I have posted a proof below and would appreciate it if someone could review my solution. Thanks!
Prove that if $A = \{1,2,3,…\}$, where $A$ is the natural numbers, then S$_A$ is an infinite group.
Proof:
Let $A = \{1,2,3,…\}$.
Then $S_A$ is the set of all permutations of $A$.
$S_A$ is clearly non empty since it is a group.
Then suppose $S_A$ was finite.
Let $P = \{p_1, p_2, \cdots , p_n \}$ where $p_i$ is an element of $S_A$ and $P$ contains all elements of $S_A$.
Then let $x = 1$. Let $J = \{ p_i(x)\ |\ p_i \in P\}$. Then $J$ is finite since $P$ is finite and so $J$ has a max value that is attained in $J$. Take some $y\gt j$ for all $j \in J$. Then define the bijection $f(1) = y$ and $f(y) = 1$, and the identity map for all other input values for $f$. The cycle decomposition would be: $(1\ y)$.
Then $f$ is permutation that is not contained in $P$ since no permutation in $P$ maps $1 \to y$. Hence we have a contradiction that $P$ is the set of all permutations from $A \to A$.
Hence $S_A$ cannot be finite. Hence $S_A$ must be infinite.
Best Answer
The proof is correct, but it would be more natural and simpler to note that, if $n\in\mathbb N\setminus\{1\}$, the permutation $(1\ \ n)$ belongs to $S_A$. Therefore, $S_A$ is infinite.