So, these are the questions:
$12$. Let $G$ be a nonempty set closed under associative product, which in addition satisfies:
a) There exists an $e \in G$ such that $a.e=a$ for all $a \in G.$
b) Give $a\in G$, there exists an element $y(a)\in G$ such that $a.y(a) = e$.
Prove that $G$ is a group.
I thought that this question is stupid since that's exactly the definition of a group. However, I was stumped after reading problem $13$. Question $13$ goes like this:
$13.$ Prove, by an example, that the conclusion of problem $12$ is false if we assume instead:
a') There exists an $e \in G$ such that $a.e=a$ for all $a \in G.$
b') Give $a\in G$, there exists an element $y(a)\in G$ such that $y(a).a = e$
.
After realising that I'll need to prove that $a.y(a)=y(a).a=e$ in question $12$, I thought:
In $12$ b), $a.y(a) = e \Rightarrow (y(a).a).(y(a).a) = (y(a).e).a = y(a).a$. So using $12$ a) $(a.e=a)$, we have $y(a).a = e$.
This part is certainly false since the same method proves that $13$ is also a group, and also it could be that $e$ may not be unique in $G$ so I cannot assert that $y(a).a=a$ (Am I correct about this?).
So, my question is what do I need to show here to prove that $G$ is a group. Not to ask too much, but I couldn't find any example for question $13$ either. I think that I'll need to find a closed set where $a.e=a$ does not necessarily imply $e.a=a$. But I'm hopeless.
