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?
Best Answer
Another classic proof, that doesn't rely on he existence of inverses, is this.
Let $e_1$ and $e_2$ both be two-sided identities: i.e. for both of them and for all other $x$, $e * x = x = x * e$. Then $e_1 = e_1 * e_2 = e_2$.
The axioms for groups usually specify that there is a two sided identity. The operation you give fails this: everything is an identity on the left, but nothing is an identity on the right.
(People rarely consider 'identities of particular elements'. Either something is an identity in the sense that multiplying by it leaves every element the same, or it is not an identity.)