This is a very straightforward question
Here is the problem in Pinter's book:
Let $A$ be an infinite set, and let $G$ be the subset of all permutations on $A$ (denoted as $S_A$), where $G$ consists of all the permutations $f$ of $A$ that move only a finite number of elements of $A$. Prove that $G$ is a subgroup of $S_A$.
I can see how to prove that closure is present and the inverses are present...however, I fail to see how the identity permutation $\epsilon$ is present within $G$.
The book provides the following definition for move:
We say that $f$ moves $a$ if $f(a)\neq a$
Obviously, $\epsilon$ doesn't move any elements...and therefore isn't a member of the set of moving permutations.
Am I missing something here? Does the movement of zero elements somehow "slip in" underneath the phrase finite number of elements?
The set of elements moved by the identity is empty.
The empty set has cardinality $0$.
$0$ is a finite number.
Therefore $e\in G$.