LS10-29 - S UMMARY OF 10/29 L ECTURE Before proving the...

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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 φ : G H , we showed that φ maps G surjectively onto its image, i.e. φ : G im φ . We further noted that this map need not be surjective (e.g. the homomorphism φ : Q × Q × which takes α 7→ α 2 does not inject into im φ ; 4 = φ ( ± 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 . I’ll 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: it’s 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 φ : G H be a homomorphism. Then ker φ G and G/ ker φ im φ. Proof. We proved last time that ker φ 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 π φ / im φ G μ = { { { { where G := G/ ker φ , whose elements we denote by g := g ker φ . Here π is the natural surjection g 7-→ g (why is this a surjection?), φ surjects onto im φ (as discussed above), and μ is the isomorphism we wish to construct. How on earth do we construct such a
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