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## Non-inversible monoids

It is a theorem that every commutative monoid is inversible, i.e. is isomorphic with a submonoid of a(commutative) group. It is also clear that a group contains all submonoids generated by any subset of its underlying set. but it is also known that non every monoid is inversible, i.e. cannot be isomorphically imbedded inside some group. Question:Are there simple examples of non-inversible monoids ? Gérard Lang

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An obvious obstruction is that $ab = ac$ implies $b = c$ and similarly $ba = ca$ must imply $b = c$. This is not true for the $2 \times 2$-matrices, for example. – Theo Buehler Jan 9 2011 at 19:23
Cher Gérard, believe it or not, in English they write inver*t*ible. I think we should generously let them indulge their idiosyncrasy :-) – Georges Elencwajg Jan 9 2011 at 19:32
I don't follow. The free monoid on an element a satisfying a^2 = a is not a submonoid of any commutative group. – Qiaochu Yuan Jan 9 2011 at 19:39
I agree with Qiaochu. Are you assuming the monoids are cancellative, but forgot to insert the hypothesis? – S. Carnahan Jan 9 2011 at 19:45
Scott/Qiaochu: Cancellative would make sense. From Wikipedia "A commutative monoid with the cancellation property can always be embedded in a group via the Grothendieck construction." and "However, a non-commutative cancellative monoid need not be embeddable in a group.". (en.wikipedia.org/wiki/Monoid#Properties) – George Lowther Jan 9 2011 at 19:50

A complete description of monoids embeddable into groups was given by Malcev more than 70 years ago. The description is in a form of infinitely many quasi-identities, the easiest of those are the two cancelative laws. See Chapter 12 of Clifford, A. H.; Preston, G. B. The algebraic theory of semigroups. Vol. II. Mathematical Surveys, No. 7 American Mathematical Society, Providence, R.I. 1967 xv+350 pp.

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 Thank you very much for this reference that plainly answers my question. Gérard Lang – Gérard Lang Jan 10 2011 at 9:13

An example is given here: http://mathoverflow.net/questions/29006/counterexamples-in-algebra/29076#29076

To write it out here, take the monoid $\langle a,b,c,d,x,y,u,v : ax=by, cx=dy, au=bv \rangle$.

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 Thank you very much for this interesting answer to my (very) question. Gérard Lang. – Gérard Lang Jan 10 2011 at 10:22

Let $M$ be a totally ordered set of cardinal $\ge3$ with a least element $\bot$, and we define a binary operation $x\cdot{}y=\max(x,y)$.

This is obviously a commutative monoid with $\bot$ as the identity element, but it is not invertible because if $c\lt{}b\lt{}a$ you have $a\cdot{}b=a\cdot{}c$ but $b\neq{}c$.

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