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Assignment11Part1

# Assignment11Part1 - Math 302 Modern Algebra Homework...

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Math 302 Modern Algebra Spring 2009 Homework Solutions 7.5 CONGRUENCE AND LAGRANGE’S THEOREM 1. Let K be a subgroup of a group G . Prove: For all a G , Ka = K if and only if a K . Proof: Suppose a is an arbitrary element of G . We need to prove Ka = K if and only if a K . Forward Direction. Suppose Ka = K. We need to show a K. Note Ka = { ka | k K } . Since K is a subgroup, then e K . It follows that a = ea Ka . Since Ka = K , then a Ka implies a K , as required. Backward Direction. Suppose a K. We need to show Ka = K. To prove this, we need to show ( g G ) [( g Ka ) ( g K )] . Suppose g is an arbitrary element of G . First suppose g Ka. We need to show g K. Since g Ka , then g = ka for some k K . Since K is a subgroup and a K and k K , then g = ka K , as required. Now suppose g K. We need to show g Ka. To show this, we need to show that g = ka for some k K . Find a k . The element k must satisfy g = ka. Solving this equation for k , we obtain k = ga 1 . Choose k = ga 1 . Verify k satisfies: (i) Domain: k K (ii) Propositional Function: g = ka Since a K , then a 1 K since K is a subgroup. Then g K and a 1 K imply k = ga 1 K , as required. Since k = ga 1 , then g = ka , as required. Therefore g Ka , as required. Thus Ka = K. This proves the backward direction. As an alternative proof for the backward direction, note that as in the proof of the forward direction, a Ka . Since a K also, then a K Ka . Since K = Ke , then K is a right coset of K . By Corollary 7.24, two right cosets of K are either disjoint or identical. Since a K Ka , then K and Ka are not disjoint, so must be identical. Thus K = Ka .

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