I don't think that both resolutions given are complete.
Let $\sigma_1,\cdots, \sigma_s$ be the 3-cycles from $S_n$. From the given answers, it was shown that $\langle \sigma_1,\cdots, \sigma_s\rangle \subseteq A_n.$ It remains, then, to show that $A_n \subseteq \langle \sigma_1,\cdots, \sigma_s\rangle. $
Let $\alpha \in A_n$. We know that $\alpha$ can be written as a product of transpositions, and by the parity of $\alpha$ it must be the product of an even number of transpositions. Now note that the product of two transpositions is always a product of 3-cycles: indeed, if $\tau_1 = (a_1,a_2), \tau_2 = (b_1,b_2)$ are disjoint, then $\tau_1\tau_2 = (a_1b_1a_2)(b_1b_2a_1);$ and if they have an element in common, say $a_2=b_1$, then $\tau_1\tau_2 = (b_1b_2a_1).$ Now, we've shown that $\alpha$ has an even number of transpositions and since the product of two transposition is a 3-cycle, $\alpha$ is then a product of 3-cycles. Hence, $A_n\subseteq \langle \sigma_1,\cdots, \sigma_s\rangle$
In this situation, the fact that the product of two odd numbers is odd is not relevant. What is relevant is that the sum of two odd numbers is even. Let $\sigma$
and $\tau$
be odd permutations, then $\sigma$
and $\tau$
are both products of an odd number of transpositions, that is, $\sigma=t_{1}...t_{n}$
and $\tau=s_{1}...s_{m}$
where the $t_{i}$
and $s_{i}$
are transpositions and $n$
and $m$
are odd. We have $\sigma\tau=t_{1}...t_{n}s_{1}...s_{m}$
. Note that $\sigma\tau$
is a product of $n+m$
transpositions, and $n+m$ is even because $n$
and $m$
are odd. Therefore $\sigma\tau$
is an even permutation.
Best Answer
If $n\geq5$, then the only normal subgroups of the symmetric group $S_n$ are the trivial group, the alternating group and the symmetric group itself. Since the $k$-cycles form a full conjugacy class, it follows that the subgroup they generate is normal. This determines everything if $n \geq 5$.
More specifically: the $k$-cycles in $S_n$ generate the alternating group if $k$ is odd and $k \ne 1$; they generate the full symmetric group if $k$ is even.