My professor remarked that:
There is no room in $A_5$ for anything to commute. Or certainly not for elements in different Sylows to commute.
Can someone please explain why "there is no room" in $A_5$ for elements in different Sylow subgroups to commute? Does it have to do anything with the fact that $A_5$ is simple?
In $A_5$, there's no "room" for anything to commute, in the sense that, for any $\sigma,\tau$ in different Sylow subgroups, they can't consist of disjoint cycles.
There are only five elements to permute, and everything is a product of at least two transpositions, which necessarily involves three of those five elements.
But when different transpositions are not disjoint, as $(12)$ and $(13)$, they don't commute.
Since the $3$-Sylow and $5$-Sylow subgroups consist in $3$-cycles and $5$-cycles, and the $2$-Sylow subgroups consist in products of two transpositions, one can try to argue along those lines.