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Let $E$ be a set with an internal operation $T$ associative such that there exist $a \in E$ such that :
$(∀y\in E) (\exists x\in E) \ y=aTxTa$
Prove that $(E;T)$ has an identity element.
What I have tried so far is setting $a=aTa'Ta$
and let $e=a'Ta$
and $y=aTxTa$ implies that $y=aTxTaTa'Ta$ and $y=aTa'TaTxTa$
so $y=yT(a'Ta)=(aTa')Ty$
So what I can do now?
Please help me.
You basically have it. Since $a \in E$, then by hypothesis there is some $a' \in E$ such that $a = a T a' T a$.
Let $y \in E$ be arbitrary. By hypothesis, there exists $x \in E$ such that $y = a T x T a = a T x T (a T a' T a)$. By associativity of $T$, we can say that $y = (a T x T a) T (a' T a) = y T (a' T a)$. In the same way, $y = a T x T a = (a T a' T a) T x T a$; by associativity of $T$, this becomes $y = (a T a') T (a T x T a) = (a T a') T y$.
What we have shown is that for arbitrary $y \in E$, it is true that
$$y = y T (a' T a) = (a T a') T y$$
If $y = a T a'$, then the equality of the first and second terms gives $a T a' = (a T a') T (a' T a)$. If $y = a' T a$, then the equality of the first and third terms gives $a' T a = (a T a') T (a' T a)$. This shows $a' T a = a T a'$, and that this element is an identity of $E$ follows immediately.
