$ (\mathbb{Z}, +) $ is a cyclic group. But who generates $0$ (zero) element? I know that $ (\mathbb{Z}, +) $ is a cyclic group. It is generated by $1$ or by $-1$.

But who generates the $0$ (zero) element?

Thank you!!!!
Every element in $\mathbb{Z}$ can be obtained from a generator $a$ by the groups law, which is addition. However, we also can use the inverse, which is $-a$. This way we obtain every integer for $a=1$ or $a=-1$. In particular we obtain $0=1+(-1)$. For the definition of a "generator" for a group and examples, see here. Often the group law is written $a\circ b$, but for abelian groups the conventions is usually to write addition, i.e., $a\circ b=a+b$, so that $a^{-1}$ is $-a$, and the neutral element $e$ is $0$. The multiplicative version of your question is, how to obtain $e$: as $a\circ a^{-1}$.
Remember the definition a cyclic group,
If a group $G$ is cyclic and $x$ is its generator then,
$G=\langle x\rangle=\{x^n|n\in\mathbb{Z}\}$
Similarly for an additive group,
$(G,+)=\langle x\rangle=\{nx|n\in \mathbb{Z}\}$.

Clearly, $(\mathbb{Z},+)$ is cyclic and $\mathbb{Z}=\langle1\rangle=\langle-1\rangle$
And $0\in\mathbb{Z}$ also generated by 1 or -1
since, $0\cdot1=0$ or $0\cdot(-1)=0$

Hope it's clear now.