I need help with a proof. Also determine what it means.
Let $S\times S\longrightarrow S $ be a binary and associative operation written as $(x,y) \mapsto xy$. Assume that for each $x\in S$, there exists a unique $ x^*\in S$ satisfying $ xx^*x=x$ and $x^*xx^*=x^*$.
Let $e $ and $f $ be elements of $S $ such that $e=e\cdot e = e^2$ and $ f=f^2$. Show that $e=e^*$ and $ ef = (ef)^2$.
Fact 1: For any $x \in S$ the uniqueness requirement of $x^*$ implies that if for some $y\in S$ the identities $xyx=x$ and $yxy=y$ hold, then $y=x^*$.
Using Fact 1 since $eee=e$ it immediately follows that $e^*=e$.
Let $b=ef$, then by assumption there is unique $b^*$ such that $$bb^*b=b \text{ and } b^*bb^*=b^* \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (1)$$ i.e. $$efb^*ef=ef\text{ and }b^*efb^*=b^*. $$ It follows that $$b(fb^*)b= effb^*ef = efb^*ef = ef = b\text{ and } (fb^*)b(fb^*)=fb^*effb^*=fb^*efb^*=f (b^*efb^*) = fb^*$$ and so (by Fact 1) $fb^*=b^*$. Similarly $$b(b^*e)b = efb^*eef=ef=b\text{ and }(b^*e)b(b^*e)=b^*e$$ and so (by Fact 1) $b^*e=b^*$. Therefore $$(b^*)^2 = (b^*e)(fb^*) = b^*(ef)b^*=b^*bb^*=b^*.$$ It follows that $b^*b^*b^*=b^*$ and hence (by Fact 1) ${b^*}^*=b^*$. On the other hand ${b^*}^*=b$ by Fact 1 using (1). This means that $$ (ef)^2=b^2=({b^*}^*)^2=(b^*)^2=b^*={b^*}^*=b=ef$$ as desired.