Sylow’s second theorem is as follows:

Sylow II: Let GG be a group with G=pmrG = p^mr, with pp and rr coprime, PP a Sylow pp-subgroup of GG and HH any pp-subgroup of GG. Then there exists x∈Gx \in G such that H⊆xPx−1H \subseteq xPx^{-1}. If HH is also a Sylow pp-subgroup then HH and PP are conjugate.

I want to prove the theorem using the following lemma:

Lemma: Let GG be a finite pp-group acting on a finite set XX. Then the number of fixed points in XX under the action of GG is congruent to |X|modp|X| \mod{p}

I’ve attempted to prove it, but I didn’t use the lemma. However I looked through the proof of the lemma and I did use parts of the proof implicitly. I’d like to know if my proof is correct, and how I would actually directly apply the lemma to prove it.

The pp-group HH acts on G/PG/P by h\:.\:(gP) = hgPh\:.\:(gP) = hgP. Therefore |G/P| = [G:P] = r|G/P| = [G:P] = r can be expressed as the sum of the cardinality of its orbits. I pick g_1P,…,g_nPg_1P,…,g_nP to be the elements of G/PG/P which represent orbits with cardinality more than one; that is, no g_iPg_iP is in the same orbit. All other elements which generate orbits with cardinality 11 (i.e. the fixed points) are collected into a set I’ll call FF. So I can write [G:P][G:P] as

[G:P] = |F| + |H\:.\:g_1P| + \dots + |H\:.\:g_nP|[G:P] = |F| + |H\:.\:g_1P| + \dots + |H\:.\:g_nP|

Now, by the orbit-stabiliser theorem,

|H\:.\:g_iP| = \frac{|H|}{|\text{stab}_H(g_iP)|}|H\:.\:g_iP| = \frac{|H|}{|\text{stab}_H(g_iP)|}

Since HH is a pp-group, it has cardinality p^kp^k for some k \in \mathbb{Z}k \in \mathbb{Z}. Therefore by Lagrange, |\text{stab}_H(g_iP)||\text{stab}_H(g_iP)| divides p^kp^k. Furthermore since the orbits of the g_iPg_iP’s have cardinality greater than one, there must be some elements h \in Hh \in H which don’t stabilise g_iPg_iP for i = 1,\dots ,ni = 1,\dots ,n. This means that |\text{stab}_H(g_iP)| < |H||\text{stab}_H(g_iP)| < |H|, therefore \frac{|H|}{|\text{stab}_H(g_iP)|} = p^j \: \text{ for some } 1 \leq j \leq k\frac{|H|}{|\text{stab}_H(g_iP)|} = p^j \: \text{ for some } 1 \leq j \leq k In particular this means that pp divides |H\:.\:g_iP||H\:.\:g_iP|. However since [G:P] = r[G:P] = r, pp doesn't divide [G:P][G:P]; This means that FF contains at least one element. Let xPxP be this fixed point in FF. We have hxP = xPhxP = xP for all h \in Hh \in H, so x^{-1}Hx \subseteq Px^{-1}Hx \subseteq P, which is equivalent to H \subseteq xPx^{-1}H \subseteq xPx^{-1}. if HH is also a Sylow pp-subgroup then |H| = |P||H| = |P| and so H = xPx^{-1}H = xPx^{-1}. ================= And the question is...? – DonAntonio Oct 20 at 21:07 I specified that in my post. " I'd like to know if my proof is correct, and how I would actually directly apply the lemma to prove it." – Jake Reid Browning Oct 20 at 21:09 It looks like you used the lemma in the second to last paragraph where you concluded that FF is nonempty (you reproved the lemma along the way). Anyway, this appears to be the standard proof of this statement. – David Hill 2 days ago ================= =================