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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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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**Ricky****Moderator**- Registered: 2005-12-04
- Posts: 3,791

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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