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Suppose a finite set $G$ is closed under associative product and that both cancellation laws hold in $G$. Prove $G$ must be a group.

I somehow need to prove identity, inverse, that closure holds to prove that set is a group.

How do i begin? Hints?

Please mention the ideas behind the proof?

Thanks.

EDIT

The proof as posted in link below is

Prove that this is a group

I have doubts regarding this proof

  1. There exists a element $e$ of G such that $f(e)=a$? Why does it exist?

  2. Now as the function is surjective there exists an element $aa_{R} =e$? Why does it exist? What is the role of surjectivity here?

Thanks.

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  • $\begingroup$ @JyrkiLahtonen I have edited question .Can you unmark question as Duplicate $\endgroup$ – Taylor Ted Mar 22 '15 at 9:59
  • $\begingroup$ Done. Sorry about the 1 hour delay. Deleting the comments that are no longer needed. $\endgroup$ – Jyrki Lahtonen Mar 22 '15 at 11:16
  • $\begingroup$ He chose $f(g)=ag$ because it works. But, really, what else could you even try? You are trying to show that there's an identity, so, given $a$, you want to look at all the products $ag$ to see whether you can show one of them is $a$. $\endgroup$ – Gerry Myerson Mar 22 '15 at 11:52
  • $\begingroup$ @GerryMyerson How should i think if i were to seek an alternate proof for this ? $\endgroup$ – Taylor Ted Mar 22 '15 at 16:48
  • $\begingroup$ @GerryMyerson Without defining function $\endgroup$ – Taylor Ted Mar 22 '15 at 16:48

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