Unformatted text preview: G be a group and suppose e and e are both identity elements in G ; that is, for all g 2 G , eg D ge D e g D ge D g . Then e D e . Proof. Since e is an identity element, we have e D ee . And since e is an identity element, we have ee D e . Putting these two equations together gives e D e . n Likewise, inverses in a group are unique : Proposition 2.1.2. (Uniqueness of inverses). Let G be a group and h;g 2 G . If hg D e , then h D g ± 1 . Likewise, if gh D e , then h D g ± 1 . Proof. Assume hg D e . Then h D he D h.gg ± 1 / D .hg/g ± 1 D eg ± 1 D g ± 1 . The proof when gh D e is similar. n 84...
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- Fall '08
- Algebra, Axiom, Identity element, ﬁrst results concern