Prove that every abelian cancellation semigroup can be imbedded in a group For an abelian cancellation semigroup $G$ and a group $F$, I think it is sufficient to prove that a homomorphism $f: G \to F$ must be injective. Following that line of reasoning, I want to say something like:
For $a,b,c \in G$ s.t. $a \neq b \neq c$ and $f(a)=f(b)$, 
$$
f(ac) = f(a)f(c) = f(b)f(c) = f(bc)
$$
and
$$
f(ca) = f(c)f(a) = f(c)f(b) = f(cb)
$$
implies that $ac = bc$ and $ca = cb$, a contradiction, because the cancellation laws imply that $a=b$.
However, the final step presupposes the injectivity of $f$. What's the right way to go about this proof?
 A: You want to construct the universal enveloping [abelian] group (in the commutative case, also known as the Grothendieck group of the subgroup/monoid). 
Let $S$ be an abelian semigroup, and let $F$ be the free abelian group on the underlying set of $S$. Denote the basis element of $F$ corresponding to the element $s\in S$ by $\overline{s}$, so that elements of $F$ consist of sums of the form
$$\sum_{s\in S}a_s\overline{s}$$
with $a_s\in\mathbb{Z}$, $a_s=0$ for almost all $s\in S$.
Let $N$ be the subgroup of $F$ generated by the elements of the form $\overline{st}-\overline{s}-\overline{t}$ (which give you the “multiplication table of $S$” as defining relations). Let $G(S)=F/N$.
Note also that $k\overline{s}+N = \overline{s^k}+N = (\overline{s})^k +N$. This is a bit awkward, but comes from using multiplicative notation for $S$ and additive notation for $F$.
There is a semigroup homomorphism $G\to G(S)$ given by $s\mapsto \overline{s}+N$. Indeed, $\overline{st}+N = (\overline{s}+\overline{t})+N = (\overline{s}+N)+(\overline{t}+N)$. Moreover, every element of $G(S)$ can be written as $\overline{s}-\overline{t} + N$, with $s,t\in S$. Indeed, taking an arbitrary element of $F$, write it as a sum of positive and negative terms, and then combine all positive terms into $s$ modulo $N$ and the negative ones into $t$ modulo $N$.
All of the above can be done with an arbitrary semigroup. 
Now prove that $\overline{s}+N = \overline{t}+N$ if and only if there exist $x\in S$ such that $sx = tx$. 
And now, if you are actually working in a cancellation semigroup, you see that this last condition yields that we must have $s=t$, and so that the universal map $S\to F/N$ is actually an embedding. 
You can see a discussion of this (plus references, including one to a paper of Mal’cev to deduce necessary and sufficient conditions for embeddability), in George Bergman’s Invitation to General Algebra and Universal Constructions, page 80, Section 4.11.
A: Arturo's answer is already very nice, but here's a few further references which may be of interest, especially since they tie in with the general question of embedding cancellative semigroups in groups.
First, a direct proof with all the gory details written out can be found in A. Cain's beautiful Nine Chapters on the Semigroup Art.
Alternatively, one may use Ore's Theorem: any right-reversible cancellative semigroup embeds in a group. A right-reversible semigroup $S$ is one in which any two principal left ideals intersect: $Sa \cap Sb \neq \varnothing$ for all $a, b \in S$. Now if $S$ is commutative, then $Sa \cap Sb$ will always contain $ab = ba$. It remains to see that one can actually make abelian the group in which one places one's semigroup: by [Dubreil, Paul, Sur les problèmes d'immersion et la théorie des modules. (French) C. R. Acad. Sci. Paris 216 (1943), 625–627], a cancellative right-reversible semigroup can be embedded in a group of left quotients of $S$, which is readily checked to be abelian if $S$ is, and we are done. For further discussion on this last point, see Clifford-Preston pp. 34--37.
Finally, the result can actually be strengthened to say: every ordered cancellative commutative semigroup can be embedded in an ordered abelian group. This can be found in [Kehayopulu et al, The Embedding of an Ordered Semigroup into an Ordered Group by Means of Pseudoorder, J. Math. Sci. 95:2 (1999)].
