I'm proving that $A_n$ is a subset of $S_n$. By definition, $A_n$ is the set of permutations in $S_n$ that can be decomposed into an even number of 2-cycles. Since $S_1$ has no 2-cycles, it seems that $A_1$ would have to empty.
If I'm right, then that would mean that $A_n$ is subgroup of $S_n$ only when $n\geq 2$. Since my textbook just asked to show that $A_n$ is a subgroup of $S_n$ without specifying that $n\geq 2$, I think that I'm wrong that $A_1 = \emptyset$. Some clarification would help me.
$A_1$ consists of the identity permutation, hence is equal to $S_1$. In fact, the identity permutation is contained in $A_n$ for all $n$ (this is important for the fact that $A_n$ is a subgroup of $S_n$) because it is an empty product of transpositions.