Logical Quantifiers order So for the definition of an inverse I have:
"There exists $e$ belonging to $G$ such that for all $x$ belonging to $G$,
$x*e=e*x=x$"
I'm aware the order of quantifiers matters but I can't see how the above is different from:
"For all $x$ belonging to $G$, there exists $e$ belonging to $G$ such that
$x*e=e*x=x$"
 A: Consider any $G$ with at least two elements and define $a*b=a$.
Then for all $x\in G$, there exists $e\in G$ (namely, $e=x$) such that $x*e=e*x=x$. However, there is no $e\in G$ such that for all $x\in G$, we have $e*x=x$ (because $e*x=e$)
A: First one states that for all $x$ there is an elemnt $e$ satisfing that condition but the Second one is saying any $x$ has its own $e$ with that property. These two ones can be found out through definitions of Continuity and Uniformy Continiuty as well: $$\forall x \in I \, \exists \delta > 0\, \forall y \in I \, ( \, |y-x|<\delta \, \Rightarrow \, |f(y)-f(x)|<\varepsilon,$$ $$\, \forall \varepsilon > 0\, \exists \delta > 0\, \forall x \in I\, \forall y \in I\,  ( \, |y-x|<\delta \, \Rightarrow \, |f(y)-f(x)|<\varepsilon$$
A: You've actually defined an identity element, not an inverse.
Anyway, in your second case, you're saying that the choice of $e$ is dependent upon the chosen $x$, so the identity is not unique. The first says you can find an $e$ for which any $x$ will yield $xe = ex = x$. This $e$ works for any choice of $x \in G.$
A: For example, consider the set $G=\{a,b,c\}$ with $x*y=y$ for all $x,y\in G$.
Let's first test the group axiom:

There exists an $e\in G$ such that for all $x\in G$, $x∗e=e∗x=x$

So let's check:


*

*Could it be that $e=a$? No, because $a*b=b\ne a$.

*Could it be that $e=b$? No, because $a*c=c\ne a$.

*Could it be that $e=c$? No, because $c*a=a\ne c$.


So no matter which element we choose for $e$, the condition is not fulfilled for all $x$, so the axiom is not fulfilled for my example.
Now let's check your alternative:

For all $x\in G$, there exists an $e\in G$ such that $x∗e=e∗x=x$

So we have to check all $x\in G$.


*

*For $x=a$, there exists such an $e$, namely $e=a$, as $a*a=a*a=a$.

*For $x=b$, there exists such an $e$, namely $e=b$, as $b*b=b*b=b$.

*For $x=c$, there exists such an $e$, namely $e=c$, as $c*c=c*c=c$.


So your alternative axiom is fulfilled for my example.
I hope that example clears up the difference between both statements.
