A problem about group with functions as elements 
We define $+$ on set $G=\{ f|f:Z \rightarrow Z/(2) \}$:
  $$(f+g)(x)=f(x)+g(x)$$

I want to prove that $\langle G,+ \rangle$ is commutative and nonzero element has order 2.
My Doubts:
1. What does the range of $f$, $Z/(2)$ means? How to understand such a function definition?
2. The proof of commutative seems straight forward. But how to prove the nonzero part? Or what exactly is the zero element?
Thank You :)
 A: As mentioned in the comments $Z/(2)$ probably means $\mathbb{Z}/2\mathbb{Z}$ i.e. the integers modulo $2$. It is the unique (up to isomorphism) group of order $2$.
So $G$ consists of all functions, $f$, from $\mathbb{Z}$ to $\mathbb{Z}/2\mathbb{Z}$. Thus the elements of $G$ are functions. 
Now to make $G$ into a group we have defined an operation that takes two functions, $f$ and $g$ and returns a new function $f + g$. How does this new function work? Well, on input $x \in \mathbb{Z}$ it takes the value $f(x) + g(x)$. 
It seems you have already proven this is a group. 
In order to to this you must find an identity element, i.e. an element (which is a function!) such that adding it to all other elements does nothing. This is the zero function, $zero : \mathbb{Z} \to \mathbb{Z}/2\mathbb{Z}$ which is defined by $zero(x) = 0$.
Now what is an element of order $2$ in a group? It is an element $f$ such that $f + f$ equals $zero$ (the identity).
Let $f \in G$ be non-zero (i.e. different from $zero$). Then we must show that $f+f = zero$. How does one show two functions are equal? By showing they have equal values at all points.
So we evaluate
\begin{eqnarray}
(f+f)(x) = f(x) + f(x) = 0
\end{eqnarray}
Since $f(x)$ is an element of $\mathbb{Z}/2\mathbb{Z}$ and all (non-zero) elements of $\mathbb{Z}/2\mathbb{Z}$ have order $2$. Since the is some $x \in \mathbb{Z}$ such that $f(x) \ne 0$ it follows that $f$ has order $2$.
