A strange result if $|G/\mathrm{Z}(G)|=p$

Question

I came across something strange, which I would like to share.

Let's take a group $G$ such that $|G/\mathrm{Z}(G)|=p$, where $p$ is a prime number.

Then, we can show that $G$ is abelian $\iff \mathrm{Z}(G)=G$.

But then $|G/\mathrm{Z}(G)|=|G/G|=1$ and we have a contradiction.

What do I miss?

Thanks

Answer 1

We have that if $G/Z(G)$ is cyclic then $G$ is abelian. But since $G$ abelian means $G=Z(G)$, this forces $|G/Z(G)|=1$. Now, if $|G/Z(G)|$ were a prime number then $G/Z(G)$ would be cyclic and then $|G/Z(G)|$ would be 1, which is impossible. Hence we can never have $|G/Z(G)|$ prime. We have proved that either $G$ is abelian or $|G/Z(G)|$ has at least two prime factors (which may be equal).

Answer 2

I really dislike this way of phrasing this particular problem (or rather, the typical more general statement of which this problem is a special case), precisely because it just leads to confusions because it contradicts (or seems to contradict) the initial hypothesis.

The general statement they are alluding to is:

Let $G$ be a group, and suppose that $N\leq Z(G)$ is such that $G/N$ is cyclic. Then $G$ is abelian.

The proof is: let $g\in G$ be such that $gN$ generates $G/N$, and let $x,y\in G$. Then there exist $a,b\in\mathbb{Z}$ and $n_1,n_2\in N$ such that $x=g^an_1$ and $y=g^bn_2$. Then we have: $$\begin{align*} xy &= (g^an_1)(g^bn_2)\\ &= g^ag^bn_1n_2 &&\text{(since }n_1\in N\subseteq Z(G)\text{)}\\ &= g^bg^a n_2n_1 &&\text{(since }n_2\in Z(G)\text{)}\\ &= (g^bn_2)(g^an_1) &&\text{(since }n_2\in Z(G)\text{)}\\ &= yx. \end{align*}$$ Thus, for all $x,y\in G$, $xy=yx$; hence, $G$ is abelian. $\Box$

Often, this statement is given as:

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

This is fine, but then it leads to confusion because in fact $G/Z(G)$ will be trivial. Worse is when it is stated as:

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

This is worse because of course the next step after the conclusion that $G$ is abelian is that $Z(G)=G$, so then $G/Z(G)$ is not, in fact, nontrivial (and cyclic). Thus, you end up proving that the assumption cannot hold.

The problem you have is an even worse special case of this; in fact, what you conclude is not merely that $G$ is abelian, but that the very assumption that led you to that conclusion is impossible. That is, you actually prove that

There does not exist a (finite) group $G$ such that $|G/Z(G)| = p$ where $p$ is a prime.

It’s bad form; it’s bad form because it asks you to prove something that contradicts the hypothesis on which you are proving it. It is bad form because it is part of a proof by contradiction that does not need to be a proof by contradiction (a particular pet peeve of mine). It’s bad form because it leads to confusion by anyone who dares to take a step back after finishing the proof and tries to understand the whole argument and its implications. In short, it is a disservice to good students like yourself who want to look beyond the immediate task at hand.