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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

    orbitsx1|Stab(a x)|=|A||G|\sum_{orbits x} \frac{{1}}{{|Stab(a_x)|}} = \frac{|A|}{|G|}

    it expresses the groupoid cardinality of the action groupoid of GG acting on AA.

    • 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=(Gp k)P_k = \binom{G}{p^k} be the collection of subsets SS of GG of cardinality p kp^k, and let GG act on P kP_k by taking images of left translations, SgSS \mapsto g S. For SP kS \in P_k, any hSh \in S yields a monomorphism

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

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

    |P k|=(p kmp k)=m j=1 p k1p kmjp kj{|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 mm on the right is easily seen to be prime to pp (any power of pp that divides one of the numerators p kmjp^k m - j also divides the denominator p kjp^k - j, so that powers of pp in the product are canceled). Therefore ord p(|P k|)=ord p(m)ord_p({|P_k|}) = ord_p(m); let rr be this number. Writing out the class equation

    |P k|= orbitsx|G||Stab(S x)|,{|P_k|} = \sum_{orbits x} \frac{{|G|}}{{|Stab(S_x)|}},

    not every term on the right can have pp-order greater than rr, so there is at least one orbit xx where

    ord p(|G||Stab(S x)|)=ord p(|G|)ord p(|Stab(S x)|)r=ord p(m).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)ord p(|Stab(S x)|)ord_p({|G|}/m) \leq ord_p({|Stab(S_x)|}); in other words p k=|G|/mp^k = {|G|}/m divides |Stab(S x)|{|Stab(S_x)|}. Therefore Stab(S x)Stab(S_x) has order p kp^k, which is what we wanted. =–

    diff, v10, current

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