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What I have in mind is something like the following: the natural numbers with addition form a monoid. You can imagine constructing the integers by taking the naturals, adding a set constructed by "reflecting" all of the non-zero natural numbers about $0$, and extending addition. What I mean by "reflecting" is that for each $i\in N$ the reflected element $-i$ is such that $i+j=k\iff -i+-j=-k$, and that $i+-i=0$.

My intuition is that this should be generally possible for monoids that don't have any inverses.

David Gudeman
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    There is a standard way to "extend" monoids to groups (at least, as much as possible). For commutative monoids this is sometimes called "the Grothendieck group of a monoid". In general, it is the "universal enveloping group" of the monoid. See here. – Arturo Magidin Jul 12 '22 at 21:45
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    A simple example where it doesn't work is $(\mathbb N^{\geq 0},\cdot ).$ There can't be a group where $ab=ac$ for some $a$ and $b\neq c.$ So you can't extend this monoid to a group. – Thomas Andrews Jul 12 '22 at 21:50
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    However, there might be an adjoint of the functor from groups to monoids, which would be a group $G$ and monoid homomorphism $M\to G$ which is "universal." The homomorphism just might not be an inclusion. – Thomas Andrews Jul 12 '22 at 21:55
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    For example, a group $G$ with a presentation containing generators the elements of $M$ and relations $(m_1)\cdot_G(m_2)=(m_1\cdot_M m_2)$ for all $m_1,m_2\in M$ will certainly be universal, but the resulting map $M\to G$ might not be an inclusion, so the word “extend” might not be accurate. – Thomas Andrews Jul 12 '22 at 22:23
  • @ThomasAndrews Such a functor does indeed exist, and explicit constructions are known. – Arturo Magidin Jul 13 '22 at 08:21

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This is possible for commutative, cancellative monoids (cancellative means $a+b = a+c$ implies $b=c$, which is automatic in a group but not a monoid). It's essentially the same as constructing $\mathbb{Z}$ from $\mathbb{N}$, as you said. But otherwise it doesn't quite work.

Ted
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