Is the set a group?

Question

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

1. $A \to A$ that sends $x$ to $x+a$
2. $A \to A$ 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.

Answer 1

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 $x$.

Answer 2

There must be a unique identity element:

There is a unique $e_a$ for each $a$ such that $ae_a=a$.

Now by taking the unique $c$ such that $ca=b$, we get that $cae_a=be_a$ and also that $cae_a=ca=b$, so that $be_a=b$ and thus $e_a=e_b$.

Thus we have that there is a unique right inverse. Similarly there is a unique left inverse. Now we need to show that the two are equal. But that is easy, since $e_le_r=e_r=e_l$.

Now bijectivity implies there must be a unique $x_a$ such that $ax_a=e$. And similarly there is a unique $y_a$ such that $y_aa=e$. But then $y_aax_a=x_a=y_a$.

Thus we have met the four conditions for a group, since closure and associativity are essentially given.

Answer 3

At least for finite $A$, yes, that suffices to have a group.

Call $\theta_a$ and $\gamma_a$, respectively, the left and right translation maps by a fixed element $a\in A$. Now, by assumption, $\theta_a,\gamma_a\in \operatorname{Sym}(A)$ and (associativity) $\theta_a\theta_b=\theta_{ab}$. Therefore (closure) $\{\theta_a, a \in A\}\le \operatorname{Sym}(A)$, and hence $\exists \tilde e\in A$ such that $\theta_{\tilde e}=Id_A$. Likewise, being $\gamma_a\gamma_b=\gamma_{ba}$, $\exists \hat e\in A$ such that $\gamma_{\hat e}=Id_A$; but $\tilde e=\tilde e\hat e=\hat e$ and hence the left and the right identities coincide, say $e:=\tilde e=\hat e$.

Now, since $\theta_a,\gamma_a\in \operatorname{Sym}(A)$, then $\exists(!) \tilde b,\hat b \in A$ such that $\theta_a(\tilde b)=\gamma_a(\hat b)=e$ or, equivalently, $a\tilde b=\hat ba=e$; from this latter we get e.g. $\hat ba=a\hat b$, whence $a\tilde b=a\hat b$ or, equivalently, $\theta_a(\tilde b)=\theta_a(\hat b)$, and finally $\tilde b=\hat b=:a^{-1}$.