Suppose that $\displaystyle G$ is a set with an associative binary operation such that
Given $\displaystyle a,y\in G,$there exists $\displaystyle x\in G$ such that $\displaystyle ax=y$; and
Given $\displaystyle a,w\in G$, there exists $\displaystyle u\in G$ such that $\displaystyle ua=w$
Then it is to be proven that $\displaystyle G$ is a group.
I started as follows: By (1), there exists $\displaystyle x\in G$ such that $\displaystyle ax=a$ and by (2),there exists $\displaystyle u\in G$ such that $ $$\displaystyle ua=a$.
We have $\displaystyle ax=( ua) x=u( ax)$ and now by property (1), there exists $\displaystyle t\in G$ such that $\displaystyle ( ax) t=x$.
It follows that \begin{equation} x=( ax) t=( u( ax)) t=u(( ax) t) =ux\ \ \tag{A} \end{equation} $\displaystyle \ \ $Similarly, \begin{equation} ua=u( ax) \Longrightarrow u=ux \tag{B} \end{equation} By (A) and (B), we get $\displaystyle u=x$ and we conclude that $\displaystyle \forall a\in G,\exists x\in G\ ( ax=xa=a)$
Now I claim that $\displaystyle x$ is the universal identity.
For any $\displaystyle b\in G,\exists y\in G$ such that \begin{equation} by=yb=b \tag{B'} \end{equation}
Now by using properties (1) and (2), we have \begin{gather} xb=xyb\Longrightarrow xy=xy^{2} \Longrightarrow y^{2} =y^{3} \Longrightarrow x=xy\Longrightarrow y=y^{2} \tag{C}\\ bx=byx\Longrightarrow yx=y^{2} x=y( yx) \Longrightarrow x=yx=xy \tag{D}\\ ya=yxa\Longrightarrow yx=yx^{2} =( yx) x=( xy) x=x( yx) \Longrightarrow y=xy=x \tag{E} \end{gather}
Therefore, for any arbitrary $\displaystyle b\in G$, we have $\displaystyle bx=xb=b$. Therefore, by definition of identity $\displaystyle x\in G$ is the identity. Let's denote $\displaystyle x$ by $\displaystyle 1$.
Existence of inverse: By properties (1) and (2), for any $\displaystyle b\in G,\exists ( t,s) \in G^{2}$ such that $\displaystyle bt=1$ and $\displaystyle sb=1$
\begin{equation*} t=1t=sbt=s( bt) =s( 1) =s \end{equation*} By definition of inverse elements, every $\displaystyle b\in G$ has an inverse $\displaystyle s$. Let's denote it by $\displaystyle b^{-1}$.
Explanation for (C): $\displaystyle xb=x( yb)$ (by (B'))
By property (1), there exists $\displaystyle g\in G$ such that $\displaystyle bg=y$. It follows that $\displaystyle xbg=xy=( x( yb)) g=xy^{2} \Longrightarrow xy=xy^{2}$ and then by property (2), there exists $\displaystyle g'\in G$ such that $\displaystyle g'x=y$ whence it follows that $\displaystyle y^{2} =y^{3}$ etc.
Similarly for (D) and (E).
Is my proof correct? Thanks.
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