Prove that a monoid with property $ (a * a' = e) \lor (a' * a = e) $ is a group

Question

Let $(G , *)$, $G \neq \emptyset$ be closed, associative, there is an identity element $ e $ in $ G $ and: $ (\forall a \in G)(\exists a' \in G) $ $ (a * a' = e) \lor (a' * a = e) $

Prove that G is a group.

---

Let $a * a' = e$ and $ a' * a = b \in G $ $$ b = a' * a / * b $$ $$ b * b = (a' * a) * (a' * a) $$ $$ b * b = a' * (a * a') * a $$ $$ b * b = a' * a = b $$

$$ b * e = b * b $$ $$ (a' * a) * (a * a') = (a' * a) * (a' * a) $$ $$ (a' * a) * e = (a' * a) * (a' * a) $$

Now I can see that:

$$ b * (a' * a) = (a' * a) * b = b \;\;\;\;\;\;\;\;\; b * e = e * b = b $$

But is $ (a' * a) = e $ here? I know that $ (a' * a) $ is an identity element for $\forall b \in G$, but does b represent every element in G? Is this proof valid.

Answer 1

I think that $$b * b = (′∗)∗(′∗)$$ $$ b*b = a' * e * a $$ $$b*b = b$$ Since $b \in G$ we can use the hypothesis and $\exists b' \in G$ such that $b * b' = e$, so

$$b*b* b' = b* b'$$ $$a'a = b = e$$

And this argument is completely general since $a,b$ could be every element of $G$