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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?

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  • $\begingroup$ Yes, it comes up in Sylow theory. I cannot recall the theorem off the top of my head... $\endgroup$ – Forever Mozart Oct 31 '14 at 1:28
  • $\begingroup$ @MikkoKorhonen, perhaps you can expand your comment and add it as an answer? $\endgroup$ – lhf Oct 6 '15 at 11:56
  • $\begingroup$ Well, for instance the theorem implies that $G/Z(G)$ cannot have prime order. That's useful. $\endgroup$ – Pete L. Clark Oct 10 '17 at 18:20
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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

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  • $\begingroup$ Good one, thanks. $\endgroup$ – lhf Oct 31 '14 at 2:16
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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 $G$ be is abelian.

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

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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.

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    $\begingroup$ 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. $\endgroup$ – lhf Oct 31 '14 at 2:12
  • $\begingroup$ It does perturb me too. "Let us assume $Z(G)\neq G$. Hence $Z(G)=G$. Contradiction!" Kind of mysterious. $\endgroup$ – caffeinemachine Oct 31 '14 at 2:41
  • $\begingroup$ How is this a proof by contradiction in any manner? $\endgroup$ – user98602 Oct 31 '14 at 2:48
  • $\begingroup$ @MikeMiller We assume $Z(G)\neq G$ and arrive at a contradiction. That is a proof by contradiction. I don't see my mistake. $\endgroup$ – caffeinemachine Oct 31 '14 at 3:31
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    $\begingroup$ You never assumed $Z(G) \neq G$. The proof you have in your own answer is perfectly direct. $\endgroup$ – user98602 Oct 31 '14 at 4:33

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