Statement:
  Let \((G,\circ)\) be a finite group of order \(p^{r}m\), where \(p\) be a prime and \(r,m\) be positive integers and \(p,m\) be relatively prime. Then \(G\) has subgroup of order \(p^{k}\), where \(k=0,1,2,…,r \).

  Proof:
  Given that \((G,\circ)\) is a finite group of order \(n=p^{r}m\), where \(p\) is a prime and \(r,m\) are positive integers and \(p,m\) are relatively prime.
  To prove \(G\) has subgroups of order \(p^{k}\), where \(k=0,1,2,…,r \).
  We prove this theorem by using the principle of Mathematical Induction.

• Let \(n=1\)
  Then \(r=0\), implies
  \(\set{e}\) is the required subgroup of order \(p^{k}\), where \(k=0 \).
  

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• Let the statement is true for all groups of order \(r\), where \(1\leq r \lt n\)
  Let \(G\) be finite group of order \(p^{r}m\), where \(p\) is a prime and \(r,m\) are positive integers and \(p,m\) are relatively prime.
  
  • Case-1: Let \(p\big|~\big|Z(G)\big| \)
    Then by Cauchy’s Theorem, \(\exists \) an element \(a\in Z(G)\) such that \(\big|a \big|=p \).
    Let us consider the subgroup \(H=\lang a\rang\).
    Then \(H\) is normal in \(G\).
    Implies \(G/H\) is exists and \begin{align*} \big|G/H\big| &=\frac{\big|G\big|}{\big|H\big|} \\ & =\frac{p^{r}m}{p} \\ & =p^{r-1}m \end{align*} Since \(p,m\) are relatively prime and \(\big|G/H\big|=p^{r-1}m\lt \big|G\big| \),
    then by hypothesis, \(G/H\) has subgroups of order \(p^{i},~~ i=0,1,2,…,r-1\).
    Let the subgroups of \(G/H\) are \(K_{i}/H\) such that \begin{align*} \big|K_{i}/H\big|=p^{i},~~ i=0,1,2,…,r-1 \end{align*} where \(K_{i}\) are subgroups of \(G\). Now
    \begin{align*} \big|K_{i}\big| & =\big|K_{i}/H\big|.\big|H\big|\\ & =p^{i}.p\\ & =p^{i+1},~~ i=0,1,2,…,r-1 \end{align*} Hence \(G\) has subgroups \(\set{e},K_{0},K_{1},…,K_{r-1}\) of order \(p^{0},p^{1},p^{2},….,p^{r} \).

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  • Case-2: Let \(p\cancel{\big|}~\big|Z(G)\big| \)
    Since \(\big|G\big|=p^{r}m\) then \( p\big|~\big|G\big| \)
    then from the class equation \begin{align*} \big|G\big|=\big|Z(G)\big|+\displaystyle\sum_{a\notin Z(G)}^{}\big[G:C(a)\big] \end{align*} We claim that \(p\cancel{\big|}~\big[G:C(a)\big] \) where \(a\notin Z(G) \) since if \begin{align*} & p \big|~\big[G:C(a)\big]\\ \implies & p \big|~\big|Z(G)\big| ~\because p\big|~\big|G\big| \end{align*} A contradiction since \(p\cancel{\big|}~\big|Z(G)\big| \).
    Therefore \(\exists~~a\notin Z(G) \) such that \(p\cancel{\big|}~\big[G:C(a)\big] \).
    Now we have \begin{align*} & \big|G\big|=\big[G:C(a)\big].\big|C(a)\big|\\ \implies & p^{r} \big|~\big|C(a)\big| \because p\big|~\big|G\big| \text{ and } p\cancel{\big|}~\big[G:C(a)\big] \\ \implies & \big|C(a)\big|=p^{r}q \end{align*} for some positive integer \(q\) and \(p,q\) are relatively prime.
    Also since \(a\in G-Z(G)\) implies \(C(a)\ne G \) and \(a\ne e \).
    Therefore \(1 \lt C(a)\lt \big|G\big| \).
    Since \(p,q\) are relatively prime and \(\big|C(a)\big|=p^{r}q\lt \big|G\big| \),
    Then by hypothesis, \(C(a)\) has subgroups \(T_{i}\) of order \(p^{i}\), where \(i=0,1,2,…,r \).
    Since \(C(a)\) is a subgroup of \(G\),
    Hence \(G\) has subgroups \(T_{0},T_{1},…,T_{r}\) of order \(p^{0},p^{1},p^{2},….,p^{r} \).

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  Therefore the statement is trues for \(\big|G\big|=n\).

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  Then by the principle of mathematical induction, the statement is true for all \(n\).