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1. • CommentRowNumber1.
   • CommentAuthorTodd_Trimble
   • CommentTimeJun 24th 2015
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexNew article [[class equation]], just to fill some gaps in the nLab literature. Truly elementary stuff.

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

2. • CommentRowNumber2.
   • CommentAuthorUrs
   • CommentTimeJun 24th 2015
   • PermaLink
   Author: Urs
   Format: MarkdownItexNice! 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 $A$.

   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 $A$.

3. • CommentRowNumber3.
   • CommentAuthorTodd_Trimble
   • CommentTimeJun 24th 2015
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexThanks for the addition!

   Thanks for the addition!

4. • CommentRowNumber4.
   • CommentAuthorTodd_Trimble
   • CommentTimeSep 23rd 2018
   • (edited Sep 23rd 2018)
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexReplaced 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](https://en.wikipedia.org/wiki/Sylow_theorems#Proof_of_the_Sylow_theorems).) 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. =-- <a href="https://ncatlab.org/nlab/revision/diff/class+equation/10">diff</a>, <a href="https://ncatlab.org/nlab/revision/class+equation/10">v10</a>, <a href="https://ncatlab.org/nlab/show/class+equation">current</a>

   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. =–

   diff, v10, current

5. • CommentRowNumber5.
   • CommentAuthorTodd_Trimble
   • CommentTimeJan 2nd 2023
   • PermaLink
   Author: Todd_Trimble
   Format: MarkdownItexAdded a corollary and a sample application of the Sylow theorems. <a href="https://ncatlab.org/nlab/revision/diff/class+equation/18">diff</a>, <a href="https://ncatlab.org/nlab/revision/class+equation/18">v18</a>, <a href="https://ncatlab.org/nlab/show/class+equation">current</a>

   Added a corollary and a sample application of the Sylow theorems.

   diff, v18, current