Generators of group $\mathbb{Z}_{4}$ Hello and sorry in advance for any mistakes, English isn't my first language.
I recently started studying group theory for my university and I got introduced to cyclic groups. As an example, my book provides the group $\mathbb{Z}_{4}$ and says that numbers $1$ and $3$ are it's generators. Now please correct me if I am wrong, but wouldn't number $1$ being generator of $\mathbb{Z}_{4} \}$ mean that $\{1^n \mid n \in \mathbb{Z} \} = \mathbb{Z}_{4}$ ?? 
I can't understand why $1$ is generator of $\mathbb{Z}_{4}$. I'm assuming what i wrote above is correct, if not please correct me.
 A: When we say the group $\mathbb{Z}_4$, we're actually talking about $(\mathbb{Z}_4, +)$, meaning the operation over the group is $+$, not $\times$. Thus, 1 is a generator, because every element of $\mathbb{Z}_4$ can be written as $n \times 1= 1+1+1 + \dots$
Now you may ask, why wouldnt we give $\mathbb{Z}_4$ the $\times$ operation ? Well, what would be the inverse if 0 ? Of 2 ?
A: The number $1$ generates $\mathbb{Z}_4$ because $\mathbb{Z}_4=\{1,1+1,1+1+1,1+1+1+1\}$.
A: Formally speaking, we define a group to be an ordered pair of a set $G$ coupled with a binary operation $*$ and satisfies the group axioms, which we write as $(G,*)$. In this case, our set is $\mathbb{Z}_4$ (set of residues modulo $4$) and our operation is $+$, so we can write the group as $(\mathbb{Z}_4,+)$. It is imperative to specify the binary operation on the set, as it is intrinsic to the definition of group.
In general, it is easy to see that for any additive group $\mathbb{Z}_n$ with $n \in \mathbb{Z}_{\geq 2}$, $1$ must be a generator as we can execute the addition operation $1, 2, ..., n-1$ times, where at the $(n-1)$-th iteration, we reduce modulo $n$ to get $0$. 
