Let $G$ be the set of bijective maps $\mathbb{Z} \rightarrow \mathbb{Z}$ that move finitely many integers. Then $G$ is a group under the operation of function composition.
My question is, what is this group?
It seems to me that it's a group that contains infinitely (countably) many copies of $S_n$, for each $n$.
Is this accurate? Would that make $G$ a countable, nonabelian group?
[For reference, this is example 2.2.5 in Herstein's "Topics in Algebra".]
EDIT: yes, the maps are bijections.
This group definitely contains infinitely many copies of each $S_n$, as we can view the countably many sets $$X_k = \{kn, kn+1,\cdots, kn+n-1 \}$$ and $G$ will act on this set as $S_n$ would.
Now this fact definitely implies that $G$ is nonabelian, but we still need to verify that it is countable.
To do this, we can note that since each $\sigma\in G$ moves only a finite number of integers, the integers moved by $\sigma$ will be included in some set $Z_N = \{a\in \mathbb{Z} : -N < a < N \}$ for sufficiently large $N$.
Since there are finitely many elements of $G$ which act only on each $Z_N$, we can thus write all the elements as a countable union of finite sets, and thus $G$ is countable. Formally, we'd write $$G = \bigcup_{n=0}^\infty \{\sigma\in G : \sigma\mid_{\mathbb{Z}\setminus Z_n} = \text{id}_{\mathbb{Z}\setminus Z_n} \}$$ And since each set in the union is finite, $G$ is countable, thus verifying your claims!