Group of order $p^2$.

Question

$G$ is a group of order $p^2$, $p$ is a prime. If $Z(G)=p$ then $G/Z(G)$ has order $p$ and $G$ is cyclic. Why $G$ is cyclic?

Is it related to the Lagrange Theorem? Actually, I have no idea about it. Thanks

Answer 1

This result is false since the conclusion says that $G$ is cyclic of order $p^2$ so the center of $G$ is $G$ and the cardinal of $G$ is different of $p$ which is supposed to be the cardinal of $Z(G)$.

Answer 2

The question is not well formed. By definition, the center of a group is as follows:

$$\mathcal{Z}(G) = \{g \in G | xg = gx\ \forall x \in G \}$$

We see that in an abelian group, $\mathcal{Z}(G) = G$. All cyclic groups are abelian (this is trivial, but if you want I can give a proof) and therefore $\mathcal{Z}(G) = G$ for $G$ cyclic. But here $|\mathcal{Z}(G)| \ne |G|$. This means the group cannot be abelian and therefore cannot be cyclic.

Answer 3

$Z(G)$ cant be of order $p$ since every group of order $p^2$ is abelian. For if $G$ is not abelian this implies that $|G|>|N(a)|>|Z(G)|$ ,$N(a)$ is the normalizer of an element for every a in G. But if $|Z(G)|=p$ this leaves that $|N(a)|$ can only be $p^2$ because of the inequality above, and sice order of a subgroup divides order of the group. But if $|N(a)|=p^2=|G|$..this means the group is abelian. Hece for a group of order $p^2$ you can never get a strict inequality between these three. So the group has to be abelian.