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Let $S$ be a finite semigroup. Recall that every element $a\in S$ determines a unique pair of positive integers $\iota=\mathrm{ind}(a)$ and $\rho=\mathrm{per}(a)$, called the index of $a$ and the period of $a$, respectively. These are the smallest positive integers such that $a^{\iota}=a^{\iota+\rho}$.

In addition, given an element $a\in S$, it can be shown that the Kernel of $a$, namely the set $$ K_a=\{a^{\iota},a^{\iota+1},\ldots,a^{\iota+\rho-1}\} $$ forms a cyclic group.

My question: What are $\mathrm{ind}(x)$ and $\mathrm{per}(x)$ for $x\in K_a$? Can it be expressed in terms of $\mathrm{ind}(a)$ and $\mathrm{per}(a)$?

boaz
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1 Answers1

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The unit of $K_a$ is an idempotent $e$ and by definition, $e = a^r$ for some $r \geqslant i$. If $e$ is also the unit of $S$, then $S$ is a monoid, and the index (in $S$) of every element $a^j$ of $K_a$ is $0$, since $(a^j)^0 = e$ (by definition of $x^0$ in a monoid) and $$(a^j)^{0+r} = a^{jr} = (a^r)^j = e^j = e = (a^j)^0.$$

If $e$ is not the unit of $S$, then the index (in $S$) of every element $a^j$ of $K_a$ is $1$ since $(a^j)^0$ is undefined, but $$(a^j)^{1+r} = a^ja^{jr} = a^j(a^r)^j = a^je^j = a^je = a^j.$$

The period of $a^j$ is the smallest integer $s$ such that $(a^j)^{1+s} = a^j$ and since $K_a$ is a cyclic group of order $\rho$, it is the smallest integer such that $j(1+s) \equiv j \bmod \rho$, that is $js \equiv 0 \bmod \rho$. In particular, the period of $ea$ is $\rho$ since $ea$ generates $K_a$.

J.-E. Pin
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