College Algebra Exam Review 104

# College Algebra Exam Review 104 - injective homomorphisms W...

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114 2. BASIC THEORY OF GROUPS regardless of whether f has an inverse function, and the notation f ± 1 .B/ is not supposed to suggest that f has an inverse function. In case that f does have an inverse function, then f ± 1 .B/ D f f ± 1 .y/ W y 2 B g . For example, if ' W Z ±! Z 6 is the map n 7! ŒnŁ , then ' ± 1 . f Œ0Ł;Œ3Ł g / is the set of integers congruent to 0 or to 3 mod 6 . Proposition 2.4.12. Let ' W G ±! H be a homomorphism of groups. (a) For each subgroup A ² G , '.A/ is a subgroup of H . (b) For each subgroup B ² H , ' ± 1 .B/ D f g 2 G W '.g/ 2 B g is a subgroup of G . Proof. We have to show that '.A/ is closed under multiplication and in- verses. Let h 1 and h 2 be elements of '.A/ . There exist elements a 1 ;a 2 2 A such that h i D '.a i / for i D 1;2 . Then h 1 h 2 D '.a 1 /'.a 2 / D '.a 1 a 2 / 2 '.A/; since a 1 a 2 2 A . Likewise, for h 2 '.A/ , there is an a 2 A such that '.a/ D h . Then, using Proposition 2.4.11 (b) and closure of A under in- verses, we compute h ± 1 D .'.a// ± 1 D '.a ± 1 / 2 '.A/: The proof of part (b) is left as an exercise. n The Kernel of a Homomorphism You might think at ﬁrst that it is not very worthwhile to look at non-
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Unformatted text preview: injective homomorphisms ' W G ! H since such a homomorphism loses information about G . But in fact, such a homomorphism also reveals cer-tain information about the structure of G that otherwise might be missed. For example, consider the homomorphism from the symmetry group G of the square into the symmetric group S 2 induced by the action of G on the two diagonals of the square, as discussed in Example 2.4.4 . Let N denote the set of symmetries ± of the square such that .±/ D e . You can compute that N D f e;c;d;r 2 g . (Do it now!) From the general theory, which I am about to explain, we see that N is a special sort of subgroup G , called a normal subgroup. Understanding such subgroups helps to un-derstand the structure of G . For now, verify for yourself that N is, in fact, a subgroup....
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