# Submonoid and abelian monoid

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?

• 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.) Sep 21, 2019 at 17:27
• 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 Sep 21, 2019 at 17:32
• Ah, ok. I'll write down an answer with some hints. Sep 21, 2019 at 17:33

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 $$xy$$?

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