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Aug 3, 2010 at 2:22 vote accept dave
Aug 2, 2010 at 23:01 comment added Tsuyoshi Ito Thanks for the comment! I agree that this fact is easy to prove, and there is no real need to consult a textbook on semigroups. However, because the questioner asked for a good reference, I wanted to make clear that I did not provide any reference in my answer.
Aug 2, 2010 at 22:21 comment added Pete L. Clark I don't think a reference to the literature on semigroups is required here. By definition, a semigroup homomorphism is a map such that $\varphi(xy) = \varphi(x) \varphi(y)$ for all $x,y \in G$. But this is also the definition of a group homomorphism! The point is that, for groups, that this implies that the identity gets sent to the identity and that inverses get sent to inverses. If one has not done so before, it is a good exercise to think about why for ring homomorphisms, in contrast, one must require explicitly that $1 \mapsto 1$.
Aug 2, 2010 at 20:49 history edited Tsuyoshi Ito CC BY-SA 2.5
fixed typo
Aug 2, 2010 at 20:26 history answered Tsuyoshi Ito CC BY-SA 2.5