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Unformatted text preview: Math 402B Solutions to Homework 6 Chapter 2 Section 57 Show that : G G , where is a homomorphism, is a monomorphism if and only if Ker = ( e ) . Proof. By definition, a monomorphism is an injective homomorphism. Ker = { g G  ( g ) = e } , i.e. Ker is the preimage of e . For a homomorphism we always have ( e ) = e . If a G is an element such that ( a ) = e , since is injective so a = e . Therefore the only element in Ker is e . If Ker = ( e ), we show the homomorphism is injective. Suppose ( a ) = ( b ) for some a,b G , then ( ab 1 ) = ( a ) ( b ) 1 = e , so ab 1 Ker , which implies ab 1 = e , so a = b and is an injective homomorphism. Remark : (1)The proof is only a restatement of definitions. Here is a suggestion: before trying to prove anything, make sure you know clearly the definition of each word. In mathematics definitions are always important. (2)This conclusion is so frequently used that from now on you can use it without a proof, that is, to show a homomorphism is injective, it suffices to check whether the kernel of this homomorphism contains only the unit element. To state it another way, the kernel of a homomorphism can be used to test how far this homomorphism is from an injective homomorphism. Section 521 Let S be any set having more than two elements and A ( S ) the set of all 11 mappings of S onto itself. If s S , we define H ( s ) = { f A ( S )  f ( s ) = s } . Prove that H ( s ) cannot be a normal subgroup of A ( S ) Proof. A normal subgroup H of a group G is a subgroup which is normal, i.e., which satisfies g G , g 1 Hg = H . In this problem H ( s ) is indeed a subgroup. To show H ( s ) is not a normal subgroup, it suffices to find a g A ( S ) and an h H ( s ) such that g 1 hg / H ( s ). Most of you gave the following proof, which is concise and clear. Since S has more than two elements, we can pick 3 distinct elements a,b and s . Now let g = ( as ) A ( S ) (i.e. g exchanges a and s and fixes any other element), h = ( ab ) H ( s ), then g 1 hg = ( as )( ab )( as ) = ( bs ), which is not in H ( s ). Hence H ( s ) is not a normal subgroup. 1 Remark : (1)For a subgroup H of a group G , the following are equivalent: (a) H is normal; (b) g 1 Hg H for any g G ; (c) g 1 Hg = H for any g G ; Here is a point I need to make clear: in (c) g 1 Hg = H means as two SETS they are equal, i.e. they contain exactly the same elements. This does not mean for any h H , g 1 hg = h ; the correct interpretation is: for any h H , g 1 hg = h for some h H , here h may be different from h itself....
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 Winter '11
 PhanThuongCang

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