Is there any group-like structure that doesn't have an identity, but has (non-equal) left and right identities?

Question

Is there any group-like structure that doesn't have an identity, but has a (non-equal) left and right identities?

Answer 1

Suppose you have some sort of collection of symbols and a binary operation on the collection represented by concatenation. If $L$ is a left identity and $R$ is a right identity, then $$ L = LR = R $$ and so they are actually the same.

Answer 2

If both left identity and right identity exist, they are definitely the same, as @davidlowryduda pointed out.

But we can construct a structure that doesn't have an identity, but have either left identity or right identity.

Let set $G = \{ a, b \}$, where $a \neq b$. Define$$ \def\opair#1{\langle #1 \rangle} \times = \{ \opair{\opair{a,b},b}, \opair{\opair{b,a},a}, \opair{\opair{a,a},a}, \opair{\opair{b,b},b} \}. $$

Obviously, $\times$ is a binary arithmetic operation, while both $a$ and $b$ are left identities of $G$. Because \begin{align*} (a \times a) \times a = a = a \times (a \times a), \\ (a \times a) \times b = b = a \times (a \times b), \\ (a \times b) \times a = a = a \times (b \times a), \\ (b \times a) \times a = a = b \times (a \times a), \\ (a \times b) \times b = b = a \times (b \times b), \\ (b \times a) \times b = b = b \times (a \times b), \\ (b \times b) \times a = a = b \times (b \times a), \\ (b \times b) \times b = b = b \times (b \times b), \end{align*} then $\times$ is associative.

Therefore, $(G,\times)$ is a semigroup with two left identities but no right identity.