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Suppose that $(M,*)$ is a finite Monoid.

Prove that $M$ is a group if and only if there is only a single idempotent element in $M$, namely $e$.

One direction is obvious, because if $M$ is a group then $x^2=x$ implies $x=e$, but the other direction has been challenging me for over an hour, so I decided to ask it here.

user66733
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    Had a perfectly good answer and then I realized I'd missed the word "finite". – Malice Vidrine Nov 15 '13 at 12:12
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    @MaliceVidrine, I wonder if the following holds: let $M$ denote an arbitrary (i.e. not-necessarily finite) commutative monoid. Then $M$ is cancellative iff the only idempotent is $e$. – goblin GONE Dec 23 '14 at 03:26

2 Answers2

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If $M$ is not a group, then there is an element $a \in M$ with no inverse. Since $M$ is finite, there exists $n>m>0$ with $a^n=a^m$. You can now find a power of $a$ that is idempotent, and it cannot be the identity, because $a$ has no inverse.

Suppose that $a^m=a^n$ with $n>m$, and hence $a^m = a^{m+(n-m)}$.

We claim that $a^m=a^{m + k(n-m)}$ for all $k \ge 0$, and we prove this by induction on $k$. We have seen that it is true for $k=0,1$. Then, for $k>1$, we have, using the inductive hypothesis for $k-1$, $$a^{m + k(n-m)} = a^{m + (k-1)(n-m) + (n-m)} = a^{m+(k-1)(n-m)}a^{n-m} = a^ma^{n-m} = a^n = a^m$$ as claimed.

Now choose $k$ sufficiently large that $t := m + k(n-m) \ge 2m$. Then, multiplying both sides of $a^m=a^t$ by $a^{t-2m}$, we get $a^{t-m}=a^{2(t-m)}$, so $a^{t-m}$ is idempotent.

For example, if $a^7=a^9$, then $a^7=a^{15}$ and multiplying by $a$ gives $a^8=a^{16}$, so $a^8$ is idempotent.

Derek Holt
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    I don't see how I can find a power of $a$ that is idempotent :/ Would you please go ahead with your argument? – user66733 Nov 15 '13 at 13:02
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    Thanks. I don't understand why "we can increase the $n$ while keeping $m$ constant" is implied by $a^{n}=a^{m+(n-m)}=a^{n+(n-m)}$, can't we always do that in general? and I don't see how that justifies our assumption that $n-2m \geq 0$. I see why we need $n-2m$ to be non-negative though (because $a$ is assumed to be non-invertible). I'd appreciate it if you try to explain those two statements. – user66733 Nov 15 '13 at 14:09
  • The equation I wrote shows that you can always increase $n$ by $n-m$ so you can keep doing that and thereby increase it as much as you like. You need $n-2m \ge 0$ to justify multiplying by $a^{n-2m}$, because $a$ has no inverse. – Derek Holt Nov 15 '13 at 14:54
  • Thanks. Yes, I see why we need $n-2m \geq 0$ as I said, but I don't see why "we can increase the $n$ while keeping $m$ constant" implies $n-2m \geq 0$. – user66733 Nov 15 '13 at 15:05
  • Sorry, I cannot see what it is that you don't understand. I did give an example. Let me give another one. Suppose initially that $m=23$, $n=27$. I can increase $n$ by $27-23=4$, so I can keep doing this and replace $n$ by 31, 35, 39, 43, 47. Now $n-2m=47-46=1 \ge 0$. – Derek Holt Nov 16 '13 at 09:27
  • OK, maybe I explained my confusion in a wrong a way. You say that since "we can increase the $n$ while keeping $m$ constant" we know that $n-2m \geq 0$. I agree with that because $m$ is constant and since it's finite by taking $n$ large enough finally we can find some $n$ such that $n \geq 2m$. I just don't know how to say it in a rigorous way mathematically, but it's only a minor problem I think. – user66733 Nov 16 '13 at 10:27
  • Does it using contradiction? Where's the contradiction? Since there assume that $M$ is not a group. – lap lapan Jan 18 '21 at 21:59
  • @user795084 I have written out the proof that some power of $a$ is idempotent more formally. Is it clear now? – Derek Holt Jan 19 '21 at 08:10
  • @DerekHolt Yes, it is clear. My question is, in your proof, there using if M is not group. We want to show M is group. Does there using a contradiction? – lap lapan Jan 19 '21 at 08:18
  • Yes, it is a proof by contradiction. I assumed that $M$ is not a group in the first line of the proof, and then I deduced that there is an idempotent element that is not equal to $e$. In other words, if $e$ is the only idempotent, then $M$ is a group. – Derek Holt Jan 19 '21 at 08:56
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Perhaps a little clearer way to put it. Assume $x^a = x^{a+b}$ with $b >0$. Then for any integers $u$, $k$, we have $x^{a+u} = x^{a+u + kb}$

So we need to find $u$ and $k$ such that $bk = a+u$. That's easy. Take $b$ such that $bk \geq a$ and let $u = bk-a$. Then $x^{a+u}$ is idempotent.

Eric Wofsey
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Alain Bruguieres
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