Prove that - For any abelian monoid (M,*) ,the set of idempotent elements of M forms a submonoid ....

I can relate between idempotent elements and abelian monoid.. but how do we prove it as a submonoid?

  • $\begingroup$ Have you studied groups before? This is not too different from when you have to prove that some subset of a group is a subgroup. If not, what is the definition of submonoid? (I'll try to get you to figure out how to do it yourself instead of giving you the solution.) $\endgroup$ Sep 21, 2019 at 17:27
  • $\begingroup$ i just started learning group this week.. and about submonoids, we define it as....Let (M,*,e) be a monoid and T⊆M , if T is closed under * and e ∈ T, where e is the identity element the T is said to be the submonoid of M $\endgroup$
    – Rejani
    Sep 21, 2019 at 17:32
  • $\begingroup$ Ah, ok. I'll write down an answer with some hints. $\endgroup$ Sep 21, 2019 at 17:33

1 Answer 1


Let $M$ be a monoid with. For simplicity, I will write the operation by contatenation (as we do for multiplication), and the unit of $M$ as $1$. A submonoid of $M$ is a subset $A\subseteq M$ satisfying the following two properties:

  1. $1\in A$ (that is, $A$ contains the unit of $M$);
  2. If $x,y\in A$, then $xy\in A$ as well.

So the question is asking you to prove that the set of idempotents of $M$ is a submonoid of $M$. There are again some definitions to unravel here: An element $e$ of $M$ is idempotent if $ee=e$.

Good, now we have all definitions spelled out just in terms of the monoid structure.

Let $E=\left\{e\in M:ee=e\right\}$, be the set of idempotents of $M$. To check that $E$ is a submonoid of $M$, we need to verify that $E$ satisfies properties $1$ and $2$ above.

  1. To check that $1\in E$, we need to verify that satisfies the property defining $E$, that is, $11=1$. Do you see why this is true?

  2. Take arbitrary elements $x,y\in E$. Then, by hypothesis, $xx=x$ and $yy=y$, and $xy=yx$, because $M$ is commutative. We need to verify that $xy\in E$ as well, that is, that $(xy)(xy)=(xy)$. Let us try to write down a sequence of identities starting with the first term and ending with the second one. I'll start for you: By associativity of the operation: $$(xy)(xy)=x(yx)y$$ can you use the hypotheses above to rewrite $x(yx)y$ as $xy$?

  • 1
    $\begingroup$ (xy)(xy)=x(yx)y=x(xy)y=(xx)(yy)=xy ...? $\endgroup$
    – Rejani
    Sep 21, 2019 at 17:47
  • $\begingroup$ @Rejani Perfect $\endgroup$ Sep 21, 2019 at 17:48

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .