Given the set of the real numbers and the binary operarion defined by a*b = b, is this set with this operation a group?
This is from a book of mine. The book says that for a set with a binary operation to be a group they have to obey three rules:
1) The operation is associative; 2) There's an identity element in the set; 3) Each element of the set has an inverse.
So, the operation is indeed associative but each element have a different identity (itself!). For example, the identity of 2 is 2 itself, because 2*2 = 2 and the identity of 3 is 3, because 3*3 = 3. And the inverse of 2 is 2, and of 3 is 3. No problem with the inverse, each element has one, but I don't know if a group must have a unique identity.
The book says 'an identity', not 'an unique identity'. I searched about it and found that the uniqueness of the identity is proved from the axioms, and is not a part of the axiom. This is the standard proof:
Let a be an element of the set and e1 and e2 the identities. So $ a*e_1 = a*e_2$, so $e_1 = e_2. $ For me this proves that each element has an unique identity associated with it, but it does not prove that the identity of element a is equal to the identity of element b.
So, is it a group or not?