$p$ prime number, $n$ a non-negative integer, $G$ group.
(a) $\forall G$ with $|G| = p^{n}$ cyclic $\Leftrightarrow$ $n \leq 1$.
(b) $\forall G$ with $|G| = p^{n}$ abelian $\Leftrightarrow$ $n \leq 2$.
My try:
a) "$\Leftarrow$": $n \leq 1$, hence $G \cong \mathbb{Z}_{p}$ and hence $G$ is cyclic.
"$\Rightarrow$": Say $n = 2$. As $G$ is cyclic, $Z(G)$ is non-trivial and $\exists g\in Z(G)$, $g \not= e$.
$g$ generates the whole group $\Rightarrow$ $G \cong \mathbb{Z}_{p^{2}}$ and thus $G$ cyclic
$g$ generates a subgroup of order $p$. Then $\exists h \notin Orbit(g)$ and the Order of the Subgroup generated by $g$ and $h$ has order $p^{2}$. (Also $g$ and $h$ commute as $g \in Z(G)$ and hence $G$ is abelian.) Thus $G$ is not generated by a single element and $G \cong \mathbb{Z}_{p} \times \mathbb{Z}_{p}$. So we can find for $n>1$ groups generated by more than one element. And if we want all all groups of Order $p^{n}$ to be cyclic, $n$ has to be 1.
b) "$\Leftarrow$": Case $n=1$ and $n=2$ hold because of a) and the fact that cyclic $\Rightarrow$ abelian.
"$\Rightarrow$": Be $n = 3$. $Z(G)$ is non-trivial. So $Z(G) = p$, $p^{2}$ or $p^{3}$. For $p^{3}$ $G$ is indeed abelian. For $p^{2}$ as well. Because $|Z(G)| = p^{2} \Rightarrow |G/Z(G)| = p \Rightarrow$ $G$ is abelian.
$|Z(G)| = p$ then $Z(G) \cong \mathbb{Z}_{p}$ and $|G/Z(G)|=p^{2}$, so that $G/Z(G) \cong \mathbb{Z}_{p^{2}}$ (then $G/Z(G)$ is cyclic and hence $G$ abelian), or $G/Z(G) \cong \mathbb{Z}_{p} \times \mathbb{Z}_{p}$. Then $G/Z(G)$ is non-cyclic and $G$ non-abelian.
For $n>2$ we can always find a group $G$ with $G/Z(G)$ like this. Hence if we want all groups with order $p^{n}$ to be abelian, then $n$ needs to be $\leq 2$.
Is that OK that way? Best, Sara :)