Left and Right Identity are Equivalent

Theorem

Given any algebraic structure \((S, \ast)\), if both a left and right identity exist, then they are equal.

Proof

Let \(e_L\) be a left identity in \(S\), that is:

\[ \forall a \in G : e_L \ast a = a,\]

and \(e_R\) be a right identity in \(S\), that is:

\[ \forall b \in G : b \ast e_R = b.\]

Then, letting \(a = e_R\) and \(b = e_L\), we have that:

\[ e_R = e_L \ast e_R = e_L.\]

Corollary

If an algebraic structure \((S, \ast)\) has two distinct left identities, it has no right identity.

Let \(e_1\) and \(e_2\) be two left identities in \(S\). If \(S\) has a right identity \(e_R\), then \(e_1 = e_R = e_2\) by two applications of the theorem above. Hence the existence of a right identity implies the uniqueness of the left identity. Therefore if there are two distinct left identities, there does not exist a right identity.