# If $|G|=p^n$ where $p$ is prime and $n\geq1$, please show that $Z(G)$ is non-trivial

$|G|=p^n$ where $p$ is prime and $n\geq1$. Show that if $G$ acts on a set $X$, and $Y$ is an orbit of this action, then either $|Y| = 1$ or $p$ divides $|Y|$. Show that $|Z(G)| >1$.

By considering the set of elements of $G$ that commute with a fixed element $x\notin Z(G)$, show that $Z(G)$ cannot have order $p^{n-1}$.

for the first part of this question,I can see |Y|=1 or |Y||p follows from orbit and stabilizer theorem. And consider G acts on itself by conjugation, Z(G) is actually the stabilizer. I got this because intended to show Z(G) is non-trivial by showing orbit can't contain all elements but I didn't find a way to work out. I got stuck here and would be thankful if anyone can help.

• Please see here for how to typeset common math expressions with LaTeX, and see here for how to use Markdown formatting. – Zev Chonoles Dec 30 '12 at 3:13
• You seem to be asking a lot of questions on similar topics. You might consider taking a bit of time to digest the answers you get before posting more. – Tobias Kildetoft Dec 30 '12 at 3:14

Making $\,G\,$ act on itself by conjugation and noting that

$$x\in Z(G)\Longleftrightarrow |\mathcal Orb(x)|=1$$

we get the class equation for $\,G\,$:

$$p^n=|G|=|Z(G)|+\sum_{x\notin Z(G)}|\mathcal Orb(x)|$$

where the sum is over different, and thus pairwise disjoint, conjugation classes.

But $\,|\mathcal Orb(x)|=[G:G_x]\,\,,\,G_x=$ the stabilizer subgroup of $\,x\,$ , and thus all the summands of the sum in the class equation are multiples of $\,p\,$ (in fact, powers of it), so that it must be that $\,|Z(G)|>1\,$ .

For the second part, prove the nice

Proposition: For any group $\,G\,$ , if $\,G/Z(G)\,$ is cyclic non-trivial then $\,G\,$ is abelian, or in otherwords: the quotient $\,G/Z(G)\,$ cannot be cyclic non-trivial.

By orbit decomposition formula $\mathrm{card}(Y)=\sum_{G_s}|G:G_s|$ Since $p\mid\lvert G:G_s\rvert$, $p\mid\mathrm{card}(Y)$

For the part $|Z(G)|>1$, use class equation $$|G|=|Z(G)|+ \sum_{\substack{\text{non-trivial}\\\text{conjugacy }\\\text{classes }G_x}}|G:G_x|,$$
$p\mid\lvert G\rvert$ and $p\mid\lvert G:G_x\rvert$ hence $p\mid\lvert Z(G)\rvert$ so $\lvert Z(G)\rvert>1$

• You can find some good starting points on how to format mathematics on the site here and here. This AMS reference is very useful. If you need to format more advanced things, there are many excellent references on LaTeX on the internet, including StackExchange's own TeX.SE site. – Zev Chonoles Dec 30 '12 at 3:24
• I am trying to help you by formatting your post with LaTeX for you. Even if you want to add to your post later, please at least don't undo the work I did. – Zev Chonoles Dec 30 '12 at 3:39