Statement:
  Let \((G,\circ)\) be a finite group of order \(p^{r}m\), where \(p\) be a prime and \(r,m\) be two positive integers and \(p,m\) be relatively prime. Then the number \(n_{p} \) of Sylow p-subgroups of \(G\) is \(1+kp\) for some non-negative integer \(k\) and \(n_{p}\big|p^{r}m \).

  Proof:
  Given that \((G,\circ)\) is a finite group of order \(p^{r}m\), where \(p\) is a prime and \(r,m\) are positive integers and \(p,m\) are relatively prime.
  Let \(n_{p}= \) the number of Sylow p-subgroups of \(G\).

• To prove \(n_{p}=1+kp \) for some non-negative integer \(k\)
  Let \(S\) be the set of all Sylow p-subgroups of \(G\).
  Then \(\big|S\big|=n_{p}\).
  Let \(P\in S\).
  First we prove \(P\) acts on S.
  • Let us construct a mapping \(\cdot:P\times S\to S \) such that for all \(a\in P, Q\in S,\) \begin{align*}a \cdot Q= aQa^{-1} \end{align*}
    • To prove \(\cdot \) is well defined.
      Let \(a,b\in P\) and \(Q,R\in S\) such that \begin{align*} & \big(a,Q \big)=\big(b,R \big)\\ \implies & a=b,~ Q=R\\ \implies & aQa^{-1}=bRb^{-1}\\ \implies & a \cdot Q=b \cdot R \end{align*} Therefore \(\cdot \) is well defined.

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    • Let \(a,b\in P\) and \(Q\in S\). Now \begin{align*} \big(a\circ b\big) \cdot Q & = \big(a\circ b\big) Q \big(a\circ b\big)^{-1}\\ & = \big(a\circ b\big) Q \big(b^{-1}\circ a\big)\\ & = a\big(b Q b^{-1}\big)a \\ & = a\cdot \big(b Q b^{-1}\big)\\ & = a\cdot \big(b \cdot Q \big) \end{align*}

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    • Let \( e\) be the identity element of \(G\) and \(Q\in S\). Now \begin{align*} e \cdot Q= e Q e^{-1} = Q \end{align*}

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  Therefore \(P\) acts on S.
  Let \begin{align*} & S_{0}=\set{Q\in S: a \cdot Q= Q ~\forall~ a\in P} \\ \implies & S_{0}=\set{Q\in S: a Q a^{-1}= Q ~\forall~ a\in P} \end{align*} Since \begin{align*} & P\in S \text{ and } a P a^{-1}= P ~\forall~ a\in P\\ \implies & P\in S_{0}\\ \implies & S_{0}\ne \phi\\ \end{align*} Let \begin{align*} & R\in S_{0} \\ \implies & a R a^{-1}= R ~\forall~ a\in P\\ \implies & P \subseteq N(R) \end{align*} Therefore \(P\) and \(R\) are Sylow p-subgroups of \(N(R)\).
  Then Sylow’s Second theorem, \(\exists\) a \(\alpha \in N(R)\) such that \begin{align*} & a R a^{-1}= P \\ \implies & R= P ~\because~a R a^{-1}= R~\forall~a\in P\\ \end{align*} Since \(R\in S_{0}\) is arbitrary then \begin{align*} & S_{0}=\set{P}\\ \implies & \big|S_{0} \big|=1 \end{align*} Since \(P\) is a Sylow p-subgroup of \(G\) then \(\big|P\big|=p^{r} \). Therefore \begin{align*} & \big|S\big|=\big|S_{0}\big|\big(~mod~p\big) \\ \implies & \big|S\big|=1\big|\big(~mod~p\big) \\ \implies & \big|S\big|=1+kp ~\text{ where } k \text{ is a non-negative integer }\\ \implies & n_{p}=1+kp \\ \end{align*}
  

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• To prove \(n_{p}\big|p^{r}m \)
  Let us construct a mapping \(\cdot:G\times S\to S \) such that for all \(a\in G, Q\in S,\) \begin{align*}a \cdot Q= aQa^{-1} \end{align*} Then by previous arguments, \(G\) acts on \(S\).
  Since \(\big|G\big|=p^{r}m\) then \(G\) has Sylow p-subgroup.
  Then \(S\ne\phi\).
  Let \(P\in S\).
  
  • Let \(R\in S\).
    \(\implies R\) is a Sylow p-subgroup.
    \(\implies R\) and \(P\) are conjugates, using Sylow’s Second theorem.
    \(\implies \exists\) an element \(\beta\in G\) such that \(\beta P \beta^{-1}=R\)
    \(\implies R\in \big[P\big]\)
    \(\implies S\subseteq \big[P\big]\)
    \(\implies \big[P\big]=S~\because~\big[P\big]\subseteq S\)
    

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  Then there is only one orbit of \(S\) under \(G\).
  Let \(A\) is a subset of \(S\) containing exactly one elements from each orbit. Then \(\big|A\big|=1\).
  We have \begin{align*} \big|S\big|=\displaystyle\sum_{\alpha\in A}^{}\big[G:G_{\alpha} \big] \end{align*} where \(A\) is a subset of \(S\) containing exactly one elements from each orbit.
  \begin{align*} & \big|S\big|=\big[G:G_{\alpha} \big]~\because~\big|A\big|=1 \\ \implies & \big|S\big|~\Big|~\big|G\big|~\because~\big[G:G_{\alpha} \big]~\Big|~\big|G\big| \\ \implies & n_{p}~\big|~p^{r}m \end{align*}

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