Statement:
  If \((G,\circ)\) is a finite group of order \(n\) such that \(p\big| n\) where \(p\) is a prime then \(G\) contains an element of order \(n\) and \(G\) has a subgroup of order \(p\).

  Proof:
  Let \((G,\circ)\) be a finite group of order \(n\) such that \(p\big| n\) where \(p\) be a prime.
  To prove \(\exists\) an element in \(G\) of order \(p\) and \(G\) has a subgroup of order \(p\)
  \(\because p\big|n\implies n\geq 2\).
  We prove this theorem by the principle of Mathematical Induction.

• Let \(n=2\)
  \(\implies \big|G\big|=2\)
  \(\implies p=2\)
  Since \(\big|G\big|\) is a prime then \(G\) is cyclic.
  Then the element in \(G\), other than identity element \(e\), is of order \(2\).

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• Let \(n\) is a prime
  \(\implies \big|G\big|\) is prime
  Then every element in \(G\), other than identity element \(e\), is of order \(p\).

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• Let the statement is true for all groups of order \(r\), where \(2\leq r \lt n\)
  Let \(G\) be a group of order \(n\) such that \(p\big|n\) where \(p\) is prime.
  
  1. Case-1: Let \(Z(G)=G\)
     Then \(G\) is commutative then \(\exists\) an element in \(G\) of order \(p\).

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  2. Case-2: Let \(Z(G)\ne G\)
     Then \(Z(G)\subset G \implies G-Z(G)\ne \phi\) and \(\big|Z(G)\big|\lt \big|G\big|\).
     Let \begin{align*} & a\in G-Z(G) \\ \implies & a\in G \text{ and } a \notin Z(G) \\ \implies & C(a)\ne G \\ \implies & \big[G:C(a)\big] \gt 1 \\ \implies & 1\lt \big|C(a)\big|\lt \big|G \big|~ \big[\because a\ne e \text{ and } a\in C(a) \big] \end{align*}
     • Subcase-1: Let \(p\big|~\big|C(a)\big| \)
       Since \(1\lt \big|C(a)\big|\lt \big|G \big| \), then hypothesis, \(\exists\) an element in \(C(a)\) of order \(p\).
       Therefore \(\exists\) an element in \(G\) of order \(p\) since \(C(a)\subset G \).

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     • Subcase-2: Let \(p\cancel{\big|}~\big|C(a)\big| \)
       Now from Lagrange’s theorem, \begin{align*} & \big|G\big|=\big[G:C(a)\big].\big|C(a)\big| \\ \implies & p\big|~\big[G:C(a)\big]~ \because ~p\big|~\big|G\big| \text{ and }p\cancel{\big|}~\big|C(a)\big|\\ \implies & p\big|~\displaystyle\sum_{a\notin Z(G)}^{}\big[G:C(a)\big] \end{align*} Also from the Class Equation , \begin{align*} & \big|G\big|=\big|Z(G)\big|+\displaystyle\sum_{a\notin Z(G)}^{}\big[G:C(a)\big]\\ \implies & p\big|~\big|Z(G)\big| \because ~p\big|~\big|G\big|\text{ and }p\big|~\displaystyle\sum_{a\notin Z(G)}^{}\big[G:C(a)\big]\\ \implies &\big|Z(G)\big|\gt 1 \end{align*} Therefore we have \(1\lt \big|Z(G)\big| \lt \big|G\big| \) and \( p\big|~\big|Z(G)\big|\). Then by hypothesis, \(\exists\) an element in \(Z(G)\) of order \(p\).
       Therefore \(\exists\) an element in \(G\) of order \(p\) since \(Z(G)\subset G \).

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  Therefore the statement is true for \(n\).
  Then by the principle of mathematical induction, the statement is true for all \(n\geq 2\).
  Let the element \(\alpha \in G\) such that \(\big|\alpha \big|=p\).
  Then \(H=\lang\alpha\rang \) is the required subgroup of \(G\) such that \(\big|H\big|=p\).
  Hence theorem is proved.