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Let be $A$ a set. It is endowed with one internal composition law which is also associative, let's say $\cdot$. There are two elements $a_1, \: a_2\in A$, such that:

$$a_1x^na_2=x,\forall\: x\in A$$

Prove that $A$ is monoid.

Regarding this post, I imagine that we can prove it for every $n$ prime number.

Especially Lime
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shangq_tou
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    The linked post has no answer or meaningful progress towards a proof. What's your intuition for why this should work specifically when $n$ is prime? – Karl Nov 03 '21 at 15:38
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    What is the quantification of $n$? Is there a common $n$ for all $x$ or, for all $x$ is there an $n$ depending on $x$? – Berci Nov 03 '21 at 16:18
  • @Berci $n$ is fixed along with $a_1$ and $a_2$. – Karl Nov 03 '21 at 16:42
  • Presumably we require $n > 0$ since $x^0$ doesn't make sense without an identity element. – Mark Saving Nov 03 '21 at 17:06
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    Hint: if $A$ is in fact a monoid, it has an identity element $e$. Plug in $x = e$. – Mark Saving Nov 03 '21 at 17:16
  • @MarkSaving Do you know the solution? I'm curious. I defined $e=a_1a_2$ and $f=a_2a_1$, applied the condition for $f$ to imply $ef=fe$. I guess $a_1a_2a_1=a_1$ could be proved or just need to find any left identity for $a_1$. – Berci Nov 03 '21 at 22:18
  • Yeah, Mark's comment is enlightening as it shows the identity must be $a_1a_2$, but it's not clear to me where to go from there. Maybe we should try to construct a counterexample? – Karl Nov 03 '21 at 22:21
  • Ah, I think I got it. It's a tricky one.. : ) – Berci Nov 03 '21 at 22:38

1 Answers1

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Apply the condition for $x=a_1$ to obtain $a_1^{n+1}a_2=a_1\ $ (1) $\,$ and set $e:=a_1^na_2$ so that $$a_1e=a_1\,.$$ Then apply for $x=ea_1$: $$ea_1=a_1(ea_1)^na_2=(a_1e)^na_1a_2=a_1^{n+1}a_2=a_1$$ by (1).

So in particular, $e$ is a left identity for $a_1$, but this suffices to prove it's a left identity for any $x$: $$ex=ea_1x^na_2=a_1x^na_2=x\,.$$ Similarly we can prove that $f=a_1a_2^n$ is a right identity for $a_2$ and hence for all $x$.

By the way, applying the condition for $x=e$ shows that $$e = a_1e^na_2 = a_1a_2\,,$$ as it was known in the comments, and similarly we get that it also equals to $f$.

Going further, applying the hypothesis for $x=a_2a_1$ we obtain $a_2a_1=e^{n+1}=e$, so actually $a_2=a_1^{-1}$.
Let $a=a_1$.

It also follows that the inverse of the automorphism $x\mapsto axa^{-1}$ is the mapping $x\mapsto x^n$, so $$x^n=a^{-1}xa$$ and $a^{n-1}=e$ by (1).
Consequently, $x^{n(n-1)}=x$ for all $x\in A$ and thus $A$ is an inverse monoid, the idempotent belonging to $x$ being $x^{n(n-1)-1}$ and the inverse of $x$ being $x^{n(n-1)-2}$.

Berci
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