 2, it thusfollows that f(x) 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(r) = f(y), and byinjectivity of f, it follows that r = y. Thus, G contains exactly one element oforder 2. We are now ready to prove Theorem 1:Proof of Theorem 1. We know that an element k E Zyn 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:= 2n-1 +1belong to Z, and we havea² = (2"-I - T)² :2n x2n-2+ T = T,%3D2nand(2"-1+1)2:2" x2n-2+ ī =T.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 a 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?

Accepted Answer

The proposition 3 says, if G is a finite cyclic then G contains at most one element of order 2.

(b) Explain how Proposition 3 can be used to show that the multiplicative group Z is not cyclic.