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?
 A: property 2. is OK, indeed 1 works as $e$ here.
It's property 3 that fails: if there an $x \in \mathbb{Z}$ such that $x \cdot 2 = 1$?
A: 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 $a \cdot 3 = 3\cdot a = 1$?
A: The property $(2)$ is satisfied for $(\mathbb Z, \cdot)$, because for $e=1$, you have that $1\cdot 0 = 0\cdot 1 = 0$.
Now, if property $(1)$ and $(2)$ both hold, that leaves only one suspect...
A: 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 $1$. 
A: Property (2) indeed holds, the problem is Property (3) in the case of $a=0$. 
