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• CommentRowNumber1.
• CommentAuthorTodd_Trimble
• CommentTimeJun 24th 2015

New article class equation, just to fill some gaps in the nLab literature. Truly elementary stuff.

• CommentRowNumber2.
• CommentAuthorUrs
• CommentTimeJun 24th 2015

Nice!

I have added the following remark:

Notice that reading the class equation equivalently as

$\sum_{orbits x} \frac{{1}}{{|Stab(a_x)|}} = \frac{|A|}{|G|}$

it expresses the groupoid cardinality of the action groupoid of $G$ acting on $A$.

• CommentRowNumber3.
• CommentAuthorTodd_Trimble
• CommentTimeJun 24th 2015

Thanks for the addition!

• CommentRowNumber4.
• CommentAuthorTodd_Trimble
• CommentTimeSep 23rd 2018
• (edited Sep 23rd 2018)

Replaced a proof that I had earlier adapted from Wikipedia with a cleaner proof that I learned from Benjamin Steinberg (Theorem 3.4). For archival purposes I will post the earlier proof here.

+– {: .proof}

###### Proof

(We adapt the proof from Wikipedia.) Let $P_k = \binom{G}{p^k}$ be the collection of subsets $S$ of $G$ of cardinality $p^k$, and let $G$ act on $P_k$ by taking images of left translations, $S \mapsto g S$. For $S \in P_k$, any $h \in S$ yields a monomorphism

$Stab(S) \hookrightarrow G \stackrel{g \mapsto g h}{\to} G$

that (by definition of $Stab(S)$) factors through $S \hookrightarrow G$; this gives a monomorphism $Stab(S) \to S$, and so ${|Stab(S)|} \leq p^k$. Now we establish the reverse inequality for a suitable $S$. Writing $n = p^k m$, we have

${|P_k|} = \binom{p^k m}{p^k} = m \prod_{j = 1}^{p^k - 1} \frac{p^k m - j}{p^k - j}$

where the product after $m$ on the right is easily seen to be prime to $p$ (any power of $p$ that divides one of the numerators $p^k m - j$ also divides the denominator $p^k - j$, so that powers of $p$ in the product are canceled). Therefore $ord_p({|P_k|}) = ord_p(m)$; let $r$ be this number. Writing out the class equation

${|P_k|} = \sum_{orbits x} \frac{{|G|}}{{|Stab(S_x)|}},$

not every term on the right can have $p$-order greater than $r$, so there is at least one orbit $x$ where

$ord_p(\frac{{|G|}}{{|Stab(S_x)|}}) = ord_p({|G|}) - ord_p({|Stab(S_x)|}) \leq r = ord_p(m).$

We may rearrange this inequality to say $ord_p({|G|}/m) \leq ord_p({|Stab(S_x)|})$; in other words $p^k = {|G|}/m$ divides ${|Stab(S_x)|}$. Therefore $Stab(S_x)$ has order $p^k$, which is what we wanted. =–