Let $A$ be an infinite set, and let $H$ be the set of all $f\in \mathcal{S}(A)$ such that $f(x)=x$ for all but a finite number of elements $x$ of $A$.
Prove that $H$ is a subgroup of $\mathcal{S}(A)$.
Note: $\mathcal{S}(A)$ denotes the set of all permutations of $A$.
So clearly, $H$ is nonempty since it contains the identity function of $A$.
Next, let $f,g$ be element of $H$. So we need to prove $f\circ g^{-1}\in H$.
Here, I can't understand the keyword that is "for all but a finite number of elements $x$ of $A$". Does this means that it may happen the case where $f(x)=x $ but $g(x)\neq x$?
"For all but a finite number of elements $x$ of $A$" means that the number of elements $x \in A$ such that $f(x) \neq x$ is finite, say $x_1,\dots,x_{n(f)}$ (we could have $n(f)=0$). This holds for $f \in H$, as well as for $g \in H$.
We just have to prove that $f \circ g^{-1}(x) \neq x$ has only finitely many solutions in $A$. This will prove $f \circ g^{-1} \in H$, as desired.
We know that $f(x)=x$ for all $x \in A \setminus \{x_1,\dots,x_{n(f)}\}$ and $g(x)=x$ for all $x \in A \setminus \{y_1,\dots,y_{n(g)}\}$. Therefore, $x=f^{-1}(x)$ for all $x \in A \setminus \{x_1,\dots,x_{n(f)}\}$ and $x=g^{-1}(x)$ for all $x \in A \setminus \{y_1,\dots,y_{n(g)}\}$.
I let you prove that $f \circ g^{-1}(x) = x$ for all $x \in A \setminus \{x_1,\dots,x_{n(f)},y_1,\dots,y_{n(g)}\}$.