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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?

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  • $\begingroup$ 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
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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 .

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  • $\begingroup$ 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
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    $\begingroup$ Interesting . . . thanks. $\endgroup$ Apr 12 '15 at 22:50
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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 $x \notin Z(G)$

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  • $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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