Statement:
  Let \((G,\circ)\) be a group and \(H\) and \(K\) be two subgroups of \(G\). Then the number of distinct conjugates of \(H\) induced by the elements of \(K\) is equal to \(\big[K:N_{K}(H) \big]\).

  Proof:
  Given that \((G,\circ)\) is a group and \(H\) and \(K\) are two subgroups of \(G\).
  Let \(T\) be the set of all distinct conjugates of \(H\) unduced by the elements of \(K\).
  Then we have \(T=\set{kHk^{-1}:k\in K}\).
  Let \(S\) be the set of all left coset of \(N_{K}(H) \) in \(K\).
  Then we have \(S=\set{kN_{K}(H):k\in K}\).
  Let us construct a mapping \(f:T\to S \) such that for \(k\in K\), \begin{align*} f(kHk^{-1})=kN_{K}(H) \end{align*}   To prove \(f\) is bijective

• To prove \(f\) is well defined
  Let \(aHa^{-1},bHb^{-1}\in T\) where \(a,b\in K\) such that \begin{align*} & aHa^{-1}=bHb^{-1}\\ \implies & b^{-1}aHa^{-1}b^{-1}=H\\ \implies & \big(b^{-1}\circ a\big)H\big(b^{-1}\circ a\big)^{-1}=H\\ \implies & b^{-1}a\in N_{K}(H)\\ \implies & aN_{K}(H)=bN_{K}(H) \\ \implies & f(aHa^{-1})=f(bHb^{-1}) \end{align*} \(\therefore \) \(f\) is well defined.

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• To prove \(f\) is injective
  Let \(aHa^{-1},bHb^{-1}\in T\) where \(a,b\in K\) such that \begin{align*} & f(aHa^{-1})=f(bHb^{-1}) \\ \implies & aN_{K}(H)=bN_{K}(H) \\ \implies & b^{-1}a\in N_{K}(H)\\ \implies & \big(b^{-1}\circ a\big)H\big(b^{-1}\circ a\big)^{-1}=H\\ \implies & b^{-1}aHa^{-1}b^{-1}=H\\ \implies & aHa^{-1}=bHb^{-1} \end{align*} \(\therefore \) \(f\) is injective.

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• To prove \(f\) is surjective
  Let \(kN_{K}(H)\in S\) where \(k\in K\).
  Then \(kHk^{-1}\in T \) such that \(f(kHk^{-1})=kN_{K}(H) \). Therefore \(kHk^{-1} \) is a pre-image of \(kN_{K}(H)\) in \( T \). Since \(kN_{K}(H)\) is an arbitrary element of \(S\) ha a then each element of \(S\) has a pre-image in \( T \).
  \(\therefore \) \(f\) is surjective.

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  Therefore \(f\) is bijective.
  Hence the number of distinct conjugates of \(H\) induced by the elements of \(K\) is equal to \(\big[K:N_{K}(H) \big]\).