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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?

EmaLee
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  • 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. – Marc van Leeuwen Apr 13 '15 at 08:05

3 Answers3

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Do you know that:

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

and every group with prime order is cyclic .

Elaqqad
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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 $x \notin Z(G)$

David Holden
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  • Sorry, why is it clear that Z(G) is a proper subset of $C_{x}$? – EmaLee Apr 12 '15 at 23:12
  • because the center of G is included in all centralizers of elements of G. but also the centralizer of any element $x$ certainly includes $$ - 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 – David Holden Apr 12 '15 at 23:54
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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.

Mr.Fry
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  • 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. – Trevor Wilson Apr 13 '15 at 06:24
  • 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. – Mr.Fry Apr 13 '15 at 06:32
  • What is why? Comments? – Trevor Wilson Apr 13 '15 at 06:34
  • 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. – Mr.Fry Apr 13 '15 at 06:41
  • Why is G/Z(G) isomorphic to Z/pZ? – Jeremy Baziw Dec 06 '16 at 22:18
  • @JeremyBaziw By First Isomorphism Theorem. Notice that Z(G) is normal. –  Nov 17 '20 at 04:36