Groups - Prove that if $G/Z(G)$ is cyclic then $G$ is abelian

Question

Prove that if $G/Z(G)$ is cyclic then $G$ is abelian. Using this fact and $G$ is a nontrivial group of prime power order, deduce that a group of order $p^2$ , $p$ prime, is abelian.

Answer 1

$G$ is abelian if and only if $Z(G) = G$. If $G/Z(G)$ is cyclic then there is a coset $xZ(G)$ such that every coset $yZ(G)$ is of the form $x^{n}Z(G)$ for some $n \in \mathbb{Z}$. Try to use this to show that all elements of $G$ commute, so that $G$ is abelian.

For the second part you only need to prove that $Z(G)$ isn't trivial. If it isn't then it must have order $p$ or $p^{2}$, so $G/Z(G)$ has order $p$ or $1$. All groups or order $p$ are cyclic, and the trivial group is cyclic, so the result follows from the first part.

Answer 2

These proofs are not too tricky. For the first, take any $g\in G$. Now, since $G/Z(G)$ is cyclic, then there is some $x\in G$ such that $G/Z(G)=\langle xZ(G)\rangle,$ so in particular, $$gZ(G)=\bigl(xZ(G)\bigr)^n=x^nZ(G)$$ for some $n\in\Bbb Z,$ whence $g\in x^nZ(G)$ and so $$g=x^nz\tag{$\clubsuit$}$$ for some $z\in Z(G)$. Note that every element $g$ of $G$ can be written in the form $(\clubsuit)$ for some $n\in\Bbb Z$ and some $z\in Z(G)$. Use this fact to show directly that $G$ is abelian.

For the second, do you know that non-trivial groups of prime power order have non-trivial centers, and that the order of a subgroup must divide the order of the group? Also, what do you know about groups of prime order?

Answer 3

Hints-without-words: if $\,G/Z(G)=\langle xZ(G)\rangle\,$ ,then:

$$\forall\,g\in G\;\exists\left(\,n_g\in \Bbb Z\;\wedge\;z_g\in Z(G)\right)\;\;s.t.\;\;g=x^{n_g}z_g$$

$$\forall\,g,h\in G\;,\;\;gh=x^{n_g}z_gx^{n_h}z_h=x^{n_g}x^{n_h}z_gz_h=\ldots$$