By Randall R. Holmes

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Define ∗ on G by x ∗ y = x + y + xy. 37 Prove that (G, ∗) is a group. ) 4–4 Let G be a group and let a, b ∈ G. Prove that there exist elements x and y of G such that ax = b and ya = b. 4–5 Let G be a finite group. Prove that each element of G appears precisely once in each row and each column of the binary table of G. 6) to show that an element of G can appear at most once in a given row of the table. Then use the fact that each row has as many entries as there are elements in the group to argue that every element of the group appears in each row.

Solution The set 1 consists of all multiples of 1, so Z = 1 . Therefore, Z is cyclic and 1 is a generator. For any integer n, the set n consists of all multiples of n, so n = Z if and only if n = ±1. Therefore, ±1 are the only generators of Z. 2 Example Let n ∈ N. Prove that the group Zn (under addition modulo n) is cyclic. Solution We claim that Zn = 1 . 5, we have 1 = {0, 1, (2)1, (3)1, . . , (n − 1)1} = {0, 1, 2, 3, . . , n − 1} = Zn as claimed. Therefore, Zn is cyclic. 4 we will see that Z and Zn (n ∈ N) are essentially the only cyclic groups in the sense that any cyclic group is isomorphic to one of these.

X−2 , x−1 , e, x, x2 , x3 , . . } and the elements xi , i ∈ Z, are all distinct. (ii) If ord(x) = n, then x = {e, x, x2 , x3 , . . , xn−1 } and the elements xi , 0 ≤ i < n, are all distinct. (iii) | x | = ord(x). Proof. (i) Assume that ord(x) = ∞. By definition, x consists of all powers xi with i ∈ Z, so the equation follows once we observe that x0 = e and x1 = x. ) Suppose that xi = xj with i ≤ j. Then xj−i = xj x−i = xj (xi )−1 = xj (xj )−1 = e. Now j − i is a nonnegative integer and, since ord(x) = ∞, this number cannot be positive.

### Abstract Algebra I by Randall R. Holmes

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