$G$ abelian when $Z(G)$ is a proper subset of $G$?

Question

These two exercises confuses me:

Let $G$ be a group of order $p^3$, where $p$ is a prime

1. Show that if $|Z(G)| = p^2$ then $G/Z(G)$ is cyclic

2. Show that if $G/Z(G)$ is cyclic, then $G$ is abelian

What? Isn't a group $G$ abelian iff $|Z(G)| = |G|$, which is obviously not the case when $|Z(G)| = p^2$?

Answer 1

Sure. And therefore it follows from 1. and 2. that, if $|G|=p^3$ for some prime $p$, then $\bigl|Z(G)\bigr|\neq p^2$.

Answer 2

1. If $Z(G) = {p}^{2}: $ then $|G/Z(G)| = p$, then its only subgroups are {$e$} and $H$ with $|H|=p, \ so \ H = G$. Any non-identity element of $G$ has order $p$, then it generates all $G$, then it's cyclic.

2. If $G/Z(G)$ cyclic, it means that $ \exists \ x \ \epsilon \ G : \ <xZ(G)>=G/Z(G) $

With this in mind we know that if ${g}_{1} , {g}_{2} \ \epsilon \ G$ then ${g}_{1}Z(G) = {(xZ(G))}^{a} = {x}^{a}Z(G)$. Then there exists ${z}_{1} \epsilon Z(G)$ such that ${g}_{1} = {x}^{a}{z}_{1}$. In the same way we can assume ${g}_{2} = {x}^{b}{z}_{2}$

${g}_{1}{g}_{2}= \ {x}^{a}{z}_{1} {x}^{b}{z}_{2}=... \\$ and since ${z}_{i} \epsilon Z(G)$ $ ...= \ {x}^{a}{x}^{b}{z}_{1}{z}_{2} = {x}^{b}{x}^{a}{z}_{2}{z}_{1}={x}^{b}{z}^{2}{x}^{a}{z}^{1}={g}_{2}{g}_{1}$

this proves that $G$ is abelian.