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Unformatted text preview: H K is a subgroup of G ; see Exercise 2.1.10(a) on page 48. We now need to verify the normality condition: let g G . Then gHg1 = H and gKg1 = K , consequently g ( H K ) g1 gHg1 gKg1 = H K . Since N is a normal subgroup of G if and only if N G and gNg1 N for all g G , the result is proven. Let G be a group and let H G . Prove that the formula g ( xH ) = gxH for g , x G denes an action of G on the left cosets of H in G . We should note that the formula is well dened because if xH = yH , then g ( xH ) = gxH = g ( xH ) = g ( yH ) = gyH = g ( yH ) . We now have (i) 1 ( xH ) = 1 xH = xH . (ii) If g , k G , then g ( k xH ) = g ( kxH ) = gkxH = ( gk ) ( xH ) . This shows that we have an action, as required....
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This note was uploaded on 01/02/2012 for the course MATH 4124 taught by Professor Staff during the Spring '08 term at Virginia Tech.
 Spring '08
 Staff
 Math, Algebra

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