This is from Dummut and Foote Abstract Algebra, 4.5.45.
Find the generators for a Sylow $p$-group of $S_{2p}$, where $p$ is an odd prime. Show that this is abelian of order $p^2$.
I have shown that the order of the $p$-Sylow is $p^2$. ALso, I have proved that if $H$ is a $p$-Sylow, since $Z(H)$ is non-tirival and then $H / Z(H)$ is cyclic, and it follows that $H$ is abelian.
Now, I know my two options for the group of order $p^2$ that $H$ is isomorphic to are $\mathbb{Z}_{p^2}$ and $\mathbb{Z}_{p}$ x $\mathbb{Z}_{p}$. In both cases I can find generators.
My Questions:
A solution I found online ignored the possibiliy that $H$ is isomorphic to $\mathbb{Z}_{p^2}$. Is that not a possibility?
Why do they specify $p$ is an odd prime? is $p=2$ a different case?
The group $S_{2(2)} = S_4$ has elements of order $4$. But the group $S_{2p}$ with $p$ an odd prime does not. You would need the decomposition of the element into disjoint cycles to involve cycles of lengths $1$, $p$, and $p^2$, with at least one cycle of length $p^2$, and that is impossible if you are in $S_{2p}$ with $p$ odd.
In general, yes, Sylow subgroups can be cyclic. But your group here does not have elements of order $p^2$, so that possibility is excluded a priori.