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This question is a problem from Allan Clarke's Elements of Abstract Algebra, which has been asked about before on MSE (I'll provide a link below). When I tried solving it I stumbled upon an issue that I didn't see addressed in this or other questions about this problem.

Let $S$ be a semigroup with a finite number of elements. Suppose that the two cancellation laws hold in $S$; that is, if either $ab = ac$ or $ba = ca$, then $b = c$. Show that $S$ is a group.

Note: I'll just outline the proof idea for completeness, while skipping most of the proof. The details can be found here. Also: I know this can be proven using injection/surjection arguments, I'm asking about a particular step, which I describe below.

I approached this using the pigeonhole principle, induction, etc. to show that there is an identity $e = a^{n-m}$ for $a \in S$ and $n,m \in \mathbb N \colon n>m$. Then I showed that $a^{n-m-1}$ is the inverse of $a$ for $n > m+1$. This is where I have my issue: I didn't know what to do if $n - m = 1$? Because then $a^{n-m-1}=a^0$ and as I understand this, we don't know whether this element is in $S$.

Am I hallucinating issues, is this a non-issue or is there a simple fix for this?

Shaun
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powerline
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  • If $n-m=1$, then by assumption $e=a^1$. – Severin Schraven Jan 20 '25 at 01:14
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    It seems there are a bunch of typos. You probably want $n>m$. And $a^{n-m-1}=a^0$. – Severin Schraven Jan 20 '25 at 01:16
  • Finally, do you mean monoid, when you write semigroup? Or where is your identity element coming from? – Severin Schraven Jan 20 '25 at 01:18
  • @SeverinSchraven yes these were typos, thanks. it's meant semigroup, the linked proof shows how $a^{n-m}$ is the identity element (that is $a^{i-j}$ in their notation). – powerline Jan 20 '25 at 01:44
  • The question you link assumes that there exists an identity element (i.e. that they are dealing with a monoid). Maybe I am missing something. – Severin Schraven Jan 20 '25 at 01:45
  • @SeverinSchraven the question does assume it, but the answer uses the same reasoning I used, which works even without assuming the identity element; ie they proved it even though they didn’t have to. sorry for confusion, should’ve probably clarified that. but maybe it’s me who is missing something – powerline Jan 20 '25 at 02:12
  • Can you specify which answer you are referring to? – Severin Schraven Jan 20 '25 at 02:14
  • @SeverinSchraven it should jump to the answer when you click the link in my question above. if that doesn’t work, it’s the answer by Prahlad Vaidyanathan – powerline Jan 20 '25 at 02:17
  • If you are looking at Prahlad Vaidyanathan's answer, then please note that he uses $a_j=ea_j$, i.e. he uses the existence of a neutral element. – Severin Schraven Jan 20 '25 at 02:17
  • @SeverinSchraven you’re right, I missed that, thanks! although I managed to show successfully (I believe) that $a^{n-m}$ is the identity, without assuming its existence. I’ll edit my question with the details for that. – powerline Jan 20 '25 at 02:21
  • @SeverinSchraven on second thought, that part isn't relevant for my question so I'm hesitant to further clutter it, but here is the reasoning:

    we know $a^n = a^m$, expanding this (using standard inductive definitions of powers):

    $$ \begin{align} \underbrace {(a \dots a)}{m-1 \ as}\underbrace{(a \dots a)}{n-m+1 \ as} = \underbrace {(a \dots a)}_{m-1 \ as} a &\implies a^{n-m+1} = a\ & \iff a^{n-m}a = a \ & \iff aa^{n-m} = a \end {align} $$ hence $e=a^{n-m}$ is the identity element

     – powerline Jan 20 '25 at 02:29
  • This only shows that $e$ leaves $a$ invariant. Why is that the case for any other element in the semigroup? – Severin Schraven Jan 20 '25 at 02:31
  • @SeverinSchraven we only know that $a$ and its natural powers are in the semigroup, so are you asking why $a^k a = a$ for all $k \in \mathbb N$? it's not obvious to me that it is even true, I need to convince myself of this. – powerline Jan 20 '25 at 02:34
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    No, part of the definition of a neutral element $e$ is that $eb=b$ for all element $b$ in the semigroup. You have defined some candidate for $e$ and checked that it holds for $b=a$. You would still need to check it for all other elements in the semigroup. – Severin Schraven Jan 20 '25 at 02:37
  • @SeverinSchraven I see. I believe I've shown that every element of the semigroup has its own identity, so if I show that they are all one same element (ie uniqueness), that would essentially prove it I think? – powerline Jan 20 '25 at 02:47
  • Indeed, it would. – Severin Schraven Jan 20 '25 at 02:53

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Note that in order to talk about inverses, one needs to first establish that a neutral element exists, i.e. an element $e$ in our semigroup such that for all $b$ in our semigroup holds $eb =b=be$. In the linked question the existence of a neutral element is assumed (in fact in one of the answers of the linked question it is shown that one does not need to assume the existence of the neutral element).

If we know that such a neutral element exists, then your question becomes meaningful and easy. Namely, if $n-m=1$, then we have $e=a^{n-m}=a^1=a$. In particular, we have that $ae=a=e$ and therefore it is invertible.

Severin Schraven
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  • many thanks for the answer and guidance in the comments – powerline Jan 20 '25 at 02:56
  • @powerline You are very welcome. I do wonder now, whether the existence of the neutral element is needed or not. – Severin Schraven Jan 20 '25 at 02:57
  • The most upvoted answer in the linked question argues that one doesn't need to assume it, and provides a proof. and I found other questions that ask exactly this, not about a cancellative monoid as in the linked question – powerline Jan 20 '25 at 03:04
  • @powerline Oh, thanks! – Severin Schraven Jan 20 '25 at 03:05
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    Surely it is almost the same argument, or rather, it is Cayley's theorem and the same argument: There is a map from $\lambda \colon S \to \mathrm{Map}(S) = {f\colon S \to S}$ which is given by $\lambda(s)(x) = s.x$. Associativity shows this is a map of semigroups, and cancellation shows that it embeds $S$ into the finite group $\text{Sym}(S)$, which has a neutral element, and the argument given already shows that a subsemigroup of a finite group is a group. – krm2233 Jan 20 '25 at 03:09
  • @krm2233 The property that one can embed a semigroup into a group, does not make it a monoid. For example $(2\mathbb{N}\setminus{0},+)$ embeds into $(\mathbb{Z},+)$. I guess slightly more is needed. – Severin Schraven Jan 20 '25 at 04:04
  • It does in the finite case -- sorry, I wasn't clear about what I was assuming! If $S$ is infinite then I guess $(\mathbb Z_{>0},+)$ embedding in $(\mathbb Z,+)$ is an even more basic counterexample! I wrote things out in more detail here: https://math.stackexchange.com/a/5025333/1155637 – krm2233 Jan 20 '25 at 04:08
  • If $S\subseteq G$ is a subsemigroup of a finite group $G$, then because every element has finite order in $G$, cyclic sub-semigroups are then same thing as cyclic subgroups, hence subsemigroups are subgroups. – krm2233 Jan 20 '25 at 04:13
  • @krm2233 Smart. Thanks – Severin Schraven Jan 20 '25 at 04:14