# What can we conclude about $a$ if $a^2=a$ in a monoid or semigroup?

I found the following question in a test paper:

Suppose $G$ is a monoid or a semigroup. $a\in G$ and $a^2=a$. What can we say about $a$?

Monoids are associative and have an identity element. Semigroups are just associative.

I'm not sure what we can say about $a$ in this case other than that $a$ could be other things apart from the identity. Any idea if there's a definitive answer to this question?

• you can conclude that $a^n=a$ for all $n\geq 1$. – Surb May 23 '18 at 11:29
• @Surb For semigroup or monoid? Also, I guess they're asking what we can "conclude about $a$" in the sense that what the restrictions on $a$ are – user563280 May 23 '18 at 11:31
• Okay that one looks trivial as $a^2=a$. So we can write $a.a^2=a.a=a$ and so on....Probably they're looking for something more specific. – user563280 May 23 '18 at 11:33
• We can conclude that our monoid is not a free monoid: en.wikipedia.org/wiki/Free_monoid – Shine On You Crazy Diamond Jun 5 '18 at 17:07

## 3 Answers

If the test is for an introductory abstract algebra course, I say $a$ is an idempotent element of $G$.

• Well, that is quite obvious (almost by definition). I'm perhaps looking for the restrictions on $a$ if $a$ is idempotent and an element of a semigroup or a monoid – user563280 May 23 '18 at 11:40
• Can we draw any conclusions from associativity and existence of identity? – user563280 May 23 '18 at 11:43
• @user48929 I think we may need more contexts to say more. – scaaahu May 23 '18 at 11:56

Several things can be said on $a$:

1. $a$ is idempotent,
2. $a$ is regular,
3. the subsemigroup generated by $a$ is trivial,
4. the $\mathcal{H}$-class of $a$ is a group.

Make your choice!

$$\{1,a,a^2,a^3,...\} = \{1,a,a,aa^2,...\} = \{1,a,a,a^2,...\} = \{1,a,a,a,...\} = \{1,a\}$$