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Let $G$ a group, $p$ a prime and $X=G^p$. Let $\sigma\in S_X$ act as follows: $\sigma(x_1,...,x_p) = (x_2,...,x_p,x_1)$. Let $Y$ the subset of elements in $X$ such that $x_1x_2...x_p=1$ and let $Y^\sigma = \{y \in Y : \sigma(y) =y\}$. Prove that $|Y| = |Y^\sigma|(\bmod p)$.

I think that $Y^\sigma$ is the stabilizer but I cannot see how to proceed after that. I think this must be straightforward and relatively easy but I am just failing to see the immediate nature of the solution.

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This is just the class equation (mod p) since the size of an orbit is either $1$ or $p$. The set $Y^{\sigma}$ collects all singleton orbits. –  peoplepower Nov 21 '12 at 3:19
(1) You meant $\,s_n\,$ acts on $\,X\,$, right? If so, then (2) the action must be $\,\sigma(x_1,...,x_p):=(x_2,...,x_p,x_1)\,$ –  DonAntonio Nov 21 '12 at 3:30
@Stefan By the way, $Y^\sigma$ is not the stabilizer. Remember that stabilizers are made of elements of the group doing the acting: $\mbox{Stab}_G(x):=\{g \in G : x^g = x\}.$ Your definition of $Y^\sigma$ is made of elements of the set being acted on. What $Y^\sigma$ is is the set of points on $X$ that are fixed by $\sigma$, and for this reason this set is sometimes alternatively written $\mbox{Fix}(\sigma)$. (This type of set will show up more when you learn Burnside's lemma.) –  Alexander Gruber Nov 21 '12 at 3:51
Why would this get a negative vote!? –  user44069 Nov 21 '12 at 3:58
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3 Answers

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You can see that order of $\sigma$ is $p$. When $\sigma$ acts on $Y$, $Y$ decomposes into orbits $\mathcal{O}_x$ (for some representative $x$ of the orbit), each of which have $|O_x|$ dividing the order of $\sigma$. Since $|Y|=\sum_x|\mathcal{O}_x|$, you can write $|Y|=\left(\sum_y|\mathcal{O}_y|\right)+\left(\sum_z|\mathcal{O}_z|\right)$ where the $y$'s represent orbits of size $p$ and the $z$'s represent orbits of size $1$. The definition given of $Y^\sigma$ makes it clear that $Y^\sigma=\bigcup_z \mathcal{O}_z$, so $|Y|=\left(\sum_y|\mathcal{O}_y|\right)+|Y^\sigma|$. Taking this mod $p$, we have $$|Y|\equiv\left(\sum_y|\mathcal{O}_y|\right)+|Y^\sigma|\mod{p}\equiv |Y^\sigma|\mod{p}.$$

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This is a well known theorem:

Theorem: Let $\,G\,$ a $\,p-$group , $\,p\,$ a prime, act on a finite set $\,X\,$ , and let $\,X^G:=\{x\in X\;;\;gx=x\,\,\forall \,g\in G\}\,$ . Then $\,|X^G|=|X|\pmod p\,$

Proof: for $\,x\in X\,\,,\,|\mathcal Orb(x)|=1\Longleftrightarrow x\in X^G\,$ , so that we have

$$X=\bigcup_{x\in X^g}\{x\}\cup\bigcup_{x\notin X^G}\mathcal Orb(x)\Longrightarrow |X|=|X^G|+\sum_{x\notin X^G}|\mathcal Orb(x)|$$


$$x\notin X^G\Longrightarrow 1<|\mathcal Orb(x)|=[G:Stab(x)]\Longrightarrow p\mid|\mathcal Orb(x)|$$

and from here we get at once that

$$|X|=|X^G|\pmod p$$

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By definition $Y^{\sigma}$ is the set of points $(x_1,\dots,x_p)\in X$ such that $x_1=x_2=\cdots=x_p$. You can show that $\sigma$ acts freely on $Y\backslash Y^{\sigma}$, and $\sigma$ has order $p$, so $Y\backslash Y^{\sigma}$ splits into $\sigma$-orbits of size $p$, in particular $|Y\backslash Y^{\sigma}$| is divisible by $p$.

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