If $G/Z(G)$ is cyclic then $G$ is abelian – what is the point?

Question

The theorem "if $G/Z(G)$ is cyclic then $G$ is abelian" is a popular exercise.

But what is the point of this theorem if $G/Z(G)$ can only be cyclic if it is trivial?

Does "$G/Z(G)$ is cyclic" actually appears in other proofs or is it just a popular exercise?

Yes, it comes up in Sylow theory. I cannot recall the theorem off the top of my head... — Forever Mozart, Oct 31 '14 at 01:28
Well, for instance the theorem implies that $G/Z(G)$ cannot have prime order. That's useful. — Pete L. Clark, Oct 10 '17 at 18:20

lhf · Answer 1 · 2022-04-27T18:19:51.320

13

A nice application is in proving that the center of a non-abelian cannot be too large:

If $G$ is a non-abelian finite group, then $|Z(G)| \leq \frac {1}{4} |G|$.

The contrapositive is

If $|Z(G)| > \frac {1}{4} |G|$, then $G$ is abelian.

Indeed, $|Z(G)| > \frac {1}{4} |G|$ implies that $G/Z(G)$ has order at most $3$, and so is cyclic.

edited Apr 27 '22 at 18:19

answered Oct 06 '15 at 10:35

lhf

3

An application of your application is http://mathoverflow.net/questions/211159/why-cant-a-nonabelian-group-be-75-abelian – j.p. Oct 06 '15 at 13:00

score 10 · Accepted Answer · edited Apr 13 '17 at 12:21

10

This other question is using the fact that $G/Z(G)$ being cyclic implies that $G$ is abelian to show that non abelian groups of order $pq$ have a trivial center Verification of Proof that a nonabelian group G of order pq where p and q are primes has a trivial center

edited Apr 13 '17 at 12:21

Community

answered Oct 31 '14 at 01:53

score 7 · Answer 3 · answered Oct 31 '14 at 01:56

7

Say you want to prove the following.

Every group $G$ of order $p^2$ ($p$ is a prime) is abelian.

You can apply this result.

Since $G$ is a $p$-group, it has a non trivial center $Z(G)$.

Therefore $|G/Z(G)|=1$ or $p$.

Therefore $G/Z(G)$ is cyclic and $G$ is abelian.

There are many other applications.

answered Oct 31 '14 at 01:56

caffeinemachine

1

It still leaves a bad taste because the cases $|Z(G)|=1$ or $p$ never happens. I never really liked proofs by contradiction of the form: To prove $P$, assume not $P$, deduce $P$, and call it a contradiction. – lhf Oct 31 '14 at 02:12
It does perturb me too. "Let us assume $Z(G)\neq G$. Hence $Z(G)=G$. Contradiction!" Kind of mysterious. – caffeinemachine Oct 31 '14 at 02:41
How is this a proof by contradiction in any manner? – Oct 31 '14 at 02:48
@MikeMiller We assume $Z(G)\neq G$ and arrive at a contradiction. That is a proof by contradiction. I don't see my mistake. – caffeinemachine Oct 31 '14 at 03:31
4

You never assumed $Z(G) \neq G$. The proof you have in your own answer is perfectly direct. – Oct 31 '14 at 04:33
@ihf I am glad I found your comment because tis confused the crap out of me when I read proofs like that. – Lemon Sep 19 '19 at 04:16

3 Answers3