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Unformatted text preview: S UMMARY OF 10/29 L ECTURE Before proving the important result stated at the end of the previous lecture, we started with an easy observation: given a homomorphism : G H , we showed that maps G surjectively onto its image, i.e. : G im . We further noted that this map need not be surjective (e.g. the homomorphism : Q Q which takes 7 2 does not inject into im ; 4 = ( 2) ). Next, some notation: given a group G and a normal subgroup N , to save space we will write G := G/N . Every element of G looks like a coset, a translation of N , i.e. is of the form gN for some g G ; we will write g := gN . Ill repeat my mantra: there are elements we can distinguish in G , which can no longer be distinguished when viewed in G . Using our new notation: its possible to have g = h for two distinct elements g and h of G . Beware! We are now ready to prove the following important result: Theorem 1. Let : G H be a homomorphism. Then ker G and G/ ker im . Proof. We proved last time that ker is a normal subgroup of G , so it remains only to prove the isomorphism in the statement of the theorem. The following diagram will be helpful: (1) G / im G = { { { { where G := G/ ker , whose elements we denote by g := g ker . Here is the natural surjection g 7 g (why is this a surjection?), surjects onto im (as discussed above), and...
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 Fall '10
 GideonMaschler

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