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?
Submonoid and abelian monoid
monoidsemigroups
Related Solutions
Elaborating on Eric's comment, the exact same argument works for monoids.
Specifically, given a monoid $\mathcal{M}=(M,*)$, we have a natural action of $\mathcal{M}$ on the set $M$ given by left multiplication: each $a\in M$ corresponds to the function $f_a: b\mapsto a*b$.
The map $act: a\mapsto f_a$ is easily checked$^1$ to be an injective monoid homomorphism from $\mathcal{M}$ to $Sym(M)$, so $\mathcal{M}$ is isomorphic to the submonoid of $Sym(M)$ with domain $im(act)$.
$^1$Note that the presence of an identity element is crucial here for showing injectivity (if $a\not=b$ then $f_a(e)=a\not=b=f_b(e)$). Consider the two-element semigroup $\mathcal{S}$ with underlying set $\{0,1\}$ and multiplication given by the constant map $(x,y)\mapsto 1$. If we run the construction above we get a non-injective homomorphism from $\mathcal{S}$ to $Sym(\{0,1\})=S_2$. It's a good exercise to check whether $\mathcal{S}$ does in fact embed into any $Sym(X)$.
Associativity is also necessary, and we know it has to be: no non-associative magma can embed into any $Sym(X)$. So what goes wrong if we don't assume associativity, in the above argument?
Let $h$ be an element of $H$. If the submonoid generated by $h$ is not free, then there exist two integers $n \geqslant 0$ and $p > 0$ such that $h^n = h^{n + p}$. Thus $H$ is a torsion monoid. There exists a two-generated infinite monoid which is torsion, thus the answer to your question is negative. Consider for instance the quotient of the free monoid $\{a,b\}^*$ by the relations $x^3y = yx^3 = x^3$, for all words $x$ and $y$. This is a monoid with zero $M$ in which every $x^3 = 0$ is an identity. To prove it is infinite, consider the Prouhet-Thue-Morse infinite word. This infinite word is known to be cube-free, which means that it contains no factor of the form $uuu$, where $u$ is a nonempty word. It follows that each finite prefix of this infinite word corresponds to an element of $M$, and thus $M$ is infinite.
The corresponding problem for groups is the famous Burnside problem which also has a negative, but difficult, answer, first given by Golod and Shafarevich.
Best Answer
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:
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.
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?
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 $xy$?