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$|G|=p^n$ 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.

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2 Answers 2

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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.

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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 $\lvert Z(G)\rvert>1$

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