Statement:
  Let \((G,\circ)\) be a group and \(H\) be a subgroup of \(G\) and \(a\in G\). Then prove that \(H \cong aHa^{-1} \).

  Proof:
  Given that \((G,\circ)\) is a group and \(H\) is a subgroup of \(G\) and \(a\in G\).
  To prove \(H \cong aHa^{-1} \)
  Let us construct a mapping \(f:H\to aHa^{-1} \) such that for \(h\in H\) \begin{align*} f(h)=a\circ h \circ a^{-1} \end{align*}

• To prove \(f\) is well defined
  Let \(h_{1},h_{2}\in H \) such that \begin{align*} & h_{1}=h_{2} \\ \implies & a\circ h_{1} \circ a^{-1}=a\circ h_{2} \circ a^{-1} \\ \implies & f(h_{1})=f(h_{2}) \end{align*} \(\therefore\) \(f\) is well defined.

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• To prove \(f\) is injective
  Let \(h_{1},h_{2}\in H \) such that \begin{align*} & f(h_{1})=f(h_{2})\\ \implies & a\circ h_{1} \circ a^{-1}=a\circ h_{2} \circ a^{-1} \\ \implies & h_{1}=h_{2} \end{align*} \(\therefore\) \(f\) is injective.

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• To prove \(f\) is surjective
  Let \(a\circ h_{3} \circ a^{-1}\in aHa^{-1} \) for some \(h_{3}\in H \).
  Then \(f(h_{3})=a\circ h_{3} \circ a^{-1}\).
  \(\therefore\) \(h_{3} \) is a pre-image of \(a\circ h_{3} \circ a^{-1} \). Since \(h_{3} \) is an arbitrary element of \(aHa^{-1}\), then each element of \(aHa^{-1}\) has a pre-image in \(H\).
  \(\therefore\) \(f\) is surjective.

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• To prove \(f\) is homomorphism
  Let \(h_{4},h_{5}\in H \) then \begin{align*} f\big( h_{4}\circ h_{5}\big) & = a\circ \big( h_{4}\circ h_{5}\big) \circ a^{-1}\\ & = \big( a\circ h_{4} \big) \circ \big( a^{-1} \circ a \big) \circ \big(h_{5} \circ a^{-1}\big) \\ & = \big( a\circ h_{4} \circ a^{-1}\big) \circ \big( a \circ h_{5} \circ a^{-1}\big) \\ & = f\big( h_{4} \big) \circ f\big( h_{5} \big) \end{align*} \(\therefore\) \(f\) is homomorphism.

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  \( \therefore f\) is an isomorphism.
  Hence \(H \cong aHa^{-1} \).