Theorem
Let \( (G,\cdot) \) be a group and \( H \) be a subgroup of \( G \). Then \( \text{Aut}(H) \) is a subgroup of \( \text{Aut}(G) \).
Proof
Given that \( (G,\cdot) \) is a group and \( H \) is a subgroup of \( G \).
To prove \( \text{Aut}(H) \) is a subgroup of \( \text{Aut}(G) \).
Step 1
To prove \( \text{Aut}(H) \subseteq \text{Aut}(G) \)
Let \( \alpha \in \text{Aut}(H) \)
\( \implies \alpha:H \to H \) is an isomorphism.
Then \( \alpha:H \to G \) is a monomorphism since \( H \subseteq G \).
Let \( \beta:G \to G \) be a mapping such that
\[
\beta(x) =
\begin{cases}
\alpha(x) & \text{when } x \in H \\
x & \text{when } x \in G – H
\end{cases}
\]
\( \implies \beta|_{H} = \alpha \). That is, \( \alpha:H \to G \) is a restriction of \( \beta:G \to G \).
Clearly, \( \beta \) is bijective and a homomorphism.
\( \implies \beta \) is an isomorphism.
\( \implies \alpha \in \text{Aut}(G) \).
Therefore, \( \text{Aut}(H) \subseteq \text{Aut}(G) \).
Step 2
To prove \( \text{Aut}(H) \neq \Phi \)
Since \( I_{H} \) is the identity mapping on \( H \), and it is also bijective and a homomorphism,
Therefore, \( I_{H} \in \text{Aut}(H) \).
\( \implies \text{Aut}(H) \neq \Phi \).
Step 3
To prove \( \alpha \circ \beta^{-1} \in \text{Aut}(H) ~\forall~ \alpha, \beta \in \text{Aut}(H) \)
Let \( \alpha, \beta \in \text{Aut}(H) \)
\( \implies \alpha, \beta \) are isomorphisms.
\( \implies \alpha, \beta^{-1} \) are isomorphisms.
Since \( \alpha:H \to H \) and \( \beta^{-1}:H \to H \), therefore \( \alpha \circ \beta^{-1}:H \to H \) exists.
Step 4
To prove \( \alpha \circ \beta^{-1} \) is bijective.
Since \( \alpha \) and \( \beta^{-1} \) are bijective, then \( \alpha \circ \beta^{-1} \) is bijective.
Step 5
To prove \( \alpha \circ \beta^{-1} \) is a homomorphism.
Let \( x, y \in H \).
\[
\begin{aligned}
(\alpha \circ \beta^{-1})(x \cdot y) &= \alpha(\beta^{-1}(x \cdot y)) \\
&= \alpha(\beta^{-1}(x) \cdot \beta^{-1}(y)) \\
&= \alpha(\beta^{-1}(x)) \cdot \alpha(\beta^{-1}(y)) \\
&= (\alpha \circ \beta^{-1})(x) \cdot (\alpha \circ \beta^{-1})(y)
\end{aligned}
\]
Therefore, \( \alpha \circ \beta^{-1} \) is a homomorphism.
Conclusion
Therefore, \( \alpha \circ \beta^{-1} \in \text{Aut}(H) \).
Hence, \( \text{Aut}(H) \) is a subgroup of \( \text{Aut}(G) \).
