Let $G$ be a semigroup, prove the following statement

Question

Let G be a semigroup with the following property: For any $a \in G$, there exist a unique $a^t\in G$ such that $aa^ta = a$, show that

If $x,a \in G$ and $a^tx = a^t$, then $x = aa^t$.

My thoughts

I can't think much about this question because I can barely understand it. It says, "For any a ∈ $G$", So any element of $G$ can be thought as an $a$. If $a^t$ is an element, so I can say $a^t$ is also an $a$ element of $G$, as it is an $a$ element, There will be another $a^t$ and so on.... Therefore I can say that $G$is infinite.

But the question treats a and x are distinct elements, and then associate x by saying that $x = aa^t$.

I'm totally confused, I'm not sure whether what I'm thinking is right, but I'm new to group theory, any help is appreciated.

Answer 1

Let $aa^ta=a$ and $a^tx=a^t$.

Thus, $$(aa^t)(aa^t)(aa^t)=(aa^ta)(a^taa^t)=$$ $$=a(a^taa^t)=(aa^ta)a^t=aa^t$$ and $$(aa^t)x(aa^t)=a(a^tx)(aa^t)=aa^t(aa^t)=$$ $$=(aa^ta)a^t=aa^t.$$ Thus, since for $aa^t$ there is an unique $b$ for which $(aa^t)b(aa^t)=aa^t$, we obtain $$x=aa^t$$ and we are done!

Answer 2

The general set-up means:
For each and every element of $G$, call the element of interest $a$, there is an element of $G$, called $a^t$ (because it depends of $a$) such that $aa^ta=a$ and for each $a$, there is only one $a^t$ that works.
(This tells us that $G$ is an inverse semigroup.)

You are asked to suppose that for some $a$ in $G$, and some $x$ of $G$, we have $a^tx=a^t$, then it will turn out that $x = aa^t$.

I was taught to call $a^t$ the (von Neumann) inverse of $a$ -- it is NOT a group inverse. One way to proceed: show that $aa^t$ is its own inverse. Then show that $x$ is an inverse of $aa^t$. Use uniqueness to draw the desired conclusion.