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In doing a proof, I am on the last step, but I can't quite seem to get it:
I have a group G and operation * (note: * is any operation, not just multiplying). The definition of a group is at the bottom.
Let i ∈ G be the identity in G. You can assume the identity is unique (I can prove that easily). I know that i * b = b, and a * b = b. It seems rather obvious that i = a, and thus, a is the identity. However, I am not sure how to explicitly state that. Anyone know?
Definition of a group: A group is a set G and an operation * which is:
Associative: For all a, b, c ∈ G, a * (b * c) = (a * b) * c
Has an identity: There exist an e ∈ G, such that for all a ∈ G, a * e = a and e * a = a
Has an inverse: For all a ∈ G, there exists b ∈ G, such that a * b = e and b * a = e
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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i * b = b, a * b = b
i * b = a * b
Let b-¹ be the inverse of b, which must exist since it's a group.
(i * b) * b-¹ = (a * b) * b-¹
The operation is associative so:
i * (b * b-¹) = a * (b * b-¹)
Definition of inverse, b * b-¹ = i (identity must be unique).
i * i = a * i
i * i = i, and a * i = a
i = a
That look good?
"In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..."
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Pages: 1