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## #1 2006-01-24 05:17:27

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

### Groups, operations, and identities, oh my!

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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## #2 2006-01-24 06:12:13

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

### Re: Groups, operations, and identities, oh my!

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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