Given a non empty set together with associative binary operation $*$ on $G$ such that
$a*x=b$ and $y*a=b$ have solutions in $G$ for all $a,b$ in $G$
To prove it is a group
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In other words: "An associative quasigroup is a group". – kahen Mar 26 '15 at 18:30
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See also this question (which has an additional assumption that the solutions for $x$ and $y$ are unique, but not all answers require that). – hmakholm left over Monica Mar 26 '15 at 18:37
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@HenningMakholm yes thanks – Taylor Ted Mar 26 '15 at 18:40
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The following equations have solutions for every $a$ and $b$ : $$\begin{align} ax&=b\tag{R_1(a,b)}\\ xa&=b \tag{R_2(a,b)} \end{align}$$
Given an arbitrary element $a$, let $e$ be a solution of $ax=a$ (solution of $R_1(a,a)$). Then for any $b$ there exists $y$ such that $ya=b$ (use $R_2(a,b)$), hence $be=yae=ya=b$. So $e$ is a right unit. Given an arbitrary element $t$ the equation $tx=e$ (use $R_1(t,e)$) has always solution, so $G$ has a right unit and every element has a right inverse. Then it is a group.
Elaqqad
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@K. Dutta Since $a,b$ are arbitrary, you can choose both of them to be the same object. – Alp Uzman Mar 26 '15 at 18:33
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@Elaqqad I am finding this convincing ,i dont know is their any other method – Taylor Ted Mar 26 '15 at 18:35
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@Uzman: the question says "binary operation $*$ on $G$" -- this wording implies an assumption that the output of the operation is always in $G$. – hmakholm left over Monica Mar 26 '15 at 18:35
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@K.Dutta yes for the first one I took $b=a$ and after that I used the second equation and to prove that any element has a right inverse I took $b=e$ – Elaqqad Mar 26 '15 at 18:42
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@K.Dutta, (I did not really understand why it should be not unique , sometimes you must explain more by showing where the contradiction comes) In the answer there is no contradiction, we defined $e$ and if you consider that an another element $e'$ verify the same assertions then we can prove that $e=e'$ (in fact the first thing we can prove in a group is there is only one unit) – Elaqqad Mar 26 '15 at 18:48
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@Elaqqad i meant why b=a=e ,doesn't show that identity is not unique ? – Taylor Ted Mar 26 '15 at 18:50
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@K.Dutta can you read the solution now ,in fact they are not the same $a$ and $b$, you have a solution which is true for every elements $a$ and $b$ and I used it for some elements – Elaqqad Mar 26 '15 at 19:04
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