A group of order $p^3$ and its quotient with the center of the group

Question

I'm having trouble understanding the following example:

Let $p$ be prime and $G$ a group of order $p^3$. Then either $G$ is abelian or $G/Z(G)$ is abelian (or both). The reason for this is that the centre of a $p$ group is always nontrivial. Therefore $G/Z(G)$ has order $1, p$ or $p^2$ and so is abelian.

I understand that $G/Z(G)$ has order $1, p$ or $p^2$ but I don't understand why this means it is abelian? Of course if it has order one it is abelian and this means $G$ is also abelian. Also for $p^2$ this means that $G/Z(G)$ is abelian as all groups of order $p^2$ are abelian, but what does this tell us about $G$? That $G$ is not abelian?

What I am most confused about is how if $G/Z(G)$ has order $p$, how does this imply it is abelian? I understand that if $G/Z(G)$ has order $p$ then since it is cyclic, $G$ is abelian.. but this doesn't imply the whole quotient is abelian does it?? is it because the cyclic order of the quotient implies $G$ is abelian and therefore the quotient is too? If so does this mean we can change the proposition:

"Let $G$ be a group such that $G/Z(G)$ is cyclic. Then $G$ is abelian."

to the following:

"Let $G$ be a group such that $G/Z(G)$ is cyclic. Then $G$ and $G/Z(G)$ is abelian."

Answer 1

What I am most confused about is how if $G/Z(G)$ has order $p,$ how does this imply it is abelian?

Any group of prime order is abelian. Let $P$ be any group of prime order $p.$ Then for all $x\in P\setminus\{1\},$ $|x|=p,$ since if it were equal to anything other number less than $p$, say $n,$ then $\langle x\rangle$ would be a subgroup of that order, and $n$ would therefore divide the order of the group by Lagranges theorem. However, this is impossible since $p$ is prime, so $n \nmid p.$

I understand that if $G/Z(G)$ has order $p$ then since it is cyclic, $G$ is abelian. but this doesn't imply the whole quotient is abelian does it??

It does. Since $G/Z(G)$ is cyclic, then let $\langle \widetilde{a} \rangle =G/Z(G).$ Let $a$ be any representative of the coset $aZ(G)=\widetilde{a}.$ Then $G=\bigcup_{g\in G}gZ(G),$ and $ab=ba$ for all $b\in Z(G),$ and $a$ commutes with all powers of itself. Thus $G$ is abelian.

If so does this mean we can change the proposition: Let $G$ be a group such that $G/Z(G)$ is cyclic. Then $G$ and $G/Z(G)$ is abelian.

Yes, this is true as a more general theorem: https://proofwiki.org/wiki/Quotient_of_Group_by_Center_Cyclic_implies_Abelian

Answer 2

Suppose $(1)$ $o(G)=p^3$ and $G$ is abelian group (this is possible because for every positive integer '$n$', there is a cyclic group of order $n$, hence abelian), so in this case, $G/Z(G)$ as a quotient of an abelian group is abelian and so $G$ and $G/Z(G)$ both are abelian.

(2) G is not Abelian, now the order of $G/Z(G)$ can be $1$, $p$, $p^2$

$(i)$ If the order is $1$, trivial group and hence $G/Z(G)$ is abelian

$(ii)$ If the order is $p$, we know every prime order group is cyclic and hence abelian.

$(iii)$ Now assume that the order is $p^2$, take $G/Z(G)=H$, then $H$ is a $p$-group of order $p^2$, so $Z(H)$ is of order $p$ or $p^2$, if $Z(H)$ is of order $p^2$ then $H=Z(H)$, so $H$ is abelian.

Next suppose that $Z(H)$ is of order $p$, in this case, $H/Z(H)$ is of order $p$, hence cyclic, which means that $H$ is abelian, by this If $G/Z(G)$ is cyclic, then $G$ is abelian .

So, finally in each case $Z(G)$ is abelian.