What's "$\mathbb Z = (1)$ is cyclic"? As I understand it, $(1)$ is a cyclic group with $1$ being its generator. So, $1^n$ with $n \in \mathbb N$ generates $\mathbb Z^+$ , but what about $\mathbb Z^{-1}$ ? Do we say $(1)$ is a cyclic group with two operations $(+, -)$ defined on it so that $\mathbb Z = (1)$? 
 A: When we say that a set generates a group, it means that every element of the group is a product of these elements and their inverse. Otherwise it generally won't be a group.
A: In algebra, when you say that a set $S$ generates some structure $A$, you mean that any element of $A$ can be represented as repeated applications of operations of the structure in question to elements of $S$.
In your case, the structure is a group. A group has not only the binary operation, but also the inverse (and, technically, the identity element, considered as a nullary operation). So, $\{1\}$ indeed generates $\mathbb{Z}$, since the inverse of it (which is $-1$) and powers of inverse are also allowed.
A: Thinking about an abelian group $G$ as a $\mathbb{Z}$-module, (intuitively "as a vector space over $\mathbb{Z}$"), then cyclic means that $G$ is generated by one element $g\in G$ (as an abelian group) and this translates to $G$ being generated by $g$ as a $\mathbb{Z}$-module, which means simply
$ G = \langle \;g \;\rangle_{\mathbb{Z}} = \{n\cdot g \,; \,n \in \mathbb{Z} \}$
where $n\cdot g$ is defined as $g$ composed with itself $n$-times with respect to the group operation of $G$. If G is written additively as in the case $G=\mathbb{Z}$, then  $n\cdot g = g+\ldots + g$ exactly $n$-times.
This approach helps to understand why every group of the form $C_m = \mathbb{Z}/m\mathbb{Z}$ is cyclic with generator 1, since for all $g\in C_m$ we have $g=n\cdot 1$ for some $n\in\mathbb{Z}$. 
