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G is a monoid and satisfies the right inverse law. Show that G is a group. I tried next: It is obviously sufficient to show that G satisfies left inverse law. (I use $a'$ notation for $a$ inverse.) We have $\forall a\in G,\exists a'\in G,a*a'=e$.$$ae=ea$$ $$ae=(aa')a$$ $$ae=a(a'a)$$ $$e=a'a$$ But unfortunately the last step requires left inverse law. Actually I doubt that there are no counterexamples for the statement that it has to be a group. I tried to find them. I created the table for this operation which satisfies right inverse law, identity law and do not satisfies left inverse. But it's not interesting to check for associativity because for the set $G=\{a,b,e\}$ one need to check 27 cases of $(xy)z=x(yz)$

Mihail
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Let $aa'=e$; consider $a''$ such that $a'a''=e$. Then $$ a=ae=a(a'a'')=(aa')a''=ea''=a'' $$ and therefore $a'$ is both a left and right inverse of $a$.

egreg
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  • Why do solutions seem so easy in theory of group(at least at beginning) when it is not your solution. Although I tried to solve it by myself and I couldn't. – Mihail Dec 20 '14 at 23:20
  • @Mihail Been there, done that. ;-) – egreg Dec 20 '14 at 23:27