10 CHAPTER 1 (1.5) LEMMA. Let Gbea group. Then, for a,b € G, there exist unique elements JC, y € G such that ax = b and ya = b. In particular, the element e is unique, and for each x € G, the element y of Definition 1.4(H) is unique. Proof. Choose z such that az = e = za. Now a(zb) = eb = b, and so we can take x = z£. For uniqueness, if ax = ax\ we have * = ex = zajc = zax' = ex' = JC7, as required. The existence and uniqueness of y are proved similarly. • In a permutation group, the unique element satisfying condition (ii) of Definition 1.4 is, of course, the identity map /. By analogy, this special element in an abstract group is called the identity element of the group and it is customarily denoted 1. The reader should note that the identity of a permutation group is defined by what it is (a particular mapping), whereas the identity of an abstract group is defined by how it behaves with respect to the group operation. Similarly, the element y of a permutation group that satisfies condition (ii) with respect to x is the inverse map, JC-1, and by analogy, in an abstract group, y is said to be the inverse element of x and the notation x~l is used in this case, too. In fact, the conditions of Definition 1.4 are more stringent than they really need to be. (1.6) THEOREM. Let Gbea set with an associative multiplication and suppose there exists e e G with the following properties: i. xe = x for all x € G and ii. for each x € G, there exists y € G with xy = e. Then G is a group. Proof. Let x e G and choose y according to property (ii). It suffices to show that ex = x and yx = e. Use property (ii) to find z € G with yz = e. We have x = xe = x(yz) = {xy)z = ez, and so yx = y(ez) = (ye)z = yz = e9 as required. Now ex = (xy)x = x(yx) = xe = x , and the proof is complete. •

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