# Index of a finite group

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?

• 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. 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 .

• I did not know that. How do you prove it? Apr 12 '15 at 22:46
• you can for example see this post Apr 12 '15 at 22:48
• Interesting . . . thanks. 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 $x \notin Z(G)$

• Sorry, why is it clear that Z(G) is a proper subset of $C_{x}$? 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 $<x>$ - 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 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.

• 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. Apr 13 '15 at 6:24
• 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. Apr 13 '15 at 6:32
• What is why? Comments? Apr 13 '15 at 6:34
• 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. Apr 13 '15 at 6:41
• Why is G/Z(G) isomorphic to Z/pZ? Dec 6 '16 at 22:18