Why does $1 \cdot 0=0$ not stand? A set $G$  together with an operation $*$ is called group when it satisfies the following properties:


*

*$a*(b*c)=(a*b)*c,  \forall a,b,c \in G$

*$ \exists e \in G: e*a=a*e=a, \forall a \in G$

*$\forall a \in G \exists a' \in G: a'*a=e=a*a'$
$$$$
$$(\mathbb{Z}, \cdot ) \text{ is not a group}$$


The property $(1)$ is satisfied.
For the property$(2)$ we take $e=1$. But while $1 \cdot a = a \cdot 1 =a, \forall a \in \mathbb{Z} \setminus \{0 \}$, it does not stand that $1 \cdot 0=0$.
I haven't understood why $1 \cdot 0=0$ does not stand...Could you explain it to me?

There isn't any failure in terms in property $(2)$. $1$ is certainly the identity, and it does stand that $0\times 1 = 1\times 0 = 0$.
But, zero creates another problem:  
Consider property $(3)$ asserting that for every element in a group, its inverses exists and is in the group, too. This is where things "go bad" for $0$, and essentially every element in $\mathbb Z$ that is not $-1$ or $1$:
Is there any $a$ such that $a \cdot 0 = 0\cdot a = e = 1$?
Is there any $a \in \mathbb Z$ such that $a \cdot 3 = 3\cdot a = 1$?
