In a group, the identity of the binary operation must be unique?

Question

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?

Answer 1

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.)

Answer 2

To say that $e_1$ and $e_2$ are identity element means that: $$ ae_1=e_1a=a \quad \forall a \in G $$ and $$ ae_2=e_2a=a \quad \forall a \in G $$ (note the ''for all'').

As a consequence we have $$ e=e_1 e_2=e_1 \quad \mbox{and} \quad e=e_1 e_2=e_2 $$

Answer 3

Yes, it should be unique. Suppose a group has two identities $e$ and $f$. Then we must have $$e*f = e$$ when we consider the identity $f$ and $$e*f = f$$ when we consider the identity $e$.

Therefore, we have $e = f$.

Answer 4

As the comment pointed out, identity $e$ is an element in the set such that for any elements $a$ in the set $a*e=a=e*a$. And by the proof you provided, such element $e$ is unique. Hence, each element has the same identity.

The examples you provided is not correct. The set of real numbers is a group under addition i.e. $\mathbb{R}$ is an additive group.

For example when you see $\mathbb{R} $ as an additive group ($\mathbb{R} $, +), $0$ is the additive identity of $\mathbb{R} $ since for any real number $a$, it satisfies $a+0=a=0+a$. And additive inverse of $a$ is $-a$ since $a+(-a)=0$.

Answer 5

To be an identity element we must have $a*e = e*a = a$ for all $a$.

If $*$ is defined by $a*b = b$ then it is CAN'T have an identity because the only way that $a*b = b*a = b$ is if $a = b$, and that only applies to $b$; and not to any element in the set.

It is not part of the definition of a group that the identity must be unique but it is easily proven to be a property that must follow from the definition.

Pf: If $e$ and $f$ are both identies then $e*f = f$ and $e*f = e$ so $e=f$. So if an identity element exists it must be unique.

Likewise to be an inverse element it must be that $a*a' = a'*a = e$.

It is easy to prove such an element must be unique.

Pf: If $a'*a= a*a' = e$ and if $b*a = e$ then $b*a*a' = b*(a*a')=b*e = b$ and $b*a*a' = (b*a)a' = e*a' = a'$ so $b = a'$. So if an inverse exists for $a$ it must be unique.