Let $G$ be a nonempty set closed under an associative product, which in addition satisfies :
A. There exists an $e$ in G such that $a⋅e=a$ for all $a∈G$.
B. Given $a∈G$, there exists an element $y(a)∈G$ such that $a⋅y(a)=e$.
Prove that $G$ must be a group under this product.
This is a question from I N Herstein. It was already asked before. Here is the solution
https://math.stackexchange.com/a/1200617/581242
Let, $ab=e\land bc=e\tag {1}$ for some $b,c\in G$. And, $ae=a\tag{2}$ From $(2)$, $$eae=ea\implies(ab)a(bc)=ea\implies ((ab)(ab))c=ea\implies ec=ea\tag{3}$$
Similarly, $$ae=a\implies a(bc)=a\implies (ab)c=a\implies ec=a\tag{4}$$
$(3)$ and $(4)$ implies, $$ea=a$$
Also from $(3)$ and $(1)$, $$(bab)(bca)=e\implies b((ab)(bc)a)=e\implies ba=e$$
I don't understand the notation used in first line. What does $$ab=e\land bc=e$$ mean?