Sylow Theorems

remark:

Useful facts:

  • Counting contributions to G from Sylp(G): writing G=pkm so that Sp=pk, using that every order p element is in some Sp one gets at least np(1) for some constant >1.
    • Warning: every Sp is the same size, so it’s tempting to take :=Sp=pk. But this only works if one knows the Sp intersect trivially, e.g. if k=1. Otherwise, the best one can do without more information =p, i.e. the Sp all intersect trivially or in subgroups of order p.
    • Warning: This isn’t quite a count of elements of order p, since elements in Sp can have orders pk for other kk.
  • When counting: just leave the identity out of every calculation, and add it back in as a +1 for the final count.
definition:

A p-group is a group G such that every element is order pk for some k. If G is a finite p-group, then |G|=pj for some j.

lemma (Congruences for fixed points):

If G for G a p{\hbox{-}}group, then letting \mathrm{Fix}(G) \coloneqq\left\{{x\in X {~\mathrel{\Big\vert}~}gx=x}\right\}, one has \begin{align*} {\sharp}X \equiv {\sharp} \mathrm{Fix}(G) \operatorname{mod}p .\end{align*}

proof (?):

    
  • Use the fixed-point count trick: \begin{align*} {\sharp}X = {\sharp} \mathrm{Fix}(G) + \sum_{x}' {\sharp}{\mathrm{Orb}}(x) .\end{align*} Note that the result follows immediately by reducing \operatorname{mod}p if the sum is zero \operatorname{mod}p.
  • Letting x be an element with a nontrivial orbit, we have {\sharp}{\mathrm{Orb}}(x) > 1, so {\operatorname{Stab}}(x) \neq G since orbit-stabilizer would yield {\sharp}{\mathrm{Orb}}(x) = [G:{\operatorname{Stab}}(x)] = 1.
  • Now use that {\sharp}{\mathrm{Orb}}(x) = {\sharp}G/ {\sharp}{\operatorname{Stab}}(x) = p^k/p^\ell where 0< \ell < k with strict inequalities. So {\sharp}{\mathrm{Orb}}(x) = p^{k-\ell} \neq 1, and p divides its size.

Statements of Sylow

For full proofs (some of which I’ve borrowed), see Keith Conrad’s notes: https://kconrad.math.uconn.edu/blurbs/grouptheory/sylowpf.pdf

remark:

Some setup and notation: assume

  • {\left\lvert {G} \right\rvert} = p^k m where (p, m) = 1,
  • S_p a Sylow{\hbox{-}}p subgroup, and
  • n_p the number of Sylow{\hbox{-}}p subgroups.

Sylow 1 (Cauchy for Prime Powers)

theorem (Sylow 1):

\begin{align*} \forall p^n \text{ dividing } {\left\lvert {G} \right\rvert} \text{, there exists a subgroup of size } p^n .\end{align*}

slogan:

Sylow p{\hbox{-}}subgroups exist for any p dividing {\left\lvert {G} \right\rvert}, and are maximal in the sense that every p{\hbox{-}}subgroup of G is contained in a Sylow p{\hbox{-}}subgroup. If {\left\lvert {G} \right\rvert} = \prod p_i^{\alpha_i}, then there exist subgroups of order p_i^{\beta_i} for every i and every 0 \leq \beta_i \leq \alpha_i. In particular, Sylow p{\hbox{-}}subgroups always exist.

proof (of Sylow 1: left translation):

    
  • Let {\sharp}G = p^k m. Idea: Induct up by showing that if {\sharp}H = p^i for i\leq k, one can product a bigger subgroup \tilde H \supseteq H with [\tilde H : H] = p. This makes {\sharp}\tilde H = p^{i+1}.
  • Let H\leq G so that H is a p{\hbox{-}}group.
  • Let H\curvearrowright G/H by left-translation.
  • Use the lemma that {\sharp}(G/H)\equiv \mathrm{Fix}_H(G/H)\operatorname{mod}p
  • Identify \mathrm{Fix}_H(G/H) = N_G(H), since fixing xH means gxH = xH \implies gHg^{-1}\subseteq H \implies gHg^{-1}= H for all g\in G. \begin{align*} xH \in \mathrm{Fix}_H(G/H) &\iff gxH &= xH \forall g\in H\\ &\iff x^{-1}g x H \in H \forall g\in H\\ &\iff x^{-1}H x^{-1}= H \\ &\iff x\in N_G(H) ,\end{align*} so \mathrm{Fix}_H(G/H) = \left\{{gH {~\mathrel{\Big\vert}~}g\in N(H)}\right\} = N_G(H)/ H are cosets whose representatives are normalizers of H.
  • Since H{~\trianglelefteq~}N_G(H), these cosets form a group.
  • We have [G:H] = {\sharp}(N_G(H) / H), and if i<k then p divides [G: H].
  • So N_G(H)/H is a p{\hbox{-}}group and has a subgroup L of order p by Cauchy.
  • Use the subgroup correspondence: L\leq N_G(H)/H corresponds to some L' \leq G with H \subseteq L' \subseteq N_G(H) and L = L'/H. Now use that {\sharp}L = p implies {\sharp}(L' / H) = [L':H] = p, so {\sharp}L' = [L':H]{\sharp}H = p{\sharp}H = p^{i+1} as desired.

Sylow 2 (Sylows are Conjugate)

theorem (Sylow 2):

All Sylow{\hbox{-}}p subgroups S_p are conjugate, i.e. \begin{align*} S_p^i, S_p^j \in \mathrm{Syl}_p(G) \implies \exists g \text{ such that } g S_p^i g^{-1}= S_p^j \end{align*}

corollary:

\begin{align*} n_p = 1 \iff S_p {~\trianglelefteq~}G .\end{align*}

proof (of Sylow 2):

    
  • Let S_1, S_2\in {\operatorname{Syl}}_p(G), and let S_1\curvearrowright G/S_2 by left-translation.

  • Use the lemma: \begin{align*} {\sharp}(G/S_2) \equiv \mathrm{Fix}_{S_1}(G/S_2) \operatorname{mod}p .\end{align*}

  • [G:S_2] = m is coprime to p, so there is a fixed point, say xS_2 where gxS_2 = xS_2 for all g\in S_1. \begin{align*} gxS_2 = xS_2 \forall g\in S_1 \\ \implies gx \in xS_2 \forall g\in S_1 \\ \implies S_1 x \subseteq xS_2 \\ \implies S_1 \subseteq xS_2 x^{-1} ,\end{align*} where we now get equality since these sets have the same cardinality.

Sylow 3 (Numerical Constraints)

theorem (Sylow 3):

    
  • n_p \divides m~, and in particular, n_p \leq m.

  • n_p \equiv 1 \operatorname{mod}p.

  • n_p = [G : N_G(S_p)] where N_G is the normalizer.

proof (of Sylow 3):

    
  • n_p\equiv 1 \operatorname{mod}p:
    • Fix a P\in {\operatorname{Syl}}_p(G), and let P\curvearrowright{\mathcal{S}}\coloneqq{\operatorname{Syl}}_p(G) by conjugation.
    • Apply the lemma to get n_p \equiv \mathrm{Fix}_{{\mathcal{S}}}(P) \operatorname{mod}p. The claim is that there is just one fixed point.
    • If Q\in \mathrm{Fix}_{{\mathcal{S}}}(P), then pQp^{-1}= Q for all p\in P, so P normalizes Q and P \subseteq N_G(Q) \leq G.
    • Then P, Q \in {\operatorname{Syl}}_p(N_G(Q)), which by Sylow II are conjugate.
    • Since Q{~\trianglelefteq~}N_G(Q), there is only one conjugate of Q, and P=Q.
    • So P is the only fixed point.
  • n_p \divides m:
    • Let G\curvearrowright X\coloneqq{\operatorname{Syl}}_p(G) by conjugation; this is transitive by Sylow II and there is one orbit.
    • Then {\sharp}X must divide {\sharp}G, so n_p divides {\sharp}G = p^k m.
    • Using n_p \equiv 1 \operatorname{mod}p, we can’t have n_p\divides p^k, and so n^p must divide m.
  • n_p = [G: N_G(P)] for any P\in {\operatorname{Syl}}_p(G):
    • Let G\curvearrowright{\operatorname{Syl}}_p(G) by conjugation and apply orbit-stabilizer to get n_p = [G: {\operatorname{Stab}}(P)]
    • Identify {\operatorname{Stab}}(P) = N_G(P).

Corollaries and Applications

corollary:

By Sylow 3, p does not divide n_p.

proposition:

Every p{\hbox{-}}subgroup of G is contained in a Sylow p{\hbox{-}}subgroup.

proof:

Let H \leq G be a p{\hbox{-}}subgroup. If H is not properly contained in any other p{\hbox{-}}subgroup, it is a Sylow p{\hbox{-}}subgroup by definition. Otherwise, it is contained in some p{\hbox{-}}subgroup H^1. Inductively this yields a chain H \subsetneq H^1 \subsetneq \cdots, and by Zorn’s lemma H\coloneqq\cup_i H^i is maximal and thus a Sylow p{\hbox{-}}subgroup.

Exercises

exercise (?):

    
  • Let G be a group of order p with v and e positive integers, p prime, p > v, and v is not a multiple of p. Show that G has a normal Sylow p-subgroup.