Lecture 5 - 5 Basic Properties of Groups Lemma 5.1 Let G be...

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5. Basic Properties of Groups Lemma 5.1. Let G be a group. (1) G contains exactly one identity element. (2) Every element of G contains exactly one inverse. (3) Let a and b be any two elements of G . Then the equation ax = b has exactly one solution in G , namely x = a - 1 b . (4) Let a and b be any two elements of G . Then the equation ya = b has exactly one solution, namely y = ba - 1 . (5) For every a G , ( a - 1 ) - 1 = a. In words the inverse of the inverse of a is a . (6) For every a and b in G , ( ab ) - 1 = b - 1 a - 1 . That is, the inverse of a product is the product of the inverses, in the opposite order. Proof. We first prove (1). By definition G has to contain at least one identity element. Suppose that both e and f are identity elements in G . We compute the product ef . As e is an identity in G , ef = f. On the other hand as f is an identity in G , ef = e. Thus e = ef = f . Thus the identity is unique. Hence (1). Now we prove (2). Suppose that g is an element of G . Then g has at least one inverse by definition. Suppose that there were two elements h and k that were both inverses of g . We compute hgk (by associativity we can drop the parentheses). On the one hand we get
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