Cayley's Theorem for Semigroups I've read and fully understand Cayley's theorem for groups, however when I get to the theorem for semigroups I come to a complete stop.
I've figured that the identity and cancellative properties are important but I'm struggling to link together all the different conditions. 
So if I have a semigroup that already contains an identity (a monoid) will the difference between a left and right representation change the type of homomorphism, for instance will a right representation produce a isomorphism or just a monomorphism (I have been able to show the monomorphism for an example of right representation).
Then when it comes to a semigroup that doesn't contain an identity already is the only option to fix an identity to the semigroup or does it still work without one. Again does the choice between left or right representation have any impact on the theorem? I read somewhere that the right translation may not always be one-to-one, why is this so? Is there a simple example?
 A: in essence this matter is not particularly mysterious, but it is basic, so worth taking the trouble to get a good understanding of. 
the main point to keep in mind is that the endomorphisms of a set $W$, which as a set you may denote by $Hom(W,W)=End(W)$, can be composed, and this composition is naturally associative. 
if you restrict attention to automorphisms of $W$ (endomorphisms which are monic and epic) then the structure is the symmetric group $S_W$ and any other group structure defined on $W$ can be naturally identified with the corresponding subgroup of $W_S$.
for a semigroup you have associativity, but the actions may fail to be injective or surjective. and you may or may not have an element which acts as an identity.
maybe try to restate your question in more specific terms
A: Let $S$ be a semigroup. For each element $s \in S$, let $\rho_s$ be the right translation by $s$, defined by $\rho_s(x) = xs$. To see the problem you can encounter with right translations, consider the semigroup $S_n = \{a_1, \dotsm, a_n\}$, with multiplication given by $xy = x$ for all $x, y \in S_n$.
Here you have $\rho_s = Id$ for all $s \in S_n$, since $\rho_s(x) = xs = x$ for all $x \in S_n$. Thus each right translation is the identity and thus the map $s \to \rho_s$ is not injective if $n > 1$. Moreover the semigroup generated by all right translations is the trivial semigroup.
