This is the first part of proof given by Lang's Algebra, pg 32.
(b) Is the statement that $A_n$ is generated by the 3-cycles.
What I don't understand in the proof:
(i) Why does it follow that "Since $\sigma$ is even, there are at least two such orbits."
(ii) "On their union, $\sigma$ is represented as a product of two transpositions."
I hope one doesn't mind checking my arguments:
My thoughts on (i): The decomposition of $J_n$ into disjoint orbits of $\langle \sigma \rangle $ is basically the cyclic decomposition of $\sigma$. If there is only one single orbit of $2$ elements, all else $1$ element, then $\sigma = (ab)$, hence odd.
My thoughts on (ii): Suppose the orbits are $\{i,j\}$ and $\{r,s\}$ then these two must be distinct. So $\sigma= (ij)(rs)\tau$. Now $\sigma$ doesn't fix $i,j,r$ or $s$, but $\sigma'$ fixes $i,j$ and possibly not $r,s$ and $k$, so in all $\sigma'$ fixes at least one more element that $\sigma$.
You're exactly right on (i), the assumption is that $\sigma$ has at least one orbit of size $2$ and the orbits not of size $2$ have size $1$. Having only one would imply that $\sigma$ is a transposition and is therefore odd.
On (ii), what you've written is a little confusing, so I think you're right, but I'll reword it. $\sigma$ must swap $i$ with $j$ and $r$ with $s$, so restricting $\sigma$ to the union $\{i,j\}\cup\{r,s\}$, $\sigma$ is represented by $(i,j)(r,s)$. $\sigma'$ fixes everything that $\sigma$ fixes except possibly $k$, but it also fixes $i$ and $j$ so has more fixed points $\sigma$.