Algebra by Michael Artin Exer 2.M.4

M.4. A semigroup S is a set with an associative law of composition and with an identity. Elements are not required to have inverses, and the Cancellation Law need not hold. A semigroup S is said to be generated by an element s if the set $\{1, s, s^2, . . .\}$ of nonnegative powers of s is equal to S. Classify semigroups that are generated by one element.

It looks like what Artin calls semigroups are what Wiki calls monoids.

How does one go about doing this? Based on Exer 2.M.3 (*), I think we have to take cases about the possible properties $s$ might have. Based on Wiki, I think monoids generated by 1 element are only the trivial monoid. Is this right? I think the issue is that if some element doesn't have an inverse, then we get $\{s,s^2,...\}$ with no identity element $1$.

(*) Exer 2.M.3

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    $\begingroup$ The definition used on the Wiki and the one given in M.4. are different. Saying $s$ generates a semigroup $S$ in the sense of M.4. is equivalent to saying $\lbrace 1, s \rbrace$ generates the monoid $S$ in the sense of the Wiki. $\endgroup$ – Theo Bendit Oct 17 '18 at 5:08
  • $\begingroup$ @TheoBendit wait I know semigroup in wiki is different from semigroup in artin but monoid in wiki is also different from semigroup in artin? $\endgroup$ – BCLC Oct 17 '18 at 5:17
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    $\begingroup$ It is explained in the question what it means for a monoid to be generated by a single element. In general, the submonoid $S$ of a monoid $M$ generated by a subset of $M$ contains the identity element by definition - it has to in order to be a submonoid. Why not just try and answer the question as it is written rather than speculating over different terminology? Roughly, either such a monoid is infinite with all non-negative powers of $s$ distinct, or you choose $n>m$ minimal with $s^n=s^m$. $\endgroup$ – Derek Holt Oct 17 '18 at 7:58
  • $\begingroup$ @DerekHolt I actually don't know how to go about doing this. I was hoping an answer would help me understand the question. $\endgroup$ – BCLC Oct 17 '18 at 8:47
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    $\begingroup$ I have more or less answered it in my comment. I will write an answer later. $\endgroup$ – Derek Holt Oct 17 '18 at 12:20

If the $s^i$ are distinct for all $i \ge 0$, then $S$ is the infinite monoid $\{1,s,s^2,s^3,\ldots \}$.

Otherwise there exists a smallest $n$ such that $s^n = s^m$ for some $m$ with $0 \le m < n$. In that case, $S$ is finite of order $n$, and $S = \{ 1,s,s^2,\ldots,s^{n-1}\}$. Then, for non-negative integers $a<b$, we have $s^a = s^b$ if and only if $a \ge m$ and $(n-m)|(b-a)$. So $1,s,\ldots,s^{m-1}$ are uniquely determined as powers of $s$, but then the higher powers of $s$ repeat cyclically.

Notice that distinct values of $n$ and $m$ give rise to nonisomorphic monoids. In the case $m=0$, we get the cyclic group of order $n$. When $m>0$, $s$ is not equal to a power of $s^k$ for any $k>0$, so $s$ is the unique generator.

  • $\begingroup$ Thanks Derek Holt! I'll analyse more later. For now, would you say Exer 2.M.5 is kind of a hint towards Exer 2.M.4? I notice part of your answer is part of one possible proof to Exer 2.M.5 (which is similar to proving multiplicative prime fields are groups or something while other proofs involve surjective or injective functions with left or right inverses) ? $\endgroup$ – BCLC Oct 17 '18 at 23:05

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