Problem

Let G be a group, and let 0\G) be the set of all subsets of G. For any A, B ∈ ℘(G), let us...

Let G be a group, and let 0\G) be the set of all subsets of G. For any A, B ∈ ℘(G), let us define the product subset AB = [ab| a ∈ A. b ∈ B).

a. Show that this multiplication of subsets is associative and has an identity clement, but that ℘(G) is not a group under this operation.

b. Show that if N is a normal subgroup of G, then the set of cosets of N is closed under the above operation on -y(G), and that this operation agrees with the multiplication given by the formula in Corollary 14.5.

c. Show (without using Corollary 14.5) that the cosets of N in G form a group under the above operation. Is its identity element the same as the identity element of ℘(G)?

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Solutions For Problems in Chapter S.14