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 Quote by mnb96 Now this discussion made me feel even more confused about a wikipedia article on the Grothendieck Group (the group which embeds an abelian semigroup or monoid). Here it is said that a commutative monoid must have the property "for all a,b a+k=b+k for some k" in order to be embedded in a group: http://en.wikipedia.org/wiki/Grothen...t_construction Instead, in this other article it is said that a necessary and sufficient condition for embedding a commutative semigroup in a group is that the semigroup must be cancellative: http://en.wikipedia.org/wiki/Cancell...lity_in_groups This is very confusing.
I couldn't find the quote you referred to in the first link, the nearest I found was the specification of the equivalence relation used for the construction viz.

"We say that (m1, m2) is equivalent to (n1, n2) if, for some element k of M, m1 + n2 + k = m2 + n1 + k."

Note that the article doesn't say that the commutative monoid M is necessarily embedded in the constructed group (though this will be the case if it's a cancellation monoid - in this case the equivalence relation reduces to m1 + n2 = m2 + n1).

For example if you look at the table for a monoid $M$ below (I've made it commutative and switched back to $0$ as the identity):

$$$\left( \begin{array}{ccccc} & | & a & b & 0\\ -& | & -& -&-\\ a & | & a & a&a\\ b & | &a & b&b\\ 0 & |&a&b&0\end{array} \right)$$$

this is not a cancellation monoid because e.g. $b+a=0+a$ and $b\neq 0$. For any ${(m_1,m_2),(n_1,n_2)\in M\times M}$ we have ${m_1+n_2+a=m_2+n_1+a}$ so ${(m_1,m_2)\sim(n_1,n_2)}$ and there is only one equivalence class, hence only one member in the Grothendieck group. So $M$ (with 3 members) obviously can't be embedded in this case.