# Show that a semi-group (E,T) (I.e., with T associative) satisfying a certain property is a monoid (i.e., possesses a neutral element)

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?

• Hello, welcome Math.SE. Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be closed. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers. Mar 18 at 19:11
• What have you tried? One way to show that $E$ has an identity is to construct it. Mar 18 at 19:11
• I have taken the liberty to change your title using in particular the "technical" terms "semi-group" and "monoid" which will attract more readers, now and in the future. Mar 18 at 21:57

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.