Is the set a group? Let $A$ be a non-empty set, $+$ is an associative binary operation and for each $a$ in $A$, the two maps

*

*$A \to A$ that sends $x$ to $x+a$

*$A \to A$ that sends $x$ to $a+x$
are bijective maps.
The question is determine whether $(A, +)$ a group or not?
My understanding that $A$ is a group if it has an identity and every element has inverse, but I am not quite sure how use these bijective maps to prove/disprove that $A$ is a group.

Neutral element (identity)
Fix an arbitrary element $a$. Since the map $x \to a + x$ is bijective, the element $a$ has exactly one preimage under this map, i.e. there exists a unique element $e$ such that $a + e = a$.
The next step is to prove $\forall y: y + e = y$. Pick an arbitrary $y$. By bijectivity of the map $x \to x + a$ there exists an $x$ such that $x + a = y$. Now, adding $x$ on the left to the equality $a + e = a$ (and using associativity) we get $y + e = y$, qed.
So, $e$ is a right neutral element. Then note that $e + e = e$, and by the same argument as above $e$ is also a left neutral element.
Inverses
Finally, we need to prove the existence of inverses. Pick an arbitrary $x$. By the surjectivity of left and right addition, there exist elements $y_1$ and $y_2$ such that $y_1 + x = e$ and $x + y_2 = e$. Now note that
$$
y_1 = y_1 + e = y_1 + (x + y_2) = (y_1 + x) + y_2 = e + y_2 = y_2.
$$
Therefore, $y_1$ (which is also $y_2$) is an inverse for $x$.
