Let $(E,⋅)$ be an associative algebraic structure. Prove that if $\forall a,b\in E$ you have $ab^2a=b$ then $|E|=1$ Let $(E,⋅)$ be an associative algebraic structure. Prove that:
If   $∀a,b∈E$ you have $ab^2a=b$ then $|E|=1$

Is my answer correct?
$$a b^2 a = b $$
$$(ab)(ba) = b$$
if $(ab)$ is inverse of $(ba)$
$$1_e = b$$
Which implies that $$1_e = a$$
The structure is formed by only the identity element, therefore $|E|=1$
Thanks in advance!
[EDIT]
I understand, what I can think of now is:
Take $a=b$
Later, $$a  a^2 a = a$$
$$(aa)(aa) = a$$
$$aa = a$$
$$a=a$$
so $a$ is idempotent
Then, I don't know if I'm on the right track.
 A: No, your proof is wrong. You do not need to have a unique unit and inverses in an associative structure. But consider:
If $ab^2a=b$ for all $a,b$ then also $b=a^2b^2a^2 = aab^2aa = aba$. Then
$$ b^2 = abaaba = baab = aa = a^2 $$
Then
$$ b = ab^2a = aa^2a = a$$
Thus either $E=\emptyset$ or $|E|=1$.
A: With a little bit of semigroup theory involving Green's relations, one can prove the following more general result.
Theorem. If a nonempty semigroup satisfies, for some $n \geqslant 0$, the identity $xy^{2n}x = y$, then it is trivial.
Proof. Let $S$ be a semigroup satisfying the identity $xy^{2n}x = y$.
Let $a, b \in S$. Since $ab^{2n}a = b$ and $ba^{2n}a = b$, $a \mathrel{\cal H} b$. Thus all elements of $S$ are $\cal H$-equivalent. Moreover, since $aa^{2n}a = a$, one gets
$$a^{2n+1}a^{2n+1} = a^{2n}a^{2n+2} = a^{2n}a = a^{2n+1}$$ and thus $a^{2n+1}$ is idempotent. Since an $\cal H$-class containing an idempotent is a group, $S$ is a group. Let $1$ be its identity. Taking $x = 1$ and $y = a$ one gets $a^{2n} = a$ and taking $x = a$ and $y = 1$, one gets $a^2 = 1$. It follows that $a = a^{2n} = (a^2)^n = 1$ and thus $S$ is trivial.
