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It is true in general that the set of all invertible elements of a Monoid form a subgroup. The proof is trivial.

However, after some thought, I feel that if we restrict invertible to left or right invertible only, then it does not form a group. It seems so because I cannot imagine a way to prove otherwise.

I am looking for examples of Monoids whose left invertible elements do not form a subgroup or else proof that it does form a group.

J.-E. Pin
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user89596
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    I think that the following is the correct generalisation of your result. Remarkably, don't need monoids. Suppose $X$ is a set of elements of a semigroup $S$ such that for all $(x,y) \in X\times X$ there exists $r, s\in S$ where $xr=y$ and $sx=y$, and suppose also that there exists $x\in X$ such that $x^2\in X$. Then $X$ forms a group. The $x^2\in X$ is required, because under the other supposition either $X^2\cap X$ is empty of $X^2\cap X=X$. You are looking for an example of when $X^2\cap X$ is empty, there $X$ is the set of left-invertible elements. – user1729 Aug 08 '13 at 18:29

2 Answers2

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Take the space $V$ of polynomials in $x$ over the reals (any other field would also work). Let us look at the monoid $End(V)$.

Derivation w.r.t. $x$, call it $D$, is an endomorphism of $V$. It has a right inverse gotten by integration. More precisely, if $P(x)$ is a polynomial, let's define $$ I(P):x\mapsto\int_0^xP(t)\,dt. $$

But because $D$ is not injective (as $D1=0$) it cannot have a left inverse. Thus the right invertible elements of $End(V)$ don't form a group.

Similarly we see that $I$ has no right inverse, because it is not surjective (the constant polynomial is not in its image). So the left invertible elements don't form a group either.

Jyrki Lahtonen
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  • More simply, we could let $V$ be the space of sequences $f:\mathbb{N}\to K$, and look at the operators shifting the sequences by one position: either dropping the first entries or tagging on an extra zero. – Jyrki Lahtonen Aug 09 '13 at 04:02
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Consider the bicyclic monoid. It is the quotient of the free monoid on two letters $a$ and $b$ by the relation $ab = 1$. The elements of this monoid can be written as $b^ma^n$ for some $m, n \geqslant 0$. The product of such words is given by $(b^ma^n)(b^pa^q) = b^ra^s$ where \begin{align} r &= m - n + \max(n,p) = m + p - \min(n,p)\\ s &= q - p + \max(n,p) = n + q - \min(n,p) %\label{bicyclic} \end{align} In particular $a^nb^n = 1$ and hence every element of the form $a^n$ has a right inverse and every element of the form $b^n$ has a left inverse. However, the only inversible element is $1$.

You can also describe this monoid as $\mathbb{N} \times \mathbb{N}$ with the product given by $$ (m, n) (p, q) = (r, s) $$ where $r$ and $s$ given as above.

J.-E. Pin
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    Thanks. I had been introduced to bicyclic monoids through the $\mathbb{N*N}$ that you mentioned. I failed to understand it. In your example, you mention, "However, the only inversible element is 1 .". Since $(b^ma^n)(b^na^m)=1$, not just that but also $(b^na^m)(b^ma^n)=1$, why are they not invertible? – user89596 Aug 09 '13 at 15:52
  • Thanks for pointing out this mistake. It is hopefully correct now. – J.-E. Pin Aug 11 '13 at 19:57
  • $(b^na^m)(b^ma^n) = b^na^n \not= 1$, since only the relation $ab = 1$ holds, but $ba \not= 1$. – J.-E. Pin Aug 11 '13 at 20:03