Sylow Theorems

Useful facts:

  • Counting contributions to \({\sharp}G\) from \({\operatorname{Syl}}_p(G)\): writing \({\sharp}G = p^k m\) so that \({\sharp}S_p = p^k\), using that every order \(p\) element is in some \(S_p\) one gets at least \(n_p(\ell - 1)\) for some constant \(\ell > 1\).
    • Warning: every \(S_p\) is the same size, so it’s tempting to take \(\ell \coloneqq{\sharp}S_p = p^k\). But this only works if one knows the \(S_p\) intersect trivially, e.g. if \(k=1\). Otherwise, the best one can do without more information \(\ell = p\), i.e. the \(S_p\) all intersect trivially or in subgroups of order \(p\).
    • Warning: This isn’t quite a count of elements of order \(p\), since elements in \(S_p\) can have orders \(p^{k'}\) for other \(k'\leq k\).
  • When counting: just leave the identity out of every calculation, and add it back in as a \(+1\) for the final count.

A \(p{\hbox{-}}\)group is a group \(G\) such that every element is order \(p^k\) for some \(k\). If \(G\) is a finite \(p{\hbox{-}}\)group, then \({\left\lvert {G} \right\rvert} = p^j\) for some \(j\).

If \(G\curvearrowright X\) 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*} 


    
    
      
        
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          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 \(\operatorname{mod}p\).
        
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          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 \({\sharp}{\mathrm{Orb}}(x) = [G:{\operatorname{Stab}}(x)] = 1\).
        
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          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.
        
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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
    

  
    

  

    

      Some setup and notation: assume
    

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

    

      

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

    

  

    

      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.
    

  

    
    
      
        
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          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 \({\sharp}\tilde H = p^{i+1}\).
        
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          Let \(H\leq G\) so that \(H\) is a \(p{\hbox{-}}\)group.
        
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          Let \(H\curvearrowright G/H\) by left-translation.
        
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          Use the lemma that \({\sharp}(G/H)\equiv  \mathrm{Fix}_H(G/H)\operatorname{mod}p\)
        
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          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 \(H\).
        
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          Since \(H{~\trianglelefteq~}N_G(H)\), these cosets form a group.
        
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          We have \([G:H] = {\sharp}(N_G(H) / H)\), and if \(i<k\) then \(p\) divides \([G: H]\).
        
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          So \(N_G(H)/H\) is a \(p{\hbox{-}}\)group and has a subgroup \(L\) of order \(p\) by Cauchy.
        
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          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.
        
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Sylow 2 (Sylows are Conjugate)

    

      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*}

    

  

    

      

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

    

  

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

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

    
    
  
        
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      \([G:S_2] = m\) 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.
    
    
  
        
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Sylow 3 (Numerical Constraints)

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

    

      By Sylow 3, \(p\) does not divide \(n_p\).
    

  

    

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

  

    

      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

    
    
      
        
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          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.
        
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