Element under Left Operation is Right Identity

Theorem
Let $\left({S, \leftarrow}\right)$ be an algebraic structure in which the operation $\leftarrow$ is the left operation.

Then no matter what $S$ is, $\left({S, \leftarrow}\right)$ is a semigroup all of whose elements are right identities.

Proof

 * It has been established that $\leftarrow$ is associative.


 * It is also immediately apparent that $\left({S, \leftarrow}\right)$ is closed, from the nature of the left operation:
 * $\forall x, y \in S: x \leftarrow y = x \in S$

whatever $S$ may be.

So $\left({S, \leftarrow}\right)$ is definitely a semigroup.


 * From the definition of left operation:
 * $\forall x, y \in S: x \leftarrow y = x$

from which it can immediately be seen that all elements of $S$ are indeed right identities.

From More than One Left Identity then No Right Identity, it also follows that there is no left identity.

Also see

 * Right Operation All Elements Left Identities
 * Left Operation All Elements Left Zeroes