Let X be a set. Prove that the collection of all subsets of X under the operation of symmetric difference is a group.
Be concise showing your steps!
Thanks ever so much for the help
(1) Closure under Δ
Given A, B in X, A Δ B = (A \ B) U (B \ A); A \ B and B \ A are two subsets of X, so their union is a subset of X. Hence A Δ B is in X.
(2) Associativity
I have a solution in the following link.
http://ca.answers.yahoo.com/question/index;_ylt=Av...
(3) Existence of an Identity
The identity in X is Ø since, given A in X,
Ø Δ A = A Δ Ø = A \ Ø U Ø \ A = A U Ø = A.
(4) Existence of Inverses
Given A in X, its inverse is A. This is because A Δ A = A \ A U A \ A = Ø U Ø = Ø.
Since properties (1), (2), (3), (4) are satisfied, (X, Δ) is a group.
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Verified answer
(1) Closure under Δ
Given A, B in X, A Δ B = (A \ B) U (B \ A); A \ B and B \ A are two subsets of X, so their union is a subset of X. Hence A Δ B is in X.
(2) Associativity
I have a solution in the following link.
http://ca.answers.yahoo.com/question/index;_ylt=Av...
(3) Existence of an Identity
The identity in X is Ø since, given A in X,
Ø Δ A = A Δ Ø = A \ Ø U Ø \ A = A U Ø = A.
(4) Existence of Inverses
Given A in X, its inverse is A. This is because A Δ A = A \ A U A \ A = Ø U Ø = Ø.
Since properties (1), (2), (3), (4) are satisfied, (X, Δ) is a group.