Empty function finite support

Question

If $I=\emptyset$ is the empty set and $G$ a group with trivial element 1. Then $G^I=\lbrace f:I\to G \rbrace$ is the trivial group with the empty function as an element. Am I right that $G^{(I)}=\lbrace f:I\to G \mid f \, \textrm{finite support} \rbrace$ is just $G^I$ itself?

I want to describe the condition "$f \, \textrm{finite support}$" via quantors. How to do that? One idea was $\exists J\subset I: J \, \textrm{finite} : f|_{I\setminus J} = 1$. But this is not compatible with $I=\emptyset$, since $f|_{I\setminus J} = 1$ is not satisfied (if I am not wrong).

Answer 1

Like EH2019 did, I would like to reinterpret my attempt in the beginning. We will do the following:

$\exists J\subset I : J \, \text{is finite} : \forall x \, (x\in I\setminus J \implies f(x)=1)$.

Since $\forall x\in \emptyset$ is always false, the implication is always true.

Answer 2

How about:

A function $f:I\to G$ has finite support if there exists finite $J\subseteq I$ such that for every $x\not\in J: f(x)=1$.