College Algebra Exam Review 105

College Algebra Exam Review 105 - 2.4. HOMOMORPHISMS AND...

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Unformatted text preview: 2.4. HOMOMORPHISMS AND ISOMORPHISMS 115 Definition 2.4.13. A subgroup N of a group G is said to be normal if for all g 2 G , gNg 1 D N . Here gNg 1 means fgng 1 W n 2 N g. Definition 2.4.14. Let ' W G ! H be a homomorphism of groups. The kernel of the homomorphism ' , denoted ker.'/, is ' 1 .eH / D fg 2 G W '.g/ D eH g. According to Proposition 2.4.12 (b), ker.'/ is a subgroup of G (since feH g is a subgroup of H ). We now observe that it is a normal subgroup. Proposition 2.4.15. Let ' W G ! H be a homomorphism of groups. Then ker.'/ is a normal subgroup of G . Proof. It suffices to show that g ker.'/g 1 D ker.'/ for all g 2 G . If x 2 ker.'/, then '.gxg 1 / D '.g/'.x/.'.g// 1 D '.g/e.'.g// 1 D e . Thus gxg 1 2 ker.'/. We have now shown that for all g 2 G , g ker.'/g 1  ker.'/. We still need to show the opposite containment. But if we replace g by g 1 , we obtain that for all g 2 G , g 1 ker.'/g  ker.'/; this is equivalent to ker.'/  g ker.'/g 1 . Since we have both g ker.'/g 1  ker.'/ and ker.'/  g ker.'/g 1 , we have equality of the two sets. I If a homomorphism ' W G ! H is injective, then its kernel ' contains only eG . The converse of this statement is also valid: Proposition 2.4.16. A homomorphism ' W G only if, ker.'/ D feG g. 1 .e / H ! H is injective if, and Proof. If ' is injective, then eG is the unique preimage of eH under ' . Conversely, suppose that ker.'/ D feG g. Let h 2 H and suppose that g1 ; g2 2 G satisfy '.g1 / D '.g2 / D h. Then '.g1 1 g2 / D '.g1 / 1 '.g2 / D h 1 h D eH , so g1 1 g2 2 ker.'/. Therefore, g1 1 g2 D eG , which gives g1 D g2 . I Example 2.4.17. The kernel of the determinant det W GL.n; R/ ! R is the subgroup of matrices with determinant equal to 1. This subgroup is called the special linear group and denoted SL.n; R/. ...
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This note was uploaded on 12/15/2011 for the course MAC 1105 taught by Professor Everage during the Fall '08 term at FSU.

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