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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

Chris
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    I don't think you're missing anything. You've shown that this ($G/Z(G)$ having prime order) cannot happen. – James Jun 20 '19 at 21:33
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    Seems to be a fine proof that $G$ is not abelian. – Ruben Jun 20 '19 at 21:36
  • Thank you for your comments. My main trouble is the following. How could a group be abelian if $|G/\mathrm{Z}(G)|=p$ and not $|G/\mathrm{Z}(G)|=1$? – Chris Jun 20 '19 at 21:40
  • How do you prove that $|G/Z(G)|=p$ implies $G$ abelian? – Jose Brox Jun 20 '19 at 21:41
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    @JoseBrox Here you are https://math.stackexchange.com/questions/63087/if-g-zg-is-cyclic-then-g-is-abelian :) (It's the same proof) – Chris Jun 20 '19 at 21:42
  • @Chris You've (presumably) shown that the assumption that $G/Z(G)$ is of prime order leads to a contradiction. Therefore, that assumption must be incorrect. The less restrictive assumption in the linked question that $G/Z(G)$ is cyclic (possibly trivlal) leads, by what should be essentially the same argument, to the conclusion that $G$ is abelian, so the central quotient is trivial (which is cyclic). – James Jun 20 '19 at 21:51

2 Answers2

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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.

Arturo Magidin
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  • Madigin Thank you for you excellent answer. I knew the first result you stated, but I was really confused with this I saw. Now it's much better and, as you said, we should always take back steps in order to study Mathematics in depth. – Chris Jun 24 '19 at 23:31
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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).

Nex
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Jose Brox
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  • Dear Jose, thanks for the answer. What do you mean by "$G/Z(G)$ has at least two prime factors"? – Chris Jun 20 '19 at 22:05
  • @Chris Well, if a number is not $1$ and cannot be a prime either, then it must either be a prime power or have at last two different prime factors. – Jose Brox Jun 20 '19 at 23:13
  • Jose, thank you for your answer. So you mean the order of the group should be $p^2$ for some prime $p$? – Chris Jun 24 '19 at 23:26
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    @Chris No, the order just must not be a prime: so it can be either 1, a power of a prime (like $p^2, p^3, p^4$, etc.) or a product of at least two distinct primes (like $pq$). For example, a priori it could be $1,4,6,27,75$ but it cannot be $2,3,5,7$. – Jose Brox Jun 24 '19 at 23:39