Minimal Set of group axioms Consider the following definition of a group:
A group $(G,\bullet)$ is
1.) A non-empty set $G\neq \emptyset$.
2.) A map $\bullet: G\times G\to G$ such that


*

*For all $g_1, g_2,g_3\in G$ we have 
$g_1\bullet(g_2\bullet g_3)=(g_1\bullet g_2)\bullet g_3$.

*There exist an $e\in G$ for all $g\in G$, such that: $$e\bullet g= g$$ $$g\bullet e= g$$

*For all $g\in G$ there exists a $f\in G$, such that: $$g\bullet f=e$$ $$f\bullet g=e$$   
Is this definition of a group "minimal" in the sense, that we can not forget about any of the previously stated properties? In other words: Are these properties independend or are some really consequences of the others? In particular I'm not exactly sure, weather or not we really need both $e\bullet 
 g= g$ AND $g\bullet e=g$.
 A: Firstly I think it is more sensible to think of a group as a triple where $e$ is part of the structure, since otherwise (before proving $e$ is unique) it is not clear which $e$ one refers to in the axiom defining inverses.
After that, we can show that the following is a minimal set of axioms for a triple $(G,\cdot,e)$, where $G$ is a set and $\cdot$ is a map $G\times G\to G$ and $e$ is an element of $G$, to be a group:


*

*For all $a,b,c$ in $G$, $a\cdot(b\cdot c)=(a\cdot b)\cdot c$;

*For all $g$ in $G$, $g\cdot e=g$;

*For all $g$ in $G$ there exists $h$ in $G$ such that $g\cdot h=e$.
Proof: Let $g$ be an element of $G$. By assumption there exists $h$ in $G$ such that $g\cdot h=e$. But since $h$ is $G$ there exists $k$ such that $h\cdot k=e$. We have $h\cdot g=h\cdot g\cdot e= h\cdot g\cdot (h \cdot k)= (h\cdot e)\cdot k= h\cdot k=e$ and hence $e\cdot g= (g\cdot h) \cdot g=g\cdot (h \cdot g)=g \cdot e=g$. This proves $(G,\cdot,e)$ is a group. It remains to establish minimality. For any set $S$ with at least two elements one of which we denote by $e$ we have: 


*

*$(S,*,e)$, where $*$ is defined for $a,b$ in $S$ by $a*b=e$, satisfies 1 and 3 above, but not 2.

*$(S,\cdot,e)$, where $\cdot$ is define for $a,b$ in $S$ by $a\cdot b=a$, satisfies 1 and 2, but not 3.

*$(S,@,e)$, where $@$ is defined by for $a,b$ in $S$ by $a@b=
\begin{cases}
a & \text{if $b=e$}\\
e & \text{otherwise}
\end{cases}$, satisfies 2 and 3 but not 1.
A: Th briefest axiom system I ever saw for a group is an associative binary operation (written as  multiplication) on a non-empty set $G$ such that
for all $x,y \in G$ $$\exists! z\, (xz=y)$$ and $$\exists! z'\, (z'x=y).$$
A: The axiom $g\bullet e=g$  (or $e \bullet g = g$) can be deduced from the others.
$$g\bullet e= g\bullet (f\bullet g)==(g \bullet f) \bullet g= e \bullet g = g.$$
But we cannot drop the whole axiom 2.2 because the structure
*|g e
-+---
g|e e
e|e e

is associative and an inverse exists but $e$ does not satisfy (2.2).
A: You can actually remove both 2.2.2 and 2.3.1 (or symmetrically 2.2.1 and 2.3.2), and this is optimal.

A group is a pair $(G,\cdot)$ such that:
  
  
*
  
*$\cdot : G\times G\to G$
  
*For all $x,y,z\in G$, $(x\cdot y)\cdot z=x\cdot(y\cdot z)$
  
*There exists $e\in G$ such that:
  
  
*
  
*For all $x\in G$, $e\cdot x=x$
  
*For all $x\in G$ there exists $y\in G$ such that $y\cdot x=e$.
  
  

In this case, $e$ and the inverse operation are unique and satisfy all the standard axioms of groups. Note that the quantifier over $e$ extends also to the definition of inverses, so it need not be in the structure. Let $e$ be given as above.
First a lemma: If $z\cdot z=z$ then $z=e$. Let $w\cdot z=e$. Then
$$z=e\cdot z=(w\cdot z)\cdot z=w\cdot (z\cdot z)=w\cdot z=e.$$
Now let $x\in G$. Let $y$ be such that $y\cdot x=e$. We will show that $x\cdot y=e$ and $x\cdot e=x$, which are the missing rules from this axiomatization.
To show $x\cdot y=e$ it suffices by the lemma that $(x\cdot y)\cdot(x\cdot y)=x\cdot y$:
\begin{align}
(x\cdot y)\cdot(x\cdot y)&=x\cdot (y\cdot(x\cdot y))\\
&=x\cdot ((y\cdot x)\cdot y)\\
&=x\cdot (e\cdot y)\\
&=x\cdot y\\
\end{align}
Thus $y$ is a two-sided inverse, and we can show $x\cdot e=x$:
$$x\cdot e=x\cdot (y\cdot x)=(x\cdot y)\cdot x=e\cdot x=x.$$
