Prove that if G is a finite group, the index of Z(G) cannot be prime.

What I have so far:

-Suppose G is Abelian, then G = Z(G). In this case the order of G:Z(G) would just be 1 which isn't prime.

What next?

  • $\begingroup$ Given the negative nature of what is to be proved, the only natural starting point would seem to be to suppose $G:Z(G)$ is a prime number, and try to derive a contradiction from this. What you have done (and any other special cases incompatible with $G:Z(G)$ being a prime number) is irrelevant for the proof. $\endgroup$ Apr 13 '15 at 8:05

Do you know that:

If $G/Z(G)$ is a cyclic group then $G$ is abelian.

and every group with prime order is cyclic .

  • $\begingroup$ I did not know that. How do you prove it? $\endgroup$ Apr 12 '15 at 22:46
  • $\begingroup$ you can for example see this post $\endgroup$
    – Elaqqad
    Apr 12 '15 at 22:48
  • 1
    $\begingroup$ Interesting . . . thanks. $\endgroup$ Apr 12 '15 at 22:50

suppose $x \notin Z(G)$, and let $C_x$ be the centralizer of $x$. clearly $$ Z(G) \subset C_x \subset G $$ the first inclusion is proper, by construction, and the second cannot be, because $[G:Z(G)]$ is a prime. thus $C_x = G$, contradicting the assumption that $x \notin Z(G)$

  • $\begingroup$ Sorry, why is it clear that Z(G) is a proper subset of $C_{x}$? $\endgroup$
    – EmaLee
    Apr 12 '15 at 23:12
  • $\begingroup$ because the center of G is included in all centralizers of elements of G. but also the centralizer of any element $x$ certainly includes $<x>$ - the subgroup generated by $x$. since $x$ was assumed outside the center of G, ipso facto the centralizer of $x$ must contain at least one element $(x)$ not in the center of G $\endgroup$ Apr 12 '15 at 23:54

Let $|G|=n< \infty$ and suppose $[G:Z(G)] = |G|/|Z(G)| = p\ $ where $p$ is prime. Then $G/Z(G) \cong \mathbb{Z}/p\mathbb{Z} \Rightarrow$ cyclic $\Rightarrow G$ is abelian; hence $|Z(G)|=|G| \Rightarrow [G:Z(G)]=1$ which is not prime. Thus you have your claim.

  • $\begingroup$ It seems to me like this argument takes unnecessary detours in order to use the hypothesis that $G$ is finite (which isn't really necessary.) For example, the fact that $[G : Z(G)] = |G|/|Z(G)|$ doesn't seem to get used. $\endgroup$ Apr 13 '15 at 6:24
  • $\begingroup$ This is why I don't get on math exchange often. No proof is perfect, if you have an alternative, just post it. I really don't care. $\endgroup$
    – Mr.Fry
    Apr 13 '15 at 6:32
  • $\begingroup$ What is why? Comments? $\endgroup$ Apr 13 '15 at 6:34
  • $\begingroup$ This is how I seen the proof. I don't care what you see as necessary and unnecessary, especially for a 2 line proof. Please I really don't want to say anymore. $\endgroup$
    – Mr.Fry
    Apr 13 '15 at 6:41
  • $\begingroup$ Why is G/Z(G) isomorphic to Z/pZ? $\endgroup$ Dec 6 '16 at 22:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.