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Problem

Suppose a finite set $G$ is closed under an associative product and that both cancellation law hold in $G$. Prove that $G$ must be a group. Also show by an example that if one just assumed one of the cancellation laws holds, then the conclusion need not follow.

This is from Topics in Algebra by Herstein

Specific question

How can I prove the properties of group from the both cancellation laws? In short, the identity and invertibility. If $a$ and $b$ belong to $G$ then $ab$ belongs to $G$ (closure property), but does it imply that it has an identity element.

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niladri
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    Where are you having trouble? It's best if you indicate what you tried and where you ran into problems so that people can best determine how to help. – James Aug 06 '14 at 19:28
  • How can I prove the properties of group from the both cancellation laws. In short the identity and invertibility – niladri Aug 06 '14 at 19:30
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    Since $G$ is finite, for any $a \in G$, we have ${ ab : b \in G } = G$, so in particular there exists $e \in G$ with $ae = a$. Now prove that $xe = x$ for all $x \in G$. – Derek Holt Aug 06 '14 at 19:38
  • If a and b belong to G then a.b belong to G. (closure property).but does it imply that it has an identity element. – niladri Aug 06 '14 at 19:45
  • I just gave you hint about how to do that. What do you not understand? – Derek Holt Aug 06 '14 at 19:49
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    Hint: Cancellation is true if and only if, for all $a\in G$, $L_a:G\to G$ and $R_a:G\to G$, defined as $L_a(g)=ag$ and $R_a(g)=ga$, are $1-1$. A function from a finite set to itself is $1-1$ if and only if it is onto... – Thomas Andrews Aug 06 '14 at 19:55
  • Hint for the second part: Take a set with at least two elements and define $x\cdot y = x$, for all $x,y$. You need to show it is associative, satisfies just one cancellation rule, and that it is not a group. – James Aug 06 '14 at 20:00
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    A minor quibble. This is false; the empty set is not a group. I checked and this is Herstein's fault, at least in the edition I have. – James Aug 06 '14 at 20:06
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    @James Well spotted! I am surprised Herstein made that mistake, because it's a common pitfall. – Derek Holt Aug 06 '14 at 20:20

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Hints:

Cancellation is true if and only if, for all $a\in G$, $L_a:G\to G$ and $R_a:G\to G$, defined as $L_a(g)=ag$ and $R_a(g)=ga$, are $1-1$.

And:

A function from a finite set to itself is $1-1$ if and only if it is onto...

Thomas Andrews
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