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College Algebra Exam Review 75

# College Algebra Exam Review 75 - the inverse map possibly...

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2.1. FIRST RESULTS 85 Corollary 2.1.3. Let g be an element of a group G . We have g D .g 1 / 1 . Proof. Since gg 1 D e , it follows from the proposition that g is the in- verse of g 1 . n Proposition 2.1.4. Let G be a group and let a; b 2 G . Then .ab/ 1 D b 1 a 1 Proof. It suffices to show that .ab/.b 1 a 1 / D e . But by associativ- ity, .ab/.b 1 a 1 / D a.b.b 1 a 1 // D a..bb 1 /a 1 / D a.ea 1 / D aa 1 D e . n Let G be a group and a 2 G . We define a map L a W G ! G by L a .x/ D ax . L a stands for left multiplication by a . Likewise, we define R a W G ! G by R a .x/ D xa . R a stands for right multiplication by a . Proposition 2.1.5. Let G be a group and a 2 G . The map L a W G ! G defined by L a .x/ D ax is a bijection. Similarly, the map R a W G ! G defined by R a .x/ D xa is a bijection. Proof. The assertion is that the map L a has an inverse map. What could
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Unformatted text preview: the inverse map possibly be except left multiplication by a ± 1 ? So let’s try that. We have L a ± 1 .L a .x// D a ± 1 .ax/ D .a ± 1 a/x D ex D x; so L a ± 1 ı L a D id G . A similar computation shows that L a ı L a ± 1 D id G . This shows that L a and L a ± 1 are inverse maps, so both are bijective. The proof for R a is similar. n Corollary 2.1.6. Let G be a group and let a and b be elements of G . The equation ax D b has a unique solution x in G , and likewise the equation xa D b has a unique solution in G ....
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