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 any two Sylow p-subgroups of \(G\) are conjugate, then therefore isomorphic.

  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.
  To prove any two Sylow p-subgroups of \(G\) are conjugate.
  Let \(H\) and \(K\) are two Sylow p-subgroups of \(G\).
  Let \(S\) be the set of all lefts cosets of \(H\) in \(G\).
  Then \(\big|S\big|=\big[G:H\big] \)
  Since \(H\) is a Sylow p-subgroup of \(G\) then \(\big|G\big|=p^{r} \).
  Then we have \begin{align*} &\big|G\big|=\big[G:H\big].\big|H\big|\\ \implies & p^{r}m=\big[G:H\big].p^{r}\\ \implies & p ~\cancel{\big|}~ \big[G:H\big]\\ \implies & p ~\cancel{\big|}~ \big|S\big|~\because ~\big|S\big|=\big[G:H\big] \end{align*}   First we prove \(K\) acts on S.

• Let us construct a mapping \(\cdot:K\times S\to S \) such that for all \(k\in K, a\in G,\) \begin{align*}k \cdot aH= (k\circ a)H \end{align*}
  • To prove \(\cdot \) is well defined.
    Let \(k_{1},k_{2}\in K\) and \(aH,bH\in S\) where \(a,b\in G\) such that \begin{align*} & \big(k_{1},aH \big)=\big(k_{2},bH \big)\\ \implies & k_{1}=k_{2},~ aH=bH\\ \implies & k_{1}=k_{2},~ b^{-1}\circ a\in H\\ \implies & b^{-1}\circ k^{-1}_{2}\circ k_{1}\circ a\in H\\ \implies & \big(k_{2}\circ b\big)^{-1}\circ\big(k_{1}\circ a\big)\in H\\ \implies & \big(k_{1}\circ a\big)H=\big(k_{2}\circ b\big)H\\ \implies & k_{1} \cdot aH =k_{2} \cdot bH \end{align*} Therefore \(\cdot \) is well defined.

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  • Let \(k_{1},k_{2}\in K\) and \(aH\in S\) where \(a\in G\). Now \begin{align*} \big(k_{1}\circ k_{2}\big) \cdot aH & = \big(\big(k_{1}\circ k_{2}\big)\circ a\big)H\\ & = \big(k_{1}\circ \big( k_{2}\circ a\big)\big)H \\ & = k_{1}\cdot \big( k_{2}\circ a\big)H \\ & = k_{1}\cdot \big( k_{2}\cdot aH\big) \\ \end{align*}

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  • Let \( e\) be the identity element of \(G\) and \(aH\in S\) where \(a\in G\). Now \begin{align*} e \cdot aH= \big( e\circ a\big)H = aH \end{align*}

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  Therefore \(K\) acts on S.
  Let \(S_{0}=\set{aH\in S: k \cdot aH= aH ~\forall~ k\in K} \).
  Since \(K\) is a Sylow p-subgroup of \(G\) then \(\big|K\big|=p^{r} \). Therefore \begin{align*} & \big|S\big|=\big|S_{0}\big|\big(~mod~p\big) \\ \implies & p ~\cancel{\big|} \big|S_{0}\big|~\text{ and }~\big|S_{0}\big|\ne 0~\because~p ~\cancel{\big|}~ \big|S\big| \\ \implies & S_{0} \ne \phi \end{align*}    Let \(\alpha H\in S_{0}\) where \(\alpha \in G\). Then we have \begin{align*} & k\cdot \alpha H =\alpha H~\forall~k\in K \\ \implies & \big(k\circ \alpha \big) H =\alpha H~\forall~k\in K \\ \implies & \alpha^{-1}\circ\big(k\circ \alpha \big) \in H~\forall~k\in K \\ \implies & \big(\alpha^{-1} K \alpha \big) \subseteq H \end{align*}   Since \(H\) and \(K\) are Sylow p-subgroups of \(G\) then \(\big|H\big|=\big|K\big| \). Therefore \begin{align*} & \big|\alpha^{-1} K \alpha\big|=\big|K\big| \\ \implies & \big|\alpha^{-1} K \alpha\big|=\big|H\big|\\ \implies & \alpha^{-1} K \alpha=H~\because~\big(\alpha^{-1} K \alpha \big) \subseteq H \end{align*}   Therefore \(H\) and \(K\) are conjugates.
  Since \(H\) and \(K\) are any two Sylow p-subgroups of \(G\) then any two Sylow p-subgroups of \(G\) are conjugates.