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?

Not everything has an inverse in $\mathbb{Z}$ (property 3). For example, there is no element of $\mathbb{Z}$ which you can multiply with $2$ to get the multiplicative identity $1$. 
