The question is to prove index of $Z(G)$ in $G$ is not a prime number.

We know that $|G:Z(G)|=|G:C_G (x)||C_G (x):Z(G)|$ where $C_G (x)$ means centralizer of $x \in G$

I want to mention that we do not want to use $G⁄Z(G)$ theorem.

We may assume that $|G:Z(G)|=p$ where $p$ is a prime.

  • 3
    $\begingroup$ For doing this without the result about $G/Z(G)$ not being cyclic, here is a hint: We can assume that $Z(G)\neq G$, so we can take an $x\in G\setminus Z(G)$. What can you say about $C_G(x)$? $\endgroup$ – Tobias Kildetoft Oct 29 '13 at 7:55

As you said, we know that $|G:Z(G)|=|G:C_G(x)|\cdot|C_G(x):Z(G)|$ for any $x\in G$.

Let's suppose that $|G:Z(G)|$ is prime. This means that $G$ is non-abelian. So we can take an element $x\in G\setminus Z(G)$. Using the above formula, we know that either $$|G:C_G(x)|=1\quad\text{and}\quad|C_G(x):Z(G)|=p$$ or $$|G:C_G(x)|=p\quad\text{and}\quad|C_G(x):Z(G)|=1$$ since $p$ is prime.

We cannot be in the first state of affairs because $|G:C_G(x)|=1$ implies that $x$ is in the center contrary to our choice that $x\in G\setminus Z(G)$.

In addition, we cannot be in the second state of affairs because $|C_G(x):Z(G)|=1$ implies that $C_G(x)=Z(G)$. The contradiction arises from the fact that $x$ commutes with itself; and since the elements that commute with $x$ are precisely the elements of the center, we have that $x\in Z(G)$. This is the same contradiction as before.


I'm going to elucidate the $G/Z(G)$ argument, since it's much more general and I believe it's important for you to understand. Let $Z=Z(G)$ to save on space.

Suppose $G$ is a group such that $G/Z$ is cyclic. Let $aZ$ be a generator for $G/Z$ for some $a\in G$ (recall all elements of $G/Z$ are cosets of $Z$). Since $Z$ is normal, we have that for all $a,b\in G$ that $aZ\cdot bZ=(ab)Z$ (this is the property that allows us to give a natural binary operation on $G/Z$).

In particular, this means that all elements of $G/Z$ are of the form $a^nZ$ for some $n\in\Bbb Z$. Now let $x, y\in G$. The projection map takes these elements to $xZ$ and $yZ$ respectively. From what we just learned, we know that $xZ=a^nZ$ and $yZ=a^mZ$ for some $n$ and $m\in\Bbb Z$. This means that $x$ and $a^n$ lie in the same coset of $Z$; the same applies to $y$ and $a^m$.

This in turn implies that there is an $h_1$ and an $h_2\in Z$ such that $x=a^nh_1$ and $y=a^mh_2$.

Now we just compute: $$xy=a^nh_1a^mh_2=a^na^mh_2h_1=a^ma^nh_2h_1=a^mh_2a^nh_1=yx$$

Since $x$ and $y$ were arbitrary, $G$ is abelian.

  • $\begingroup$ Very nice solution! Thank you for sharing it! $\endgroup$ – Bman72 Jan 9 '15 at 9:32

Hint: Every group of prime order is cyclic, so the quotient

$$G / Z(G)$$ is cyclic. It's a general fact if $G / Z(G)$ is cyclic, then $G$ is abelian, which leads to a contradiction. If you haven't shown this general fact, try it.

  • $\begingroup$ And if $G$ is abelian so we'll be faced to another contradiction. $\endgroup$ – mrs Oct 29 '13 at 8:39

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