4
$\begingroup$

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.

$\endgroup$
  • 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
9
$\begingroup$

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.

ADDENDUM

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.

$\endgroup$
  • $\begingroup$ Very nice solution! Thank you for sharing it! $\endgroup$ – Bman72 Jan 9 '15 at 9:32
6
$\begingroup$

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.

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

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.