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I was studying group theory and got into thinking about "groups without inverse elements", which are apparently called monoids. In every definition of monoid (and group) that I was able to find, the existence of the identity element is stated in this way:

Definition 1: There is $e \in G$ s.t. $e a = a e = a$ for all $a \in G$.

Now suppose we kept the other monoid axioms and changed this statement to:

Definition 2: There is $e \in G$ s.t. $e a = a$ for all $a \in G$.

Are definitions 1 and 2 equivalent (when taken together with the other two monoid axioms)? I wasn't able to show that they are (if an inverse of $a$ exists, i.e. we are talking about groups, then they clearly are). If not, is there a name for the construction that comes out using definition 2? Some kind of monoid with a non-commuting identity element? Are they useful for something?

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2 Answers 2

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Consider a set $S$ with $|S|>1$ bestowed with the binary operation $\circ$ where:

$$\forall x, y\in S: x \circ y=y$$

Then every element of $S$ is a left identity, and there are no right identities.

(Here $(S,\circ)$ is a semigroup but not a monoid, as associativity holds.)

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  • $\begingroup$ I'm sorry, but what is the connection to my question here? In both of the definitions I wrote, only the existence of an identity is stated, not that every element acts as a (left) identity. $\endgroup$
    – vjk
    Commented Sep 26, 2020 at 15:37
  • $\begingroup$ I proved existence of a left identity. It satisfies your condition. Moreover, there is no two-sided identity, which is also apparent, which shows that the two conditions are indeed distinct. $\endgroup$
    – player3236
    Commented Sep 26, 2020 at 15:49
  • $\begingroup$ @vjk your question said nothing about the left identity being unique. Saying “there is” does not mean “there is only one.” $\endgroup$
    – Randall
    Commented Sep 26, 2020 at 16:12
  • $\begingroup$ @player3236, yes, I'm sorry, it's been a while since I've been thinking in mathematics. In essence, your answer shows that there exists a construction that satisfies definition 2 but not definition 1, which means that the definitions are not equivalent. I got confused, because for groups the identity element is unique, which I now realize doesn't hold for the construction I defined. It would be interesting to see a less trivial counter-example, but since that is not what I originally asked, I will accept your answer. $\endgroup$
    – vjk
    Commented Sep 28, 2020 at 8:52
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The semigroups in which the identity $xy = y$ holds is called a right-zero band in the semigroup literature. In such a semigroup, every element is idempotent ($x^2 = x$) and a left identity, but is an identity only if the semigroup has only one element.

To answer your question "Are they useful for something?", the answer is yes. For instance, they are useful to describe the fine structure of the minimal ideal of a semigroup. They also play a role in the wreath product decomposition theory of finite semigroups.

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  • $\begingroup$ Since my definition 1 is stronger than definition 2 (in the sense that 1 guarantees more properties than 2), everything that is true for definition 2 should also be true with definition 1. For this reason, I don't believe that the idempotency of every element is true with definition 2 since that would imply that this holds for every monoid and group. If I understand the Wikipedia definition of bands correctly, it says that every element of a band has this property, my definitions only state that there is one element that has it. $\endgroup$
    – vjk
    Commented Sep 26, 2020 at 15:42
  • $\begingroup$ You are right. I was actually referring to player3236's answer. $\endgroup$
    – J.-E. Pin
    Commented Sep 27, 2020 at 5:10

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