Monoids and Idempotents

Question

If an element of a commutative Monoid M is said to be idempotent for some $k \ge 1$ then I am trying to show that $M^+$ the set of elements that are idempotent is closed. What I understand is that if $a^k=a$ and $b^j=b$ then I need to find some integer $q$ such that $(a*b)^q=a*b$. I am having difficulty finding this integer or perhaps i have misunderstood the question.

Answer 1

I claim that $(ab)^{1+r} = ab$ where $r$ is the least common multiple of $k-1$ and $j-1$. To prove this, let us first observe that $(a^{k-1})^2 = a^{k-1}$ since $$ (a^{k-1})^2 = a^{2k-2} = a^ka^{k-2} = a^{}a^{k-2} = a^{k-1}. $$ Similarly, $(b^{j-1})^2 = b^{j-1}$. Now, let $c$ and $d$ be positive integers such that $r = c(k-1)$ and $r = d(j-1)$. Then $$ a^{1+r} = a^{1 + c(k-1)} = a^{}a^{c(k-1)} = a^{}a^{k-1} = a^k = a^{} $$ and similarly $b^{1+r} = b^{1 + d(j-1)} = b^{}b^{j-1} = b^{}$. Finally, since $M$ is commutative, we get $(ab)^{1+r} = a^{1+r}b^{1+r} = a^{}b^{}$.

Answer 2

Suppose $a^k = a$ and $b^k = b$. Then $(ab)^k = a^k b^k = ab$.

A few comments:

• Commutativity is used to get from $(ab)^k$ to $a^kb^k$.
• I don't think this works if the "idempotency" of $a$ and $b$ are different values ($j$ and $k$, if you will.)