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Martin Rattigan
May25-10, 07:41 AM
P: 312
Quote Quote by mnb96 View Post
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:

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:

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 [itex]M[/itex] below (I've made it commutative and switched back to [itex]0[/itex] as the identity):

[tex]\[ \left( \begin{array}{ccccc}
& | & a & b & 0\\
-& | & -& -&-\\
a & | & a & a&a\\
b & | &a & b&b\\
0 & |&a&b&0\end{array} \right)\][/tex]

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